# BMO1 2016 Q5 – from areas to angles

For the second year in a row Question 5 has been a geometry problem; and for the second year in a row I presented the video solution; and the for the second year in a row I received the question(s) while I was abroad. You can see the video solutions for all the questions here (for now). I had a think about Q5 and Q6 on the train back from a day out at Lake Balaton in Western Hungary, so in keeping with last year’s corresponding post, here are some photos from those sunnier days.

I didn’t enjoy this year’s geometry quite as much as last year’s, but I still want to say some things about it. At the time of writing, I don’t know who proposed Q5, but in contrast to most geometry problems, where you can see how the question might have emerged by tweaking a standard configuration, I don’t have a good intuition for what’s really going on here. I can, however, at least offer some insight into why the ‘official’ solution I give on the video has the form that it does.

The configuration given is very classical, with only five points, and lots of equal angles. The target statement is also about angles, indeed we have to show that a particular angle is a right-angle. So we might suspect that the model approach might well involve showing some other tangency relation, where one of the lines AC and BC is a radius and the other a tangent to a relevant circle. I think it’s worth emphasising that throughout mathematics, the method of solving a problem is likely to involve similar objects to the statement of the problem itself. And especially so in competition problems – it seemed entirely reasonable that the setter might have found a configuration with two corresponding tangency relations and constructed a problem by essentially only telling us the details of one of the relations.

There’s the temptation to draw lots of extra points or lots of extra lines to try and fit the given configuration into a larger configuration with more symmetry, or more suggestive similarity [1]. But, at least for my taste, you can often make a lot of progress just by thinking about what properties you want the extra lines and points to have, rather than actually drawing them. Be that as it may, for this question, I couldn’t initially find anything suitable along these lines [2]. So we have to think about the condition.

But then the condition we’ve been given involves areas, which feels at least two steps away from giving us lots of information about angles. It doesn’t feel likely that we are going to be able to read off some tangency conditions immediately from the area equality we’ve been given. So before thinking about the condition too carefully, it makes sense to return to the configuration and think in very loose terms about how we might prove the result.

How do we actually prove that an angle is a right-angle? (*) I was trying to find some tangency condition, but it’s also obviously the angle subtending by the diameter of a circle. You could aim for the Pythagoras relation on a triangle which includes the proposed right-angle, or possibly it might be easier to know one angle and two side-lengths in such a triangle, and conclude with some light trigonometry? We’ve been given a condition in terms of areas, so perhaps we can use the fact that the area of a right-angled triangle is half the product of the shorter side-lengths? Getting more exotic, if the configuration is suited to description via vectors, then a dot product might be useful, but probably this configuration isn’t.

The conclusion should be that it’s not obvious what sort of geometry we’re going to need to do to solve the problem. Maybe everything will come out from similar triangles with enough imagination, but maybe it won’t. So that’s why in the video, I split the analysis into an analysis of the configuration itself, and then an analysis of the area condition. What really happens is that we play with the area condition until we get literally anything that looks at all like one of the approaches discussed in paragraph (*). To increase our chances, we need to know as much about the configuration as possible, so any deductions from the areas are strong.

The configuration doesn’t have many points, so there’s not much ambiguity about what we could do. There are two tangents to the circle. We treat APC with equal tangents and the alternate segment theorem to show the triangle is isosceles and that the base angles are equal to the angle at B in ABC. Then point Q is ideally defined in terms of ABC to use power of a point, and add some further equal angles into the diagram. (Though it turns out we don’t need the extra equal angle except through power of a point.)

So we have some equal angles, and also some length relations. One of the length relations is straightforward (AP=CP) and the other less so (power of a point $CQ^2 = AQ\cdot BQ$). The really key observation is that the angle-chasing has identified

$\angle PAQ = 180 - \angle \hat C,$

which gives us an alternative goal: maybe it will be easier to show that PAQ is a right-angle.

Anyway, that pretty much drinks the configuration dry, and we have to use the area condition. I want to emphasise how crucial this phase in for this type of geometry problem. Thinking about how to prove the goal, and getting a flavour for the type of relation that comes out of the configuration is great, but now we need to watch like a hawk when we play with the area condition for relations which look similar to what we have, and where we might be going, as that’s very likely to be the key to the problem.

We remarked earlier that we’re aiming for angles, and are given areas. A natural middle ground is lengths. All the more so since the configuration doesn’t have many points, and so several of the triangles listed as having the same area also have the same or similar bases. You might have noticed that ABC and BCQ share height above line AQ, from which we deduce AB=BQ. It’s crucial then to identify that this is useful because it supports the power of a point result from the configuration itself. It’s also crucial to identify that we are doing a good job of relating lots of lengths in the diagram. We have two pairs of equal lengths, and (through Power of a Point) a third length which differs from one of them by a factor of $\sqrt{2}$.

If we make that meta-mathematical step, we are almost home. We have a relation between a triple of lengths, and between a pair of lengths. These segments make up the perimeter of triangle APQ. So if we can relate one set of lengths and the other set of lengths, then we’ll know the ratios of the side lengths of APQ. And this is excellent, since much earlier we proposed Pythagoras as a possible method for establish an angle is a right-angle, and this is exactly the information we’d need for that approach.

Can we relate the two sets of lengths? We might guess yes, that with a different comparison of triangles areas (since we haven’t yet used the area of APC) we can find a further relation. Indeed, comparing APC and APQ gives CQ = 2PC by an identical argument about heights above lines.

Now we know all the ratios, it really is just a quick calculation…

[1] – I discussed the notion of adding extra points when the scripts for the recording were being shared around. It was mentioned that for some people, the requirement to add extra points (or whatever) marks a hard division between ‘problems they can do’ and ‘problem they can’t do’. While I didn’t necessarily follow this practice while I was a contestant myself, these days the first thing I do when I see any angles or an angle condition in a problem is to think about whether there’s a simple way to alter the configuration so the condition is more natural. Obviously this doesn’t always work (see [2]), but it’s on my list of ‘things to try during initial thinking’, and certainly comes a long way before approaches like ‘place in a Cartesian coordinate system’.

[2] – Well, I could actually find something suitable, but I couldn’t initially turn it into a solution. The most natural thing is to reflect P in AC to get P’, and Q in BC to get Q’. The area conditions [AP’C]=[ABC]=[BCQ’] continue to hold, but now P’ and B are on the same side of AC, hence P’B || AC. Similarly AQ’ || BC. I see no reason not to carry across the equal length deductions from the original diagram, and we need to note that angles P’AC, ACP’, CBA are equal and angles Q’AB and BAC are equal. In the new diagram, there are many things it would suffice to prove, including that CP’Q’ are collinear. Note that unless you draw the diagram deliberately badly, it’s especially easy accidentally to assume that CP’Q’ are collinear while playing around, so I wasted quite a bit of time. Later, while writing up this post, I could finish it [3].

[3] – In the double-reflected diagram, BCQ’ is similar to P’BA, and since Q’C=2P’C = P’A, and Q’B=AB, you can even deduce that the scale factor is $\sqrt{2}$. There now seemed two options:

• focus on AP’BC, where we now three of the lengths, and three of the angles are equal, so we can solve for the measure of this angle. I had to use a level of trigonometry rather more exotic than the Pythagoras of the original solution, so this doesn’t really serve purpose.
• Since BCQ’ is similar to P’BA and ABQ’ similar to CP’A, we actually have Q’BCA similar to AP’BC. In particular, $\angle CBP' = \angle ACB$, and thus both are 90. Note that for this, we only needed the angle deductions in the original configuration, and the pair of equal lengths.
• There are other ways to hack this final stage, including showing that BP’ meets AQ’ at the latter’s midpoint, to give CP’Q’ collinear.

# DGFF 3 – Gibbs-Markov property for entropic repulsion

In the previous post, we saw that it isn’t much extra effort to define the DGFF with non-zero boundary conditions, by adding onto the zero-BC DGFF the unique (deterministic) harmonic function which extends the boundary values into the domain. We also saw how a Gibbs-Markov property applies, whereby the values taken by the field on some sub-region $A\subset D$ depend on the values taken on $D\backslash A$ only through values taken on $\partial A$.

In this post, we look at how this property and some other methods are applied by Deuschel [1] to study the probability that the DGFF on a large box in $\mathbb{Z}^d$ is positive ‘everywhere’. This event can be interpreted in a couple of ways, all of which are referred to there as entropic repulsion. Everything which follows is either taken directly or paraphrased directly from [1]. I have tried to phrase this in a way which avoids repeating most of the calculations, instead focusing on the methods and the motivation for using them.

Fix dimension $d\ge 2$ throughout. We let $P^0_N$ be the law of the DGFF on $V_N:=[-N,N]^d\subset \mathbb{Z}^d$ with zero boundary conditions. Then for any subset $A\subset \mathbb{Z}^d$, in an intuitively-clear abuse of notation, we let

$\Omega^+(A):= \{ h_x\ge 0, x\in A\},$

be the event that some random field h takes only non-negative values on A. The goal is to determine $P^0_N ( \Omega^+(V_N))$. But for the purposes of this post, we will focus on showing bounds on the probability that the field is non-negative on a thin annulus near the boundary of $V_N$, since this is a self-contained step in the argument which contains a blog-friendly number of ideas.

We set $(L_N)$ to be a sequence of integers greater than one (to avoid dividing by zero in the statement), for which $\frac{L_N}{N}\rightarrow 0$. We now define for each N, the annulus

$W_N = \{v\in V_N: L_N\le d_{\mathbb{Z}^d}(v, V_N^c)\le 2L_N \}$

with radius $L_N$ set a distance $L_N$ inside the box $V_N$. We aim to control $P^N_0 (\Omega^+(W_N))$. This forms middle steps of Deuschel’s Propositions 2.5 and 2.9, which discuss $P^N_0(\Omega^+(V_{N-L_N}))$. Clearly there is the upper bound

$P^N_0(\Omega^+(V_{N-L_N})) \le P^N_0(\Omega^+(W_N))$ (1)

and a lower bound on $P^N_0(\Omega^+(V_{N-L_N}))$ is obtained in the second proposition by considering the box as a union of annuli then combining the bounds on each annulus using the FKG inequality.

Upper bound via odds and evens

After removing step (1), this is Proposition 2.5:

$\limsup_{N\rightarrow \infty} \frac{L_N}{N^{d-1} \log L_N} \log P^N_0(\Omega^+(W_N)) < 0.$ (2)

This is giving a limiting upper bound on the probability of the form $L_N^{-CN^{d-1}/L_N}$, though as with all LDP estimates, the form given at (2) is more instructive.

