Doob inequalities and Doob-Meyer decomposition

The first post I wrote on this blog was about martingales, way back in 2012 at a time when I had known what a martingale was for about a month. I now don’t have this excuse. So I’m going to write about a couple of properties of (discrete-time) martingales that came up while adjusting a proof which my thesis examiners suggested could be made much shorter as part of their corrections.

Doob’s submartingale inequality

When we prove that some sequence of processes converges to some other process, we typically want to show that this holds in some sense uniformly over a time-interval, rather than just at some fixed time. We don’t lose much at this level of vagueness by taking the limit process to be identically zero. Then, if the convergent processes are martingales or closely similar, we want to be able to bound $\sup_{k\le n} |Z_k|$ in some sense.

Doob’s submartingale inequality allows us to do this. Recall that a submartingale has almost-surely non-negative conditional increments. You might think of it heuristically as ‘more increasing than a martingale’. If $Z_n$ is a martingale, then $|Z_n|$ is a submartingale. This will be useful almost immediately.

The statement is that for $(Z_n)$ a non-negative submartingale,

$\mathbb{P}\left( \sup_{k\le n} Z_k \ge \lambda\right) \le \frac{\mathbb{E}\left[Z_n\right]}{\lambda}.$

The similarity of the statement to the statement of Markov’s inequality is no accident. Indeed the proof is very similar. We consider whether the event in question happens, and find lower bounds on the expectation of $Z_n$ under both possibilities.

Formally, for ease of notation, let $Z_n^*$ be the running maximum $\sup_{k\le n}Z_k$. Then, we let $T:= n\wedge \inf\{k\le n, M_j\ge \lambda\}$ and apply the optional stopping theorem for submartingales at T, which is by construction at most n. That is

$\mathbb{E}[Z_n]\ge \mathbb{E}[Z_T]=\mathbb{E}\left[Z_T\mathbf{1}_{Z_n^*<\lambda}\right] + \mathbb{E}\left[Z_T \mathbf{1}_{Z_n^*\ge \lambda}\right].$

The first of these summands is positive, and the second is at least $\lambda \mathbb{P}\left( Z_N^* \ge \lambda \right)$, from which the result follows.

We’ve already said that for any martingale $Z_n$, $|Z_n|$ is a submartingale, but in fact $f(Z_n)$ is a submartingale whenever f is convex, and $\mathbb{E}|f(Z_n)|<\infty$ for each n. Naturally, this continues to hold when $Z_n$ is itself a submartingale.

[Note that $Z_n^*$ is also a submartingale, but this probably isn’t as interesting.]

A particularly relevant such function f is $f(x)=x^p$, for p>1. If we take $Z_n$ a non-negative submartingale which is uniformly bounded in $L^p$, then by applying Holder’s inequality and this submartingale inequality, we obtain

$\mathbb{E}\left( \sup_{k\le n}Z_n^p \right) \le \left(\frac{p}{p-1}\right)^p \mathbb{E}\left[ Z_n^p \right].$

Since $Z_n^p$ is a submartingale, then a limit in n on the RHS is monotone, and certainly a limit in n on the LHS is monotone, so we can extend to

$\mathbb{E}\left( \sup_{k\le n}Z_\infty^p \right) \le \left(\frac{p}{1-p}\right)^p \mathbb{E}\left[ Z_\infty^p \right].$

Initially, we have to define $\mathbb{E}\left[ Z_\infty^p \right]$ through this limit, but in fact this result, Doob’s $L^p$ inequality, shows that $Z_\infty:= \lim Z_n$ exists almost surely as well.

Naturally, we will often apply this in the case p=2, and in the third of these three sections, we will see why it might be particularly straightforward to calculate $\mathbb{E}\left[Z_\infty^2\right].$

Remark: as in the case of Markov’s inequality, it’s hard to say much if the submartingale is not taken to be non-negative. Indeed, this effect can be seen even if the process is only defined for a single time step, for which the statement really is then Markov’s inequality.