Morally, the reason why it is unlikely that the field should be non-negative everywhere within the annulus is that the distribution at each location is centred, and even though any pair of values are positively correlated, this correlation is not strong enough to avoid this event being unlikely. But this is hard to corral into an upper bound argument directly. In many circumstances, we want to prove upper bounds for complicated multivariate systems by projecting to get an unlikely event for a one-dimensional random variable, or a family of independent variables, even if we have to throw away some probability. We have plenty of tools for tail probabilities in both of these settings. Since the DGFF is normal, a one-dimensional RV that is a linear combination (eg the sum) of all the field heights is a natural candidate. But in this case we would have thrown away too much probability, since the only way we could dominate is to demand that the sum $\sum_{x\in W_N}h^N_x\ge 0$, which obviously has probability 1/2 by symmetry. (3)

So Deuschel splits $W_N$ into $W_N^o,W_N^e$, where the former includes all vertices with odd total parity in $W_N$ and the latter includes all the vertices with even total parity in the interior of $W_N$. (Recall that $\mathbb{Z}^d$ is bipartite in exactly this fashion). The idea is to condition on $h^N\big|_{W^o_N}$. But obviously each even vertex is exactly surrounded by odd vertices. So by the Gibbs-Markov property, conditional on the odd vertices, the values of the field at the even vertices are independent. Indeed, if for each $v\in W_N^e$ we define $\bar h_v$ to be the average of its neighbours (which is measurable w.r.t to the sigma-algebra generated by the odd vertices), then

$\{h_v: v\in W_N^e \,\big|\, \sigma(h_w: w\in W_N^o)\},$

is a collection of independent normals with variance one, and where the mean of $h_v$ is $\bar h_v$.

To start finding bounds, we fix some threshold $m=m_N\gg 1$ to be determined later, and consider the odd-measurable event $A_N$ that at most half of the even vertices v have $\bar h_v\ge m$. So $A_N^c\cap \Omega^+(W_N)$ says that all the odd vertices are non-negative and many are quite large. This certainly feels like a low-probability event, and unlike at (3), we might be able to obtain good tail bounds by projection into one dimension.

In the other case, conditional on $A_N$, there are a large number of even vertices with conditional mean at most m, and so we can control the probability that at least one is negative as a product

$(1-\varphi(m))^{\frac12 |W_N^e|}$. (4)

Note that for this upper bound, we can completely ignore the other even vertices (those with conditional mean greater than m).

So we’ll go back to $A_N^c \cap \Omega^+(W_N)$. For computations, the easiest one-dimensional variable to work with is probably the mean of the $\bar h_v$s across $v\in W_N^e$, since on $A_N^c\cap \Omega^+(W_N)$ this is at least $\frac{m}{2}$. Rather than focus on the calculations themselves involving

$\bar S^e_N:= \frac{1}{|W_N^e|} \sum\limits_{v\in W_N^e} \bar h_v,$

let us remark that it is certainly normal and centered, and so there are many methods to bound its tail, for example

$P^0_N \left( \bar S^e_N \ge \frac{m}{2} \right) \le \exp\left( \frac{-m^2}{8\mathrm{Var}(\bar S^e_N)} \right),$ (5)

as used by Deuschel just follows from an easy comparison argument within the integral of the pdf. We can tackle the variance using the Green’s function for the random walk (recall the first post in this set). But before that, it’s worth making an observation which is general and useful, namely that $\bar S^e_N$ is the expectation of

$S^e_N:= \sum{1}{|W_N^e|}\sum\limits_{v\in W_N^e} h_v$

conditional on the odds. Directly from the law of total variance, the variance of any random variable X is always larger than the variance of $\mathbb{E}[X|Y]$.

So in this case, we can replace $\mathrm{Var}(\bar S^e_N)$ in (5) with $\mathrm{Var}(S^e_N)$, which can be controlled via the Green’s function calculation.

Finally, we choose $m_N$ so that the probability at (4) matches the probability at (5) in scale, and this choice leads directly to (2).

In summary, we decomposed the event that everything is non-negative into two parts: either there are lots of unlikely local events in the field between an even vertex and its odd neighbours, or the field has to be atypically large at the odd sites. Tuning the parameter $m_N$ allows us to control both of these probabilities in the sense required.

Lower bound via a sparse sub-lattice

To get a lower bound on the probability that the field is non-negative on the annulus, we need to exploit the positive correlations in the field. We use a similar idea to the upper bound. If we know the field is positive and fairly large in many places, then it is increasingly likely that it is positive everywhere. The question is how many places to choose?

We are going to consider a sub-lattice that lives in a slightly larger region than $W_N$ itself, and condition the field to be larger than $m=m_N$ everywhere on this lattice. We want the lattice to be sparse enough that even if we ignore positive correlations, the chance of this happening is not too small. But we also want the lattice to be dense enough that, conditional on this event, the chance that the field is actually non-negative everywhere in $W_N$ is not too small either.

To achieve this, Deuschel chooses a sub-lattice of width $\lfloor\epsilon L_N^{2/d}\rfloor$, and sets $\Lambda_N(\epsilon)$ to be the intersection of this with the annulus with radii $[N-\frac{5}{2}L_N, N-\frac{1}{2}L_N]$, to ensure it lives in a slightly larger region than $W_N$ itself. The scaling of this sub-lattice density is such that when a random walk is started at any $v\in W_N$, the probability that the RW hits $\Lambda_N(\epsilon)$ before $\partial V_N$ is asymptotically in (0,1). (Ie, not asymptotically zero or one – this requires some definitely non-trivial calculations.) In particular, for appropriate (ie large enough) choice of $\epsilon$, this probability is at least 1/2 for all $v\in W_N$. This means that after conditioning on event $B_N:=\{h_v\ge m : v\in \Lambda_N(\epsilon)\}$, the conditional expectation of $h_w$ is at least $\frac{m}{2}$ for all $w\in W_N\backslash \Lambda_N(\epsilon)$. Again this uses the Gibbs-Markov property and the Gaussian nature of the field. In particular, this conditioning means we are left with the DGFF on $V_N\backslash \Lambda_N(\epsilon)$, ie with boundary $\partial V_N\cup \Lambda_N(\epsilon)$, and then by linearity, the mean at non-boundary points is given by the harmonic extension, which is linear (and so increasing) in the boundary values.

At this point, the route through the calculations is fairly clear. Since we are aiming for a lower bound on the probability of the event $\Omega^+(W_N)$, it’s enough to find a lower bound on $P^0_N(\Omega^+(W_N)\cap B)$.

Now, by positive correlation (or, formally, the FKG inequality) we can control $P^0_N(B)$ just as a product of the probabilities that the field exceeds the threshold at each individual site in $\Lambda_N(\epsilon)$. Since the value of the field at each site is normal with variance at least 1 (by definition), this is straightforward.

Finally, we treat $P^0_N(\Omega^+(W_N) \,\big|\, B)$. We’ve established that, conditional on B, the mean at each point of $W_N\backslash \Lambda_N(\epsilon)$ is at least $\frac{m}{2}$, and we can bound the variance above too. Again, this is a conditional variance, and so is at most the corresponding original variance, which is bounded above by $\sigma_N^2:=\mathrm{Var}(h^N_0)$. (This fact that the variance is maximised at the centre is intuitively clear when phrased in terms of occupation times, but the proof is non-obvious, or at least non-obvious to me.)

Since each of the event $h_v^N\ge 0$ for $v\in W_N\backslash \Lambda_N(\epsilon)$ is positively correlated with B, we can bound the probability it holds for all v by the product of the probabilities that it holds for each v. But having established that the conditional mean is at least $\frac{m_N}{2}$ for each v, and the variance is uniformly bounded above (including in N), this gives an easy tail bound of the form we require.

Again it just remains to choose the sequence of thresholds $m_N$ to maximise the lower bound on the probability that we’ve found in this way. In both cases, it turns out that taking $m_N= \sqrt{C\log N}$ is sensible, and this turns out to be linked to the scaling of the maximum of the DGFF, which we will explore in the future.

References

[1] – J-D Deuschel, Entropic Repulsion of the Lattice Free Field, II. The 0-Boundary Case. Available at ProjectEuclid.

# DGFF 2 – Boundary conditions and Gibbs-Markov property

In the previous post, we defined the Discrete Gaussian Free Field, and offered some motivation via the discrete random walk bridge. In particular, when the increments of the random walk are chosen to be Gaussian, many natural calculations are straightforward, since Gaussian processes are well-behaved under conditioning and under linear transformations.

Non-zero boundary conditions

In the definition of the DFGG given last time, we demanded that $h\equiv 0$ on $\partial D$. But the model is perfectly well-defined under more general boundary conditions.

It’s helpful to recall again the situation with random walk and Brownian bridge. If we want a Brownian motion which passes through (0,0) and (1,s), we could repeat one construction for Brownian bridge, by taking a standard Brownian motion and conditioning (modulo probability zero technicalities) on passing through level s at time 1. But alternatively, we could set

$B^{\mathrm{drift-br}}(t) = B(t)+ t(s-B(1)),\quad t\in[0,1],$

or equivalently

$B^{\mathrm{drift-br}}(t)=B^{\mathrm{br}}(t)+ st, \quad t\in[0,1].$

That is, a Brownian bridge with drift can be obtain from a centered Brownian bridge by a linear transformation, and so certainly remains a Gaussian process. And exactly the same holds for a discrete Gaussian bridge: if we want non-zero values at the endpoints, we can obtain this distribution by taking the standard centred bridge and applying a linear transformation.

We can see how this works directly at the level of density functions. If we take $0=Z_0,Z_1,\ldots,Z_{N-1},Z_N=0$ a centred Gaussian bridge, then the density of $Z=\mathbf{z}\in \mathbb{R}^{N+1}$ is proportional to

$\mathbf{1}\{z_0=z_N=0\}\exp\left( -\frac12 \sum_{i=1}^N (z_i-z_{i-1})^2 \right).$ (3)

So rewriting $z_i= y_i- ki$ (where we might want $k=s/N$ to fit the previous example), the sum within the exponent rearranges as

$-\frac12 \sum_{i=1}^N (y_i-y_{i-1} - k)^2 = -\frac12 \sum_{i=1}^N (y_i-y_{i-1})^2 - 2k(y_N-y_0)+ Nk^2.$

So when the values at the endpoints $z_0,z_n,y_0,y_N$ are fixed, this middle term is a constant, as is the final term, and thus the density of the linearly transformed bridge has exactly the same form as the original one.