Doob-Meyer decomposition

Unfortunately, most processes are not martingales. Given an discrete-time process $X_n$ adapted to $\mathcal{F}=(\mathcal{F}_n)$, it is a martingale if the conditional expectations of the increments are all almost surely zero. But given a general adapted process $X_n$ which is integrable (so the increments have well-defined finite expectation), we can iteratively construct a new process $M_n$, where the increments are centred versions of $X_n$‘s increments. That is,

$M_{n+1}-M_n:= X_{n+1}-X_n - \mathbb{E}\left[ X_{n+1}-X_n \,\big|\, \mathcal{F}_n\right] = X_{n+1}-\mathbb{E}\left[X_{n+1} \,\big|\, \mathcal{F}_n\right].$ (*)

Then it’s immediately clear from the definition that $M_n$ is a martingale.

There’s a temptation to tie oneself up in knots with the dependence. We might have that increments of the original process $X_n$ depend on the current value of the process. And is it necessarily clear that we can recover the current value of the original process from the current value of $M_n$? Well, this is why we demand that everything be adapted, rather than just Markov. It’s not the case that $M_n$ should be Markov, but it clearly is adapted.

Now we look at the middle expression in (*), and in particular the term we are subtracting, namely the conditional expectation. If we define, in the standard terminology, $A_0=0$ and

$A_{n+1}-A_n:= \mathbb{E}\left[ X_{n+1}-X_n \,\big|\, \mathcal{F}_n\right],$

then we have decomposed the original process $X_n$ as the sum of a martingale $M_n$, and this new process $A_n$. In particular, note that the increment $A_{n+1}-A_n$ given above is adapted to $\mathcal{F}_n$, which is a stronger condition than being adapted to $\mathcal{F}_{n+1}$ as we would expect a priori. This property of the process $(A_n)$ is called predictability (or possibly previsibility).

This decomposition $X_n=X_0+M_n+A_n$ as just defined is called the Doob-Meyer decomposition, and there is a unique such decomposition where $M_n$ is a martingale, and $A_n$ is predictable. The proof of uniqueness is very straightforward. We look at the equalities given above as definitions of $M_n,A_n$, but then work in the opposite direction to show that they must hold if the decomposition holds.

I feel a final heuristic is worthwhile, using the term drift, more normally encountered in the continuous-time setting to describe infinitissimal expected increments. The increments of $A_n$ represent the drift of $X_n$, and the increments of $M_n$ are what remains from $X_n$ after subtracting the drift. In general, the process to be subtracted to turn a non-martingale into a martingale is called a compensator, and the existence or otherwise of such processes is important but challenging for some classes of continuous-time processes.

In particular, note that when $X_n$ is itself a martingale, then $A_n\equiv 0$. However, probably the most useful case is when $X_n$ is a submartingale, as then the drift is always non-negative, and so $A_n$ is almost surely increasing. The converse holds too.

This is relevant because this Doob-Meyer decomposition is obviously only a useful tool for treating $X_n$ if we can handle the two processes $M_n,A_n$ easily. We have tools to bound the martingale term, but this previsible term might in general be tricky, and so the case where $X_n$ is a submartingale is good, as increasing processes are much easier than general processes, since bounding the whole process might involve only bounding the final term in many contexts.

A particularly relevant example is the square of a martingale, that is $X_n=M_n^2$, where $M_n$ is a martingale. By the convexity condition discussed earlier, $X_n$ is a submartingale (provided it is integrable, ie $M_n$ is square-integrable), and so the process $A_n$ in its Doob-Meyer decomposition is increasing. This is often called the (predictable) quadratic variation of $(X_n)$.

This predictable quadratic variation is sometimes denoted $\langle X_n\rangle$. This differs from the (regular) quadratic variation which is defined as the sum of the squares of the increments, that is $[X_n]:= \sum_{k=0}^{n-1} (X_{k+1}-X_k)^2$. Note that this is adapted, but obviously not previsible. The distinction between these two processes is more important in continuous time. There, they are almost surely equal for a continuous local martingale, but not for eg a Poisson process. (For a Poisson process, the PQV is deterministic, indeed linear, while the (R)QV is almost surely equal to the Poisson process itself.) In the discrete time setting, the regular quadratic variation is not relevant very often, while the predictable quadratic variation is useful, precisely because of this decomposition.