In two or more dimensions, the analogue of adding a linear function is to add a harmonic function. First, some notation. Let $\varphi$ be any function on $\partial D$. Then there is a unique harmonic extension of $\varphi$, for which $\nabla \varphi=0$ everywhere on D, the interior of the domain. Recall that $\nabla$ is the discrete graph Laplacian defined up to a constant by

$(\nabla \varphi) _x = \sum\limits_{x\sim y} \varphi_x - \varphi_y.$

If we want $h^D$ instead to have boundary values $\varphi$, it’s enough to replace $h^D$ with $h^D+\varphi$. Then, in the density for the DGFF ( (1) in the previous post), the term in the exponential becomes (ignoring the $\frac{1}{4d}$ )

$-\sum\limits_{x\sim y} \left[ (h^D_x-h^D_y)^2 + (\varphi_x-\varphi_y)^2 +2(h^D_x - h^D_y)(\varphi_x-\varphi_y)\right].$

For each $x\in D$, on taking this sum over its neighbours $y\in \bar D$, the final term vanishes (since $\varphi$ is harmonic), while the second term is just a constant. So the density of the transformed field, which we’ll call $h^{D,\varphi}$ is proportional to (after removing the constant arising from the second term above)

$\mathbf{1}\left\{h^{D,\varphi}_x = \varphi_x,\, x\in\partial D\right\} \exp\left( -\frac{1}{4d} \sum\limits_{x\sim y} \left( h^{D,\varphi}_x - h^{D,\varphi}_y \right)^2 \right).$

So $h^{D,\varphi}:= h^D + \varphi$ satisfies the conditions for the DGFF on D with non-zero boundary conditions $\varphi$.

Harmonic functions and RW – a quick review

Like the covariances in DGFF, harmonic functions on D are related to simple random walk on D stopped on $\partial D$. (I’m not claiming a direct connection right now.) We can define the harmonic extension $\varphi$ to an interior point x by taking $\mathbb{P}_x$ to be the law of SRW $x=Z_0,Z_1,Z_2,\ldots$ started from x, and then setting

$\varphi(x):= \mathbb{E}\left[ \varphi_{\tau_{\partial d}} \right],$

where $\tau_{\partial D}$ is the first time that the random walk hits the boundary.

Inverse temperature – a quick remark

In the original definition of the density of the DGFF, there is the option to add a constant $\beta>0$ within the exponential term so the density is proportional to

$\exp\left(-\beta \sum\limits_{x\sim y} (h_x-h_y)^2 \right).$

With zero boundary conditions, the effect of this is straightforward, as varying $\beta$ just rescales the values taken by the field. But with non-zero boundary conditions, the effect is instead to vary the magnitude of the fluctuations of the values of the field around the (unique) harmonic function on the domain with those BCs. In particular, when $\beta\rightarrow \infty$, the field is ‘reluctant to be far from harmonic’, and so $h^D \Rightarrow \varphi$.

This parameter $\beta$ is called inverse temperature. So low temperature corresponds to high $\beta$, and high stability, which fits some physical intuition.

A Markov property

For a discrete (Gaussian) random walk, the Markov property says that conditional on a given value at a given time, the trajectory of the process before this time is independent of the trajectory afterwards. The discrete Gaussian bridge is similar. Suppose we have as before $0=Z_0,Z_1,\ldots, Z_N=0$ a centred Gaussian bridge, and condition that $Z_k=y$, for $k\in\{1,\ldots,N-1\}$, and $y\in\mathbb{R}$. With this conditioning, the density (3) splits as a product

$\mathbf{1}\{z_0=z_N=0, z_k=y\}\exp\left(-\frac12 \sum\limits_{i=1}^N (z_i-z_{i-1})^2 \right) =$

$\mathbf{1}\{z_0=0,z_k=y\} \exp\left(-\frac12 \sum\limits_{i=1}^k (z_i-z_{i-1})^2 \right) \cdot \mathbf{1}\{z_k=y,z_N=0\} \exp\left(-\frac12 \sum\limits_{i=k+1}^N (z_i-z_{i-1})^2 \right).$

Therefore, with this conditioning, the discrete Gaussian bridge splits into a pair of independent discrete Gaussian bridges with drift. (The same would hold if the original process had drift too.)

The situation for the DGFF is similar, though rather than focusing on the condition, it makes sense to start by focusing on the sub-domain of interest. Let $A\subset D$, and take $B=\bar D\backslash A$. So in particular $\partial A\subset B$.

Then we have that conditional on $h^D\big|_{\partial A}$, the restricted fields $h^D\big|_{B\backslash \partial A}$ and $h^D\big|_A$ are independent. Furthermore, $h^D\big|_A$ has the distribution of the DGFF on A, with boundary condition given by $h^D\big|_{\partial A}$. As in the discrete bridge, this follows just by splitting the density. Every gradient term corresponds to an edge in the underlying graph that lies either entirely inside $\bar A$ or entirely inside B. This holds for a general class of Gibbs models where the Hamiltonian depends only on the sum of some function of the heights (taken to be constant in this ‘free’ model) and the sum of some function of their nearest-neighbour gradients.

One additional and useful interpretation is that if we only care about the field on the restricted region A, the dependence of $h^D\big|_A$ on $h^D\big|_{D\backslash A}$ comes only through $h^D\big|_{\partial A}$. But more than that, it comes only through the (random) harmonic function which extends the (random) values taken on the boundary of A to the whole of A. So, if $h^A$ is an independent DGFF on A with zero boundary conditions, we can construct the DGFF $h^D$ from its value on $D\backslash A$ via

$h^D_x \stackrel{d}= h^A_x + \varphi^{h^D\big|_{\partial A}},$

where $\varphi^{h^D\big|_{\partial A}}$ is the unique harmonic extension of the (random) values taken by $h^D$ on $\partial A$ to $\bar A$.

This Markov property is crucial to much of the analysis to come. There are several choices of the restricted domain which come up repeatedly. In the next post we’ll look at how much one can deduce by taking A to be the even vertices in D (recalling that every integer lattice $\mathbb{Z}^d$ is bipartite), and then taking A to be a finer sublattice within D. We’ll use this to get some good bounds on the probability that the DGFF is positive on the whole of D. Perhaps later we’ll look at a ring decomposition of $\mathbb{Z}^d$ consisting of annuli spreading out from a fixed origin. Then the distribution of the field at this origin can be considered, via the final idea discussed above, as the limit of an infinite sequence of random harmonic functions given by the values taken by the field at increasingly large radius from the origin. Defining the DGFF on the whole lattice depends on the existence or otherwise of this local limit.

# DGFF 1 – The discrete Gaussian free field from scratch

I’ve moved to Haifa in northern Israel to start a post-doc in the probability group at the Technion, and now that my thesis is finished I want to start blogging again. The past couple of weeks have been occupied with finding an apartment and learning about the Discrete Gaussian Free Field. All questions about the apartment are solved, but fortunately lots remain open about the DGFF, so I thought I’d write some background about this object and methods which have been used to study it.

Background – Random walk bridge

When we think of a random walk, we usually think of the index as time, normally going forwards. So for a random walk bridge, we might assume $Z_0=0$, and then condition on $Z_N=0$, thinking of this as a demand that the process has returned to zero at the future time. In some applications, this is the ideal intuition, but in others, it is more useful to think of the random walk bridge

$(0=Z_0,Z_1,\ldots,Z_{N-1},Z_N=0),$

as a random height function indexed by [0,N], where the probability of a given path decomposes naturally into a product depending on the N increments, up to a normalising constant.

Naturally, we are interested in the asymptotic behaviour of such a random walk bridge when $N\rightarrow\infty$. So long as the step distribution has finite variance, a conditioned version of Donsker’s theorem shows that the rescaled random walk bridge converges in distribution to Brownian bridge. Note that Brownian bridge

$(B^{\mathrm{br}}_t, t\in[0,1])$

can be constructed either by conditioning a standard Brownian motion B to return to zero at time one (modulo some technicalities – this event has zero probability), or by applying an appropriate (random) linear shift

$B^{\mathrm{br}}(t):= B(t) - tB(1).$ (*)

It is not too hard to calculate the distribution of $B^{\mathrm{br}}(t)$ for each $t\in[0,1]$, and with a bit more work, one can calculate the joint distribution of $(B^{\mathrm{br}}(s),B^{\mathrm{br}}(t))$. In particular, the joint distribution is multivariate Gaussian, and so everything depends on the covariance ‘matrix’ (which here is indexed by [0,1]).

So if we return to a random walk bridge what should the step distribution be? Simple symmetric RW is a natural choice, as then lots of the quantities we might want to consider boil down to combinatorial calculations. Cleverness and Stirling’s formula can often get us useful asymptotics. But there are lots of inconveniences, not least the requirement to be careful about parity (N has to be even for a start unless you make the walk lazy, in which case the combinatorics becomes harder), and even if these can be overcome in a given calculation, it would be better not to have this.

The claim is that the random walk with Gaussian increments is by far the easiest to analyse asymptotically. As a further heuristic, think about the statement of the central limit theorem in the case where the underlying distribution is normal: it’s true but obvious. [Indeed, it’s my favourite piece of advice to anyone taking second year probability exams to check that your proposed statement of CLT does actually work for $N(\mu,\sigma^2)$…] More concretely, if a RW has Gaussian increments, then the path $(Z_1,\ldots,Z_N)$ is a multivariate normal, or a Gaussian process with finite index set. In particular, covariances define the distribution. It remains a Gaussian process after conditioning on $Z_N=0$, and the linear tilting argument at (*) remains true here, and can indeed be applied to turn any boundary conditions into any other boundary conditions.

The discrete Gaussian free field

We know how to generalise the domain of a random walk to higher dimensions. But what generalising the index to higher dimension? So now there is definitely no arrow of time, and the notion of a random height function above $\mathbb{Z}^2$ (or a subset of it) is helpful, for which a scaling limit might be a random surface rather than Brownian motion.

Because we can’t well-order $\mathbb{Z}^d$, it’s harder to define any such random object on the entire lattice immediately, so we start with compact connected subsets, with zero boundary conditions, as in the one-dimensional case of random walk bridge. Formally, let D be a finite subset of $\mathbb{Z}^d$, and the boundary $\partial D$ those elements of $D^c$ which are adjacent to an element of D, and let $\bar D:= D\cup \partial D$.

Then, the discrete Gaussian free field on D is a random real vector $h^D=(h^D_x: x\in \bar D)$, with probability density proportional to

$\mathbf{1}\{h^D_x=0, x\in\partial D\}\exp\left ( - \frac{1}{4d} \sum_{x\sim y}(h^D_x - h^D_y)^2 \right),$ (1)

where we write $x\sim y$ if that x,y are adjacent in $\bar D$. We won’t at any stage worry much about the partition function which normalises this pdf. Note also that $\frac{1}{4d}$ is just a convenient choice of constant, which corresponds to one of the canonical choices for the discrete Laplacian. Adjusting this constant is the same as uniformly rescaling the values taken by the field.

The immediate interpretation of (1) is that the values taken by the field at vertices which are close to each other are positively correlated. Furthermore, the form of the density is Gaussian. Concretely, if the values of $h^D$ are fixed everywhere except one vertex $x\in D$, then the conditional distribution of $h^D_x$ is Gaussian. Later, or in subsequent posts, we will heavily develop this idea. Alternatively, we could if we really wanted describe the model in terms of independent Gaussians describing the ‘increment’ along each edge in D (which we should direct), subject to a very large number of conditions, namely that the sum of increments along any directed cycle is zero. This latter description might be more useful if you wanted to define a DGFF on a more sparse graph, but won’t be useful in what follows.

Note that we can rearrange the Laplacian in (1) in terms of the transition kernel p( ) of the simple random walk of D to obtain

$\exp\left( -\frac12 (h^D)^T (\mathbf{P}-\mathbf{1})h^D \right),$

where $P_{x,y}=p(y-x)$ is the transition matrix of SRW on D. In particular, this means that the free field is Gaussian, and we can extract the covariances via

$\mathrm{Cov}(h^D_x,h^D_y) = \left[ (\mathbf{1}-\mathbf{P})^{-1}\right]_{x,y}$

$= \left[\sum_{n\ge 0} \mathbf{P}^n\right]_{x,y} = \sum_{n\ge 0} \mathbb{P}_x\left[X_n=y,\tau_{\partial D}>n\right],$

where, under $\mathbb{P}_x$, $(X_0,X_1,\ldots)$ is simple random walk started from x.

This final quantity records the expected number of visits to y before leaving the domain D, for a random walk started at x, and is called the Green’s function.

In summary, the DGFF on D is the centred Gaussian random vector indexed by $\bar D$ with covariance given by the Green’s function $G_D(x,y)$.

How many of these equivalences carries over to more general D-indexed random fields is discussed in the survey paper by Velenik. But it’s worth emphasising that having the covariance given by the Green’s function as in the definition we’ve just given is a very nice property, as there are lots of pre-existing tools for calculating these. By contrast, it’s hard to think of a natural model for an integer-valued surface of this kind, as an analogue to SRW.

[Though definitely not impossible. The nicest example I’ve heard of is for height functions of large uniform domino tilings within their ‘arctic circle’, which have GFF asymptotics. See this paper by Kenyon.]

A continuous limit?

We motivated the discussion of random walk bridge by the limit object, namely Brownian bridge. Part of the reason why the DGFF is more interesting than Gaussian random walk bridge, is that the limit object, the (continuum) Gaussian free field is hard to define classically in two dimensions.

We might suppose that the DGFF in $V_N$, the square box of width N has some scaling limit as $N\rightarrow\infty$. However, for fixed $x,y\in [0,1]^2$, (and taking integer parts component-wise), well-known asymptotics for SRW in a large square lattice (more on this soon hopefully) assert that

$\mathrm{Cov}(h^{V_N}_{\lfloor Nx \rfloor},h^{V_N}_{\lfloor Ny\rfloor}) \sim \log |x-y|,$ (2)

and so any scaling limit will rescale only the square domain, not the height (since there is no N on the RHS of (2)). However, then the variance of the proposed limit is infinite everywhere.

So the GFF does not exist as a random height function on $[0,1]^2$, with the consequence that a) more care is needed over its abstract definition; b) the DGFF in 2D on a large square is an interesting object, since it does exist in this sense.

What makes it ‘free’?

This seemed like a natural question to ask, but I’ve received various answers. Some sources seem to suggest that having zero boundary condition is free. Other sources refer to the Hamiltonian (that is the term inside the exponential function at (1) ) as free since it depends only on the increments between values. If the Hamiltonian also depends on the heights themselves, for example via the addition of a $\sum_{x} \Psi(h^D_x)$ term, then for suitable choice of function $\Psi$, this is interpreted as a model where the particles have mass. The physical interpretation of these more general Gibbs measures is discussed widely, and I’m not very comfortable with it all at the moment, but aim to come back to it later, when hopefully I will be more comfortable.

# The Envelope ‘Paradox’

At the recent IMO in Hong Kong, there were several moments where the deputy leaders had to hang around, and I spent some of these moments discussing the following problem with Stephen Mackereth, my counterpart from New Zealand. He’s a mathematically-trained philosopher, so has a similar level of skepticism to me, but for different reasons, regarding supposed paradoxes in probability. Because, as we will see shortly, I don’t think this is a paradox in even the slightest fashion, I think there’s probably too much written about this on the internet already. So I’m aware that contributing further to this oeuvre is hypocritical, but we did the thinking in HKUST’s apparently famous Einstein Cafe, so it makes sense to write down the thoughts.

[And then forget about it for eight weeks. Oops.]

Here’s the situation. A cryptic friend gives you an envelope containing some sum of money, and shows you a second envelope. They then inform you that one of the envelopes contains twice as much money as the other. It’s implicit in this that the choice of which is which is uniform. You have the option to switch envelopes. Should you?

The supposed paradox arises by considering the amount in your envelope, say X. In the absence of further information, it is equally likely that the other envelope contains X/2 as 2X. Therefore, the average value of the other envelope is

$\frac12 \left(\frac{X}{2}+2X \right)= \frac54 X > X.$

So you should switch, since on average you gain money. But this is paradoxical, since the assignment of larger and smaller sums was uniform, so switching envelope should make no difference.

Probabilistic setup

This is not supposed to be a problem on a first-year probability exercise sheet. It’s supposed to be conducive to light discussion. So saying “I won’t engage with this problem until you tell me what the probability space is” doesn’t go down terribly well. But it is important to work out what is random, and what isn’t.

There are two sources of randomness, or at least ignorance. Firstly, there is the pair of values contained in the envelopes. Secondly, there is the assignment of this pair of values to the two envelopes. The second is a source of randomness, and this problem is founded on the premise that this second stage is ‘symmetric enough’ to smooth over any complications in the first stage. If we think that probability isn’t broken (and that’s what I think), then the answer is probably that the second stage isn’t symmetric enough.

Or, that the first stage isn’t very well-defined. In what follows, I’m going to make the second stage very symmetric, at the expense of setting up the first stage in what seems to me a reasonable way using the conventional language of probability theory to record our ignorance about the values in play.

So what’s the first stage? We must have a set of possible pairs of values taken by the envelopes. Let’s call this A, so

$A\subset \mathbb{A}:=\{(x,2x)\,:\, x\in (0,\infty)\}.$

Maybe we know what A is, but maybe we don’t, in which we should take $A=\mathbb{A}$, on the grounds that any pair is possible. Suppose that your friend has chosen the pair of values according to some distribution on $\mathbb{A}$, which we’ll assume has a density f, which is known by you. Maybe this isn’t the actual density, but it serves perfectly well if you treat it as *your* opinion on the likelihood. Then this actually does reduce to a problem along the lines of first-year probability, whether or not you get to see the amount in your envelope.

Suppose first that you do get to see the amount, and that it is x. Then the conditional probabilities that the pair is (x/2,x) or (x,2x) are, respectively

$\frac{f(x/2,x)}{f(x/2,x)+f(x,2x)},\quad \frac{f(x,2x)}{f(x/2,x)+f(x,2x)}.$

So you can work out your expected gain by switching, and decide accordingly. If you don’t know the value in your envelope, you can still work out the probability that it is better (in expectation) to switch, but this isn’t really a hugely meaningful measure, unless it is zero or one.

It’s worth noting that if you can view inside your envelope, and you know A has a particular form, then the game becomes determined. For example, if

$A\subset \{(n,2n), n\text{ an odd integer}\},$

then life is very easy. If you open your envelope and see an odd integer, you should switch, and if you see an even integer you shouldn’t.

We’ll return at the end to discuss a case where it is always better to switch, and why this isn’t actually a paradox.

Improper prior and paradox of resampling when $\mathbb{E}=\infty$

For now though, let’s assume that we don’t know anything about the amounts of money in the envelopes. Earlier, we said that “in the absence of further information, it is equally likely that the other envelope contains X/2 as 2X”. In the language of a distribution on $\mathbb{A}$, we are taking the uniform measure. Of course this not a distribution, in the same way that there isn’t a uniform distribution on the positive reals.

However, if this is your belief about the values in the pair of envelopes, what do you think is the mean value of the content of your envelope? Well, you think all values are equally likely. So, even though this isn’t a distribution, you pretty much think the value of your envelope has infinite expectation.

[This is where the philosophy comes in I guess. Is (expressing uniform ignorance about the content of the other envelope given knowledge of your own) the same as (expressing uniform ignorance of both envelopes at the beginning)? I think it is, even though it has a different consequence here, since the former can be turned into a proper distribution, whereas the latter cannot.]

Let’s briefly consider an alternative example. It’s fairly easy to conjure up distributions which are almost surely finite but which have infinite expectation. For example $\mathbb{P}(X=2^k)=2^{-k}$ for k=1,2,…, which is the content of the *St. Petersburg paradox*, another supposed paradox in probability, but one whose resolution is a bit more clear.

Anyway, let X and Y be independent copies of such a distribution. Now suppose your friend offers you an envelope containing amount X. You look at the value, and then you are offered the opportunity to switch to an envelope containing amount Y. Should you?

Well, if expectation is what you care about, then you definitely should. Because with probability one, you are staring at a finite value in your envelope, whereas the other unknown envelope promises infinite expectation, which is certainly larger than the value that you’re looking at.

Is this also a paradox? I definitely don’t think it is. The expectation of the content of your envelope is infinite, the expected gain is infinite with probability one, which is consistent with the expected content of the other envelope being infinite. [Note that you don’t want to be claiming that the expectation of X-Y is zero.]

An example density function

As an exercise that isn’t necessarily hugely interesting, let’s assume that f, the distribution of the smaller of the pair, is $\mathrm{Exp}(\lambda)$. So the mean of this smaller number is $1/\lambda$. Then, conditional on seeing x in my envelope, the expected value of the number in the other envelope is

$\frac{\frac{x}{2} e^{-\lambda x/2} + 2x e^{-\lambda x}}{e^{-\lambda x/2}+ e^{-\lambda x}}.$ (*)

Some straightforward manipulation shows that this quantity is at least x (implying it’s advantageous to switch) precisely when

$e^{-\lambda x/2}\ge \frac12.$

That is, when $x\le \frac{2\log 2}{\lambda}$. The shape of this interval should fit our intuition, namely that the optimal strategy should be to switch if the value in your envelope is small enough.

The point of doing this calculation is to emphasise that it ceases to be an interesting problem, and certainly ceases to be a paradox of any kind, once we specify f concretely. It doesn’t matter whether this is some true distribution (ie the friend is genuinely sampling the values somehow at random), or rather a perceived likelihood (that happens to be normalisable).

What if you should always switch?

The statement of the paradox only really has any bite if the suggestion is that we should always switch. Earlier, we discussed potential objections to considering the uniform prior in this setting, but what about other possible distributions f which might lead to this conclusion?

As at (*), we can conclude that when $f(x)+f(x/2)>0$, we should switch on seeing x precisely if

$f(x)\ge 2f\left(\frac{x}{2}\right).$

Therefore, partitioning the support of f into a collection of geometric sequences with exponent 2, it is clear that the mean of f is infinite if everything is integer-valued. If f is real-valued, there are some complications, but so long as everything is measurable, the same conclusion will hold.

So the you-should-switch-given-x strategy can only hold for all values of x if f has infinite mean. This pretty much wraps up my feelings. If the mean isn’t infinite, the statement of the paradox no longer holds, and if it is infinite, then the paradox dissolves into a statement about trying to order various expectations, all of which are infinite.

Conclusions

Mathematical summary: it’s Bayes. Things may be exchangeable initially, but not once you condition on the value of one of them! Well, not unless you have a very specific prior.

Philosophical summary: everything in my argument depends on the premise that one can always describe the protagonist’s prior opinion on the contents of the pair of envelopes with a (possibly degenerate) distribution. I feel this is reasonable. As soon as you write down $\frac12 \cdot\frac{x}{2} + \frac12 \cdot2x$, you are doing a conditional expectation, and it’s got to be conditional with respect to something. Here it’s the uniform prior, or at least the uniform prior restricted to the set of values that are now possible given the revelation of your number.

Second mathematical summary: once you are working with the uniform prior, or any measure with infinite mean, there’s no reason why

$\mathbb{E}\left[X|Y\right]>Y,$

with probability one (in terms of Y) should be surprising, since the LHS is (almost-surely) infinite while the RHS is almost surely finite, despite having infinite mean itself.

# Questionable Statistics

My four years in the Statistics Department have been characterised by a continual feeling that I don’t know enough about statistics. That said, I can’t help but notice how statistics are used in non-specialist contexts, and of course there are times when eyebrows might reasonably be raised. Andrew Gelman (who gave a very enjoyable departmental seminar on this topic in May) has a notable blog critiquing poor statistical methodology, especially in social science; and the blogosphere is packed with compilations of egregious statistical oxymorons (“All schools *must* be above average” etc).

I’m aiming for more of a middle ground here. I’ve picked three examples of articles I found interesting at some level over the past few months. In each of them, some kind of data presentation or experiment design arose, and in all cases, I think it resulted in more questions rather than more answers, not really in a good way, since in all cases I did actually want to know more about what was going on.

Extreme rounding errors

This Guardian article about inequality in career trajectories for teachers in schools serving more advantaged and more deprived areas raises a lot of interesting questions, and actually proposes answers to several of them. The problem is the double bar chart halfway down.

First a superficial moan. Rounding to the nearest integer is fine, indeed sensible in most contexts, but not when you are treating small numbers. Comparing the extreme bars, we could have 11.5 v 12.5 or we could have 10.5 v 13.5. The headlines would probably read “N% more teachers leave disadvantaged schools”. In the first case N is 9, in the second it’s 29. So it’s not really a superficial moan. Data is supposed to offer a measure of the effect under consideration, and a discussion of how treat this effect requires a veneer of reliability about the size of the effect. We don’t just need to be asking “how can we solve this”; we also need to be asking “is this worth solving?” I’m not saying the answer is no, but based on this graph alone the answer is not ‘definitely yes’. [1]

A question to ask in the context of this report is “why do more teachers leave the profession from schools in deprived areas?” But I think most people could speculate a broadly sensible answer to this. The data raises the more subtle question “why are the rates of leaving the profession so similar across all types of school?” Essentially I think this survey is a negative result – the effect just isn’t as strong as I’d have suspected, nor I imagine the authors nor the Guardian education editorship.

By contrast, even with the amateurish presentation, the rates of moving school clearly are significant and worth talking about based on this survey. Significant data demands further information though. Once an effect is significant, we need to know more details, like to what extent the teachers are moving to schools in the same band or to mostly towards the left end of the graph. Again though, there’s no comment on the magnitude of this effect. Assuming the rate is given per year [2], then among teachers who do not leave the profession, the average tenure of a given job is more than ten years, even in the most deprived category. Maybe this is all very heavy-tailed, and so the average isn’t a good thing to be using. But this doesn’t seem like an unreasonable number to me? If the claim is that this is a major problem for the teaching profession, then we should be told what the numbers are in vaguely comparable industries.

The final thing is pretty bad. Three paragraphs under the graph, it is claimed that the rate of leaving is 70% higher in most- than least-deprived categories. We decided that N was somewhere between 9 and 29. But not 70. I can only assume that this came from the ‘moving’ category instead…

[1] I’m assuming the sample pool was large enough that the effect is statistically significant, because there isn’t a direct link to the source data or even the source report unfortunately.

[2] and is it plausible that this is rate per year? Does the profession really experience ten percent annual turnover? Is it even plausible that the scale is the same? Maybe rate of leaving really is twice rate of moving school, and I have no idea about anything, but this seems questionable.

Many, many categories

The bank Halifax commissions an annual survey about pocket money, and this year’s received widespread coverage. My attention was caught by this article on BBC news. The short summary is that they surveyed about 1,200 children from about 600 families (so in many cases a direct comparison between siblings could have been possible), and it sounds like they restricted to the age range 8-15, and asked how much pocket money they received, and whether they wanted more.

The clickbait summary was that boys receive about 13% more than girls. While it doesn’t give the exact numbers, it also mentions that more boys than girls thought they deserved more money. A psychologist conjectures that this latter effect might explain the former effect, and indeed this seems plausible. The problem is that elsewhere in the article, it says that in the corresponding survey in 2015, boys receive on average only 2% more than girls.

We therefore have to consider a couple of things. 1) Is it plausible that the actual average pocket money rates (among the whole population) have fluctuated in such a gender-dependent way in 12 months? 2) If not, then can we still say something useful, even if our confidence in the methodology of the survey is reduced?

I think the answer to the first question is definitely ‘no’. So we come to the second question. Firstly, how could this have arisen? Well, there’s no indication how the children were chosen, but equally it’s hard to think of a way of choosing them that would increase artificially the boys’ pocket money. In this last sentence, I’m deliberately suggesting the null hypothesis that boys and girls are the same. This is misleading. For now, our null hypothesis is that 2015 should be the same as 2016. Indeed, it’s equally plausible that this year’s summary data was ‘more correct’, and so we are asking why the difference was excessively compressed in 2015.

This is entirely a matter of effect size, and it’s to the article’s credit that so much information is given further down the page that we can actually make the following remarks quantitatively. At the bottom, they show the average pay (across genders) in different regions. There is a lot of variation here. The rate of pay in East Anglia is only 60% what it is in London. At some level, I’m surprised it isn’t even higher. The article tells us that children between the ages of eight and fifteen were surveyed. By what rate did your pocket money change in that interval? Surely it doubled at the very least? Weekly pocket money at age eight is for sweets and an occasional packet of Pokemon cards (substitute 21st century equivalent freely…). Teenagers need to pay for outings and clothes etc, so there’s no reason at all why this should be comparable. For the purposes of definiteness in the next calculation, let’s say that fifteen-year-olds get paid four times as much as eight-year-olds on average, which I think is rather conservative.

So here’s a plausible reason for the fluctuations. Suppose between 2015 and 2016, the survey substituted 10 eight-year-old boys from the Midlands for 10 fifteen-year-old boys from London. The expectation change in the average amongst the 600 boys surveyed is

$\frac{10}{600} \times (\sim 6.5) \times (4 \times (0.6)^{-1}) \approx 0.7,$

in pounds, which is on the order of the actual change from 2015 to 2016. Choosing the participants in a non-uniform way seems like too basic a mistake to make, but mild fluctuations in the distribution of ages and locations, as well as the occasional outlier (they want to use the mean, so one oligarch’s daughter could make a noticeable difference in a sample size of 600) seems more plausible as an explanation to me than a general change in the population rates. Choosing participants in a uniform way is just hard when there are loads and loads of categories.

I’m not really saying that this survey is bad – they have to report what they found, and without a lot more information, I have no idea whether this year’s is more or less plausible than last year’s. But if you add the phrase “especially in 2016” to everything the psychologist says, it suddenly seems a bit ridiculous. So it would be worth making the remark that even if this effect is statistically significant, that doesn’t mean the effect size is large relative to lots of other less controversial effect sizes visible in the data.

Comparing precious metals

I’ve recently returning from this year’s International Mathematical Olympiad, and now we are well into the swing of its sweatier namesake in Rio. At both events, there have been many opportunities to observe how different people with different ambitions and levels of extroversion display different levels of pleasure at the same medal outcomes. In the light of this, I was drawn to this article, which has been getting quite a lot of traction online.

The basic premise is simple. Silver medallists are less happy than bronze medallists (when – unlike at the IMO – there is only one of each medal), and it’s not hard to come up with a cheap narrative: silver-winners are disappointed to have missed out on gold; bronze-winners are glad to have got a medal at all. This is all pretty convincing, so maybe there’s no need actually to collect any data, especially since asking a crowd of undergraduates to rate faces on the 1-10 agony-ecstacy scale sounds like a bit of an effort. Let’s read a couple of sentences, describing a follow-up study on judo medallists at the Athens Olympics:

Altogether, they found that thirteen of the fourteen gold medal winners smiled immediately after they completed their winning match, while eighteen of the twenty-six bronze medalists smiled. However, none of the silver medalists smiled immediately after their match ended.

You might wonder how come there are almost twice as many bronzes as golds. Well, very full details about the judo repechage structure can be found here, but the key information is that all bronze-medallists won their final match, as, naturally, did the gold-medallists. The silver-medallists lost their final match, ie the gold-medal final. So this study is literally reporting that highly-driven competitive sportsmen are happier straight after they win, than straight after they lose. This doesn’t have much to do with medals, and isn’t very exciting, in my opinion. I would, however, be interested to know what was eating the one gold medallist who didn’t smile immediately.

This isn’t really a gripe about the statistics, more about the writing. They are testing for an effect which seems highly plausible, but is branded as surprising. The study for which they give more methodological details in fact seems to be testing for a related effect, which is not just plausible, but essentially self-evident. So naturally I want to know more about the original study, which is the only place left for there to be any interesting effects, for example if they looked at athletics events which don’t have a binary elimination structure. But we aren’t told enough to find these interesting effects, if they exist. How annoying. The only thing left is to think about is my bronze medal at IMO 2008. They said eight years would be enough for the wounds to heal but I’m still not smiling.

# IMO 2016 Diary – Part Four

A pdf of this report is also available here.

Thursday 14th July

I have now spent a while thinking about square-free n in Q3 after rescaling, and I still don’t know what the markscheme should award it. I therefore request that Joe and Warren receive the same score as each other, and any other contestant who has treated this case. In my opinion this score should be at most one, mainly as a consolation, but potentially zero. However, we are offered two, and after they assure me this is consistent, I accept.

There is brief but high drama (by the standards of maths competitions) when we meet Angelo the Australian leader, who confirms that he has just accepted one mark for almost the same thing by his student Johnny. A Polish contestant in a similar situation remains pending, so we all return for a further meeting. I’m unconvinced that many of the coordinators have read all the scripts in question, but they settle on two for everyone, which is consistent if generous. The only drama on Q5 is the ferocious storm that sets in while I’m making final notes in the plaza. Again though, coordinator Gabriele has exactly the same opinion on our work as Geoff and I, apart from offering an additional mark for Lawrence’s now slightly damp partial solution.

And so we are finished well before lunch, with a total UK score of 165 looking very promising indeed. I’m particularly pleased with the attention to detail – Jacob’s 6 on Q4 is the only mark ‘dropped’, which is brilliant, especially since it hasn’t come at the expense of the students’ usual styles. We’ll have to wait until later to see just how well we have done.

It would be nice to meet the students to congratulate them in person, but they are with Jill on the somewhat inaccessible Victoria Peak, so instead I take a brief hike along the trail down the centre of HK Island, ending up at the zoo. This turned out to be free and excellent, though I couldn’t find the promised jaguar. There was, however, a fantastic aviary, especially the striking flock of scarlet ibis. A noisy group of schoolchildren are surrounding the primates, and one lemur with an evil glint in his eye swings over and languidly starts an activity which elicits a yelp from the rather harried teacher, who now has some considerable explaining to do.

With 1000 people all returning to UST at roughly 6.30, dinner is not dissimilar to feeding time at the zoo, and afterwards various leaders lock horns during the final jury meeting. Two countries have brought an unresolved coordination dispute to the final meeting, and for the first time since I became deputy leader, one of them is successful. Congratulations to the Koreans, who now have a third student with a highly impressive perfect score. Andy Loo and Geoff chair the meeting stylishly and tightly, and although there are many technical things to discuss, it doesn’t drag for too long. Eventually it’s time to decide the medal boundaries, and the snazzy electronic voting system makes this work very smoothly. I feel the gold and bronze cutoffs at 29 and 16 are objectively correct, and the 50-50 flexibility at silver swings towards generosity at 22. We can now confirm the UK scores as:

This is pretty much the best UK result in the modern era, placing 7th and with a medal tally tying with the famous food-poisoning-and-impossible-geometry IMO 1996 in India. But obviously this is a human story rather than just a 6×6 matrix with some summary statistics, and Harvey in particular is probably looking at the world and thinking it isn’t fair, while Warren’s gold is the ideal end to his four years at the IMO, two of which have ended one mark short. The American team are pretty keen to let everyone know that they’ve placed first for the second year in succession, and their remarkable six golds will hopefully allow scope for some good headlines. There is much to talk about, celebrate and commiserate, and this continues late into the night.

Friday 15th July

Our morning copy of the IMO Newsletter includes an interview with Joe, with the headline ‘Meh’. Frank Morgan has rather more to say, which is good news, since he’s delivering the IMO lecture on Pentagonal Tilings. He discusses the motivation of regular tilings where the ratio Perimeter/Area is minimised, starting from questions about honeycombs raised by the Roman author Varro! We move onto more mathematical avenues, including the interesting result of L’Huilier that given a valid set of angles, the associated polygon with minimal Perimeter/Area has an incircle, and the corresponding result for in-n-spheres in higher dimension. A brief diversion to the beach on the way home is punctuated with attempts to project the hyperbolic plane onto the sand.

The day’s main event is the closing ceremony, held at the striking Hong Kong Convention Centre. As usual, the adults and our students have been vigorously separated for the journey. As I arrive, it seems the UK boys have been directing a massed gathering behind the EU flag on stage, while the non-European teams are divided into two sides in a giant paper aeroplane dogfight. All attempts by the organisers to quash this jocularity are being ignored, and after bringing everyone here two hours early, I have minimal sympathy. Geoff sits on a secluded bench, and agrees to the many selfie requests from various teams with regal if resigned tolerance.

The ceremony is started by a fantastically charismatic school brass band, and proceeds with some brief speeches, and more astonishing drumming. Then it’s time to award the medals. Lawrence and Jacob get to go up together among the clump of 24-scorers, while Kevin from Australia does an excellent job of untangling his flag and medal while keeping hold of the ubiquitous cuddly koala. Neel has been threatened with death if he appears on stage again with an untucked shirt, but no direction is required for his and Warren’s smiles as they receive the gold medallists’ applause.

Afterwards, there is a closing banquet. We get to join British coordinators James and Joseph for a climate-defying carrot soup, followed by a rare diversion onto Western carbohydrates accompanying what is, for many of us, a first taste of caviar. Both Geoff and the American team are forced to make speeches at no notice. It is all generally rather formal, and fewer photographs are taken than usual. An attempt to capture Joe and Harvey looking miserable results in one the biggest grins of the evening. The UK and Australian teams have a thousand stickers and micro-koalas to give out as gifts, and some of the attempts at this descend into silliness. All clothing and body parts are fair game, and Jacob makes sure that Geoff is fully included. The UK and Australian leaders, variously coated, retreat from the carnage to the relative safety of our top-floor balcony as the IMO drifts to an end, until midnight, when it seems sensible to find out what the students are up to.

Saturday 16th July

This is what the students are up to. When we arrived at UST last week, everyone was given food vouchers to redeem at the campus’s various restaurants. Very very many of these are left over, and, despite the haute cuisine on offer earlier, people are hungry. They have therefore bought McDonalds. And I mean this literally. Animated by Jacob and American Michael, they have bought the entire stock of the nearest branch. If you want to know what 240 chicken nuggets looks like, come to common room IX.1, because now is your chance. Fortunately our team have made many friends and so after the Herculean task (I make no comment on which Herculean labour I feel this most resembles) of getting it to their common room, pretty much the entire IMO descends to help. Someone sets up a stopmotion of the slow erosion of the mountain of fries, while the usual card games start, and a group around a whiteboard tries to come up with the least natural valid construction for n=9 on Q2. Around 3.30am everything is gone, even the 30 Hello Kitties that came with the Happy Meals, and we’re pre-emptively well on the way to beating jetlag.

I wake up in time to wave Geoff off, but he’s been bumped to an earlier bus, so the only thing I see is Lawrence and colleagues returning from a suicidal 1500m round the seaside athletics track. Our own departure is mid-morning, and on the coach the contestants are discussing some problems they’ve composed during the trip. They’ll soon be able to submit these, and by the sounds of it, anyone taking BMO and beyond in 2018 has plenty to look forward to. Jacob has already mislaid his room key and phone, and at the airport he’s completed the hat-trick by losing one of the two essential passport insert pages. Fortunately, it turns out that he’s lost the less essential one, so we can clear security and turn thoughts towards lunch.

Jill has given me free licence to choose our dim sum, so the trip ends with pork knuckle and chicken feet. Our aim is to stay awake for the whole flight, and Neel helps by offering round copies of a Romanian contest from 2010, while I start proof-reading. By the time they finish their paper, many rogue commas have been mercilessly expunged. It should be daylight outside, but the windows are all shut, and by the ninth hour time starts to hang drowsily in a way that combinatorial geometry cannot fix, and so the mutual-waking-up pact kicks in, aided by Cathay Pacific’s unlimited Toblerone. Winding through Heathrow immigration, Joe unveils his latest airport trick of sleeping against vertical surfaces. We diverge into the non-humid night.

Reflection

There’s a great deal more to life and mathematics than problem-solving competitions, but our contestants and many other people have worked hard to prepare for IMO 2016 over the past months (and years). So I hope I’m allowed to say that I’m really pleased for and proud of our UK team for doing so well! The last three days of an IMO are very busy and I haven’t had as much time as I’d have liked to talk in detail about the problems. But I personally really liked them, and thought the team showed great taste in choosing this as the British annus mirabilis in which to produce lots of beautiful solutions.

But overall, this is really just the icing on the cake of a training progamme that’s introduced lots of smart young people to each other, and to the pleasures of problem-solving, as well as plenty of interesting general mathematics. I have my own questions to address, and (unless I’m dramatically missing something) these can’t be completed in 4.5 hours, but as ever I’ve found the atmosphere of problem discussion totally infectious, so I hope we are doing something right.

Lawrence and Warren are now off to university. I’m sure they’ll thrive in every way at this next stage, and hopefully might enjoy the chance to contribute their energy and expertise to future generations of olympiad students. The other four remain eligible for IMO 2017 in Brazil, and while they will doubtless have high personal ambitions, I’m sure they’ll also relish the position as ideal role models for their younger colleagues over the year ahead. My own life will be rather different for the next two years, but our camp for new students is held in my no-longer-home-town Oxford in a few weeks’ time, and I’m certainly feeling excited about finding some new problems and doing as much as possible of the cycle all over again!

# IMO 2016 Diary – Part Three

Sunday 10th July

I’m awake at 6am and there’s nothing to do, so take a short run along the edge of the bay. I meet an old lady singing along to a walkman (yes, really) while doing taichi. She encourages me to join and it seems rude to refuse. Suffice it to say I’m as grateful no video evidence exists as she should be that no audio recording was made. Six-hundred mathematicians queueing for powdered eggs seems like an unwelcome start to the day, so we are self-catering. The guides have been commanded to show every student how to find their place in the exam hall, and I approve of Allison’s contempt for the triviality of this task.

The main event of the day is the opening ceremony, held at the Queen Elizabeth stadium in the centre of Hong Kong Island. To no-one’s surprise, this involves a lot of time waiting around in the stifling UST plaza, which the students use to take a large number of photographs. The UK and Australian boys are smartly turned out as usual, but the polyester blazers are rather ill-suited to this tropical conditions, so we invoke Red Sea rig until air conditioning becomes available. The Iceland team are particularly keen to seek out the English members for reasons connected to a football match of which Neel proudly claims total ignorance. I picked up an EU flag for next-to-nothing last Friday, and now Jacob and Warren prove very popular as they circulate inviting our (for now) European colleagues to join us behind the stars.

The deputies are segregated in an upper tier and obliged to watch a rehearsal of the parade. Some of the organisers have a confused interpretation of the IMO roles. I still have some of the uniform with me, but an official says it is literally impossible for me to give it to the team. She is small and Joe Benton can catch flying ties as well as colds, so it turns out to be literally entirely possible, but for my trouble I get called ‘a very bad boy’.

Many hours after we left our rooms, the ceremony starts, and is actually very good, with a handful of well-chosen speeches, a mercifully quickfire parade of teams, and musical interludes from a full symphony orchestra, with various traditional and non-traditional percussion. The new IMO song Every day in love we are one involves a B section accompanied by a melange of watercooler bottles, but despite its catchy conclusion about maths, friendship and beyond, I suspect it may not trouble the top of the charts.

Monday 11th July

It’s the morning of the first IMO paper, and you can feel both the excitement and the humidity in the air. Some of our boys are looking a bit under the weather, but we know from past experience that the adrenaline from settling down in a room of 600 young contestants who’ve been preparing for exactly this can carry them through anything. I skip an excursion in order to receive a copy of the contest paper. Security is tight, and the deputies who have chosen this option are locked in a lecture theatre for two hours, and our bathroom visits monitored with commendable attention to detail. I guess that the combinatorial second problem is most likely to provoke immediate discussion, so I spend my time working through the details of the argument, just in time to meet our contestants when their 4.5 hours are up.

Q3 has been found hard by everyone, and Q2 has been found hard by other countries. Harvey’s kicking himself for drawing the wrong diagram for the geometry, an error that is unlikely to improve Geoff’s mood when he receives the scripts later today. Apart from that, we have a solid clutch of five solutions to each of the first two problems, and various nuggets of progress on the final problem, which is an excellent start. Several of the team are itching to keep trying to finish Q3, but the campus is likely to be annoying hotbed of spurious gossip all day, so Allison and I take them out. The very convenient MTR takes us under the harbour while the students and I debate the usefulness of the square-free case, and how well it is preserved under rescaling so that the circumcentre is a lattice point.

As we emerge above ground, Jacob is entranced by the live-action Finding Dory playground at Causeway Bay, and we toy with buying a pig’s trotter from a nearby market, but not even Lawrence is feeling adventurous enough with another exam tomorrow. We travel over to Kowloon via double-decker tram and ferry, and fortified by ice cream, take lots of photographs of the unique HK skyline, where even the giant waterfront office towers are dwarfed by Victoria Peak, which the contestants will visit while I’m marking. On our return journey, some of the team are impressed by the HK rush hour, indicating that they’ve clearly never tried to change line at Leicester Square around 6pm on a Friday…

Tuesday 12th July

Another morning, another trek uphill to a 4.5 hour exam. Time passes rapidly, especially now I’ve worked out how to order coffee without the ubiquitous condensed milk. The security arrangements concerning the deputies’ copies of the paper have been increased even further, but the IMO photographers have outdone themselves, and published on Instagram some pictures of the exam room with a level of crispness such that it’s clear the paper includes no geometry, and after finally getting hold of a proper hard copy, it looks like a paper which the UK team should really enjoy.

As so often after IMO papers, there is a range of reactions. Lawrence is unsure whether he presented his exemplar polynomial in a form that actually works. Joe knows and I know that he could easily have got at least 35 on these papers, but after over-meta-thinking himself on Q5, this isn’t his year. Like Aeneas gazing on the ruins of Troy, sunt lacrimae rerum, but also plans for new foundations. By contrast, Harvey has atoned for yesterday’s geometric lapse with what sounds like a perfect score today. Warren and Neel seem to be flying overall, and are doing a good job of keeping their excitement under control while the others muse. There’s plenty to think about, and Geoff has now arrived bearing yesterday’s scripts and several novels’ worth of anecdotes from the leaders’ site.

Before getting down to business, it feels sensible to walk off the Weltschmerz, and provide an outlet for joy in the nearby Clearwater Bay country park. There’s a long trail all over the New Territories, and we join it for a brief but purposeful stroll up through the light jungle and along the ridge. We’re confident we didn’t find the global maximum, but we find a couple of local maxima with great views out around the coastline, which seems to have Hausdorff dimension slightly greater than 1. We see some enormous spiders (though the Australians are substantially less impressed) before ending up an uncontroversial minimum where Jill has bedded in with merciful bottles of water on the beach. To say we are sticky doesn’t even begin to cover it but, crucially, we are no longer consumed by the morning’s events.

The UK boys are now masters of the complicated UST food court ordering process, and Warren endears himself to Geoff by producing a steaming bowl of spicy ramen as if by magic. The contestants have a ‘cultural night’, which apparently includes a greater number of hedge fund representatives than one might have expected. For me, it’s a night in with Geoff, green tea and the scripts for Q2. Joe and Neel have filled fourteen pages between them checking a construction in glorious detail, a step which Harvey has described in its entirety with the words ‘glue them together’. Overall, they are complicated but precise, and I have few concerns, so it’s only necessary to burn the candle at one end.

Wednesday 13th July

It’s time for coordination, where Geoff and I agree the UK marks with a team of local and international experts. The scheduling has assigned us the Q1 geometry early in the morning, which is a clear case of five perfect solutions, so we move to Q2. Coordinator Stephan seems very well-prepared for the UK scripts, so again we are finished in a matter of minutes. This allows us to bring forward our discussion of Q4. Jacob has made several small errors, all of which could be fixed by attacking his script with a pair of scissors and some glue. I believe the mark scheme should award this 4+2, and coordinator Juan thinks it should be 5+1. We are both open to each other’s interpretations, and have at least basic proficiency in addition, so again there is little need for debate.

The early evening brings the main challenge of the day, Q6, at which the UK has excelled. Our frogmaster Geoff has listed marks for five of our attempts, but the final script belonging to Joe has generated only the comment ‘magical mystery tour’. His solution to part a) diverges substantially from the most natural argument, and indeed involves wandering round the configuration, iteratively redirecting lines [1]. I am eventually convinced by the skeleton of the argument, though unconvinced I could complete the details in the finite time available.

We discuss the script with Lisa Sauermann, who explains some of the main challenges [2]. After a short pause for thought, we’re convinced by Lisa’s suggestion of equivalence with a point on the conventional markscheme. It would have been nice to have had more time to think about the subtleties myself, but this was some really interesting maths and we pack up for the day feeling very impressed with the quality of coordination here so far.

We and the coordinators are also very impressed with the quality of Harvey’s art. As a result, we now have an answer to the question ‘What should you do if you finish the IMO two hours early?’ Harvey’s answer at least is to draw a diagram of the Q6 configuration in the case n=3, where at each of the intersection points with the outer boundary stands a member of the current UK team. Precisely UNKs 1, 3 and 5 are wearing a frog. The real life sextet have been taken by Allison to Disneyland today, so some are potentially now wearing a princess. But while the contestants can let it go now, it’s off to work I go, as there’s still two sets of scripts left to ponder.

[1] The mechanism for this redirection is neither canonical nor explained, and even in the best setup I can come up with in an hour or so of trying a huge class of diagrams, exactly half of the indices in the resulting calculation are off by $\pm 1$. The pressure of IMO Day Two can indeed derail even the most well-prepared contestants.

[2] There is a non-trivial difficulty when the area enclosed by our path is concave, as then some intersection points on the path arise from lines which are also part of the path. Handling the parity of such points looks easy once you’ve been shown it, but is definitely not obvious.

# IMO 2016 Diary – Part Two

Wednesday 6th July

After starting the third exam, Mike, Jo and I go for a walk through some of the smaller villages on the other side of the ridge. Along the way, we pick up a bunch of local rascals who ask us, via their English-speaking henchman, many questions about basketball, and the colour of Mike’s shoes. Jo asks him why they aren’t in school, but this remains shrouded in mystery. Partly as a means of escape, we take a detour through a grove of the famous Tagaytay pineapples, which are indeed a striking crimson just before they ripen fully. I’m nervous about beard tanlines so am looking for a barber, but it seems I’m one of only two people in the Philippines with facial hair. (The other is Neel, who is adamant that his school approves of the ‘Wild man of Borneo’ look.)

We return to find that the students have been issued with cake. Its icing is impossible to manage without a fork. It is also entirely purple, and Lawrence describes it as ‘tasting of air’. None of this has distracted the UK students, who all solve the first two problems perfectly, which bodes well for the IMO itself, now less than a week away. To fill some time and provide a brief variation from the constant problem-solving, I give a talk about correlation and graphs, based on a subsubsubsection of my thesis and for now, fortunately no-one finds any logical holes.

Thursday 7th July

To add variety, today the two teams have set each other a paper, which they will spend the afternoon marking. It transpired late last night that the hotel has no means of printing or photocopying documents, and we haven’t brought copies of tomorrow’s final exam. So today’s paper has been painstakingly written on whiteboards, and some of the adults set off round Tagaytay in search of a working printer. The mode of transport is the ‘tricycle’, a small motorcycle with one place behind the rider and two in a bone-shaking pillion enclosed within a lace curtain. Availability of tricycles is infinite, availability of photocopiers is positive but small, and availability of printers is zero. We’ll be going for the handwritten, personal touch.

Both teams have chosen their papers so as to get some fiddly answers, and both teams have helped the exercise by writing some rubbish. Mostly it is all correct on close inspection, but much requires serious digestion, and gives the students at least a flavour of what Andrew and I have to endure on a daily basis. The Australians have rephrased a combinatorial problem in terms of Neel wandering through security checks at an airport, and the UK boys have proved that whatever happens here, a gold medal at the International Metaphor-Extending Olympiad seems inevitable.

Courtesy of Australian student Wilson, a penchant for fedoras has swept through the camp. Joe and Harvey look like extras in an ultra-budget production of Bugsy Malone. Our mock coordination has taken most of the afternoon, so we have not been tracking the imminent supertyphoon Nepartak as carefully as we ought, but at least the new headwear fashion offers some protection from the elements.

Friday 8th July

This morning is the final training exam, and in keeping with tradition is designated the Mathematical Ashes. Whichever team wins gets to keep an impressive urn, filled with the charred remains of some old olympiad scripts. The urn is quite heavy, so for the Cathay Pacific weight restrictions, it would be convenient for me if Australia won this year. The lack of an actual Ashes this year renders the competition all the more important in some people’s eyes, though if there were a test match here today, the covers would be on all day as the typhoon squalls.

The Ashes paper is the original Day Two paper [1] from IMO 2015. The problems are supposed to be secret until after this year’s IMO (exactly because of events like the one we are running) but the entire shortlist has been released overnight on the internet. Fortunately none of the students have been checking the relevant forum over breakfast, but ideally people will curb their admirable enthusiasm and follow the actual rules in future years. I mark the second problem, a fiddly recursive inequality, which invites many approaches, including calculus of varying rigour. Whatever the outcome, both teams have done a good job here.

For dinner, we are hosted by Dr Simon Chua, and some of his colleagues involved in the Philippines maths enrichment community, who suggested this location, and helped us set up this camp. We are treated to various Philippine dishes, including suckling pig and squid in its own ink, with a view of sunset across the lake as the storm clears. We’re very grateful to Simon, Joseph and their colleagues for tonight and their help and advice in advance.

We finish the marking after dinner, and the UK has consolidated our position on the third question, including a superb 21/21 for Warren on a genuinely hard paper, and we have won 82-74. It is late, everyone is tired, and there is packing required, so the celebrations are slightly muted, though it gives Jacob an excellent opportunity to lose his room key again, an alternative competition in which he is certainly the unique gold medallist. We transfer to the main event in Hong Kong early tomorrow, so it’s an early night all round.

[1] – Potted summary: some copies of this paper were accidentally released before they should have been, and so the paper had to be re-set.

Saturday 9th July

After a disturbed night, it has been vociferously recommended that we leave at 6am to beat the Manila traffic, with the result that we have four hours at the airport. I try to remain stoic, with difficulty. Joe practises sleeping on every available surfaces while the rest of us have a sudden enthusiasm to solve N8 from last year’s shortlist. It turns out the UKMT travel agent has outdone themselves, and booked half the group in premium economy, and half in regular economy, though the only real difference seems to be armrest width.

We are met in HK by Allison, our local guide for the week, and escorted onto coaches across from the airport on Lantau island to the University of Science and Technology in the New Territories. The check-in process is comprehensive: I sign and initial to confirm that they have correctly provided us with seven laptop sleeves, and then repeat for an infinite supply of other branded goods. Finally, we are allowed out to explore the spectacular campus, which stretches steeply down to Clearwater Bay. It is a novelty to take elevators up a total of 37 floors, and arrive on something called ‘Ground Floor’.

After a confusingly-managed dinner at the student cafeteria, a few of us head out to look at the nearby neighbourhood of Hang Hau. We pass the olympic velodrome, which gives Lawrence a good opportunity to explain gearing to those among his colleagues who do not naturally seek out applied mathematics. We return to find that Harvey decided to go to sleep before working out how to turn on his air conditioning. In humid hindsight, this was a poor strategy, as this was one of HK’s hottest days since records began. We have arrived back at the perfect time to watch the awesome thunderstorm from dry safety, which hopefully isn’t an omen of terrible things to follow in the contest, which starts on Monday.

# IMO 2016 Diary – Part One

Friday 1st July

It’s my last morning as an Oxford resident, and I have to finish the final chapter of my thesis, move out of my flat, print twenty-four boarding passes, and hurtle round town collecting all the college and department stamps on my pre-submission form 3.03 like a Pokemon enthusiast. Getting to Heathrow in time for an early evening flight seems very relaxed by comparison, even with the requirement to transport two boxes of IMO uniform. Because I wasn’t paying very much attention when I signed off the order, this year we will be wearing ‘gold’, but ‘lurid yellow’ might be a better description. Hopefully the contestants might have acquired some genuinely gold items by the time we return to this airport in two weeks.

Saturday 2nd July

Our flight passes rapidly. I proved an unusual function was locally Lipschitz, watched a film, and slept for a while. Others did not sleep at all, though I suspect they also did not prove any functions were locally Lipschitz. The airport in Hong Kong is truly enormous; for once the signs advertising the time to allow to get to each gate have a tinge of accuracy. We have plenty of time though, and there is substantial enthusiasm for coffee as we transfer. Cathay Pacific approach me with a feedback form, which turns out to include 130 detailed questions, including one concerning the ‘grooming’ of the check-in staff, while we all collectively tackle an inequality from the students’ final sheet of preparatory problems.

Before long though, we have arrived in Manila, where Jacob is uncontrollably excited to receive a second stamp in his passport, to complement his first from Albania at the Balkan Olympiad last month. As we bypass the city, we get a clear view of the skyscrapers shrouded in smog across the bay, though the notorious Manila traffic is not in evidence today. We pass through the hill country of Luzon Island, the largest of the Philippines and get caught in a ferocious but brief rainstorm, and finally a weekend jam on the lakeside approach to Tagaytay, but despite these delays, the fiendish inequality remains unsolved. I’m dangerously awake, but most of the students look ready to keel over, so we find our rooms, then the controls for the air conditioning, then let them do just that.

Sunday 3rd July

We have a day to recover our poise, so we take advantage of morning, before the daily rain sets in, to explore the area. We’ve come to Tagaytay because it’s high and cool by Philippine standards, so more conducive to long sessions of mathematics than sweltering Manila. We follow the winding road down the ridge to the shore of Taal Lake, where a strange flotilla of boats is docked, each resembling something between a gondola and a catamaran, waiting to ferry us to Taal Volcano, which lies in the centre of the lake. The principal mode of ascent from the beach is on horseback, but first one has to navigate the thronging hordes of vendors. Lawrence repeatedly and politely says no, but nonetheless ends up acquiring cowboy hats for all the students for about the price of a croissant in Oxford.

Many of us opt to make the final climb to the crater rim on foot, which means we can see the sulphurous volcanic steam rising through the ground beside the trail. From the lip we can see the bright green lake which lies in the middle of the volcano, which is itself in the middle of this lake in the middle of Luzon island. To the excitement of everyone who likes fractals, it turns out there is a further island within the crater lake, but we do not investigate whether this nesting property can be extended further. After returning across the outer lake, we enjoy the uphill journey back to Tagaytay as it includes a detour for a huge platter of squid, though the van’s clutch seems less thrilled. Either way, we end up with a dramatic view of an electrical storm, before our return to the hotel to await the arrival of the Australians.

Monday 4th July

Morning brings the opportunity to meet properly the Australian team and their leaders Andrew, Mike and Jo. We’ve gathered in the Philippines to talk about maths, and sit some practice exams recreating the style of the IMO. The first of these takes place this morning, in which the students have 4.5 hours to address three problems, drawn from those shortlisted but unused for last year’s competition.

After fielding a couple of queries, I go for a walk with Jo to the village halfway down the ridge. On the way down, the locals’ glances suggest they think we are eccentric, while on the ascent they think we are insane. About one in every three vehicles is a ‘jeepney’, which is constructed by taking a jeep, extending it horizontally to include a pair of benches in the back, covering with chrome cladding, and accessorising the entire surface in the style of an American diner. We return to find that the hotel thinks they are obliged to provide a mid-exam ‘snack’, and today’s instalment is pasta in a cream sauce with salad, served in individual portions under cloches. Andrew and I try to suggest some more appropriate options, but we’re unsure that the message has got across.

I spend the afternoon marking, and the UK have started well, with reliable geometry (it appears to be an extra axiom of Euclid that all geometry problems proposed in 2015/2016 must include a parallelogram…) and a couple of solutions to the challenging number theory problem N6, including another 21/21 for Joe. Part of the goal of this training camp is to learn or revise key strategies for writing up solutions in an intelligible fashion. At the IMO, the students’ work will be read by coordinators who have to study many scripts in many languages, and so clear logical structure and presentation is a massive advantage. The discussion of the relative merits of claims and lemmas continues over dinner, where Warren struggles to convince his teammates of the virtues of bone marrow, a by-product of the regional speciality, bulalo soup.

Tuesday 5th July

The second exam happens, and further odd food appears. Problem two encourages solutions through the medium of the essay, which can prove dangerous to those who prefer writing to thinking. In particular, the patented ‘Agatha Christie strategy’ of explaining everything only right at the end is less thrilling in the realm of mathematics. It’s a long afternoon.

We organise a brief trip to the People’s Park in the Sky, based around Imelda Marcos’s abandoned mansion which sits at the apex of the ridge. In the canon of questionable olympiad excursions, this was right up there. There was no sign of the famous shoe collection. Indeed the former ‘palace’ was open to the elements, so the style was rather more derelicte than chic, perfect for completing your I-Spy book of lichen, rust and broken spiral staircases. Furthermore after a brief storm, the clouds have descended, so the view is reminiscent of our first attempt at Table Mountain in 2014, namely about five metres visibility. A drugged parrot flaps miserably through the gloom. Even the UK team shirts are dimmed.

There is a shrine on the far side of the palace, housing a piece of rock which apparently refused to be dynamited during the construction process, and whose residual scorchmarks resemble the Blessed Virgin Mary. A suggested prayer is written in Tagalog (and indeed in Comic Sans) but there is a man sitting on the crucial rock, and it’s not clear whether one has to pay him to move to expose the vision. Eventually it clears enough to get a tolerable set of team photos. Joe tries to increase the compositional possibilities by standing on a boulder, thus becoming ten times taller than the volcano, so we keep things coplanar for now. Harvey finds a giant stone pineapple inside whose hollow interior a large number of amorous messages have been penned. He adds

$\mathrm{Geoff }\heartsuit\,\triangle \mathrm{s}$

in homage to our leader, who has just arrived in Hong Kong to begin the process of setting this year’s IMO papers.