Whenever we have random variables which we then centre, there is a standard trick to apply when treating their variance. That is

$A_{n+1}-A_n= \mathbb{E}\left[ M^2_{n+1}-M^2_n \,\big|\, \mathcal{F}_n\right]$
$= \mathbb{E}\left[ M^2_{n+1}\,\big|\, \mathcal{F}_n\right] - 2M_n^2 +M_n^2$
$= \mathbb{E}\left[ M^2_{n+1}\,\big|\, \mathcal{F}_n\right] - 2M_n \mathbb{E}\left[ M_{n+1}\,\big|\, \mathcal{F}_n\right] + M_n^2$
$= \mathbb{E}\left[ \left(M_{n+1}-M_n\right)^2\,\big|\, \mathcal{F}_n\right].$

One consequence is seen by taking an ‘overall’ expectation. Because $M_n^2-A_n$ is a martingale,

$\mathbb{E}\left[M_n^2\right] = \mathbb{E}\left[A_n\right] = \mathbb{E}\left[M_0^2\right] + \sum_{k=0}^{n-1} \mathbb{E}\left[A_{k+1}-A_k\right]$
$= \mathbb{E}\left[ M_0^2\right] + \sum_{k=0}^{n-1}\mathbb{E}\left[ \left(M_{k+1}-M_k\right)^2 \right].$ (**)

This additive (Pythagorean) property of the square of a martingale is useful in applications where there is reasonably good control on each increment separately.

We can also see this final property without the Doob-Meyer decomposition. For a martingale it is not the case that the increments on disjoint intervals are independent. However, following Williams 12.1 [1], disjoint intervals are orthogonal, in the sense that

$\mathbb{E}\left[(M_t-M_s)(M_v-M_u)\right]=0,$

whenever $s\le t\le u\le v$. Then, when we square the expression $M_n=M_0+\sum M_{k+1}-M_k$, and take expectations, all the cross terms vanish, leaving precisely (*).

References

[1] Williams – Probability with Martingales

I also followed the notes I made in 2011/12 while attending Perla Sousi’s course on Advanced Probability, and Arnab Sen’s subsequent course on Stochastic Calculus, though I can’t find any evidence online for the latter now.

Subordinators and the Arcsine rule

After the general discussion of Levy processes in the previous post, we now discuss a particular class of such processes. The majority of content and notation below is taken from chapters 1-3 of Jean Bertoin’s Saint-Flour notes.

We say $X_t$ is a subordinator if:

• It is a right-continuous adapted stochastic process, started from 0.
• It has stationary, independent increments.
• It is increasing.

Note that the first two conditions are precisely those required for a Levy process. We could also allow the process to take the value $\infty$, where the hitting time of infinity represents ‘killing’ the subordinator in some sense. If this hitting time is almost surely infinite, we say it is a strict subordinator. There is little to be gained right now from considering anything other than strict subordinators.

Examples

• A compound Poisson process, with finite jump measure supported on $[0,\infty)$. Hereafter we exclude this case, as it is better dealt with in other languages.
• A so-called stable Levy process, where $\Phi(\lambda)=\lambda^\alpha$, for some $\alpha\in(0,1)$. (I’ll define $\Phi$ very soon.) Note that checking that the sample paths are increasing requires only that $X_1\geq 0$ almost surely.
• The hitting time process for Brownian Motion. Note that this does indeed have jumps as we would need. (This has $\Phi(\lambda)=\sqrt{2\lambda}$.)

Properties

• In general, we describe Levy processes by their characteristic exponent. As a subordinator takes values in $[0,\infty)$, we can use the Laplace exponent instead:

$\mathbb{E}\exp(-\lambda X_t)=:\exp(-t\Phi(\lambda)).$

• We can refine the Levy-Khintchine formula;

$\Phi(\lambda)=k+d\lambda+\int_{[0,\infty)}(1-e^{-\lambda x})\Pi(dx),$

• where k is the kill rate (in the non-strict case). Because the process is increasing, it must have bounded variation, and so the quadratic part vanishes, and we have a stronger condition on the Levy measure: $\int(1\wedge x)\Pi(dx)<\infty$.
• The expression $\bar{\Pi}(x):=k+\Pi((x,\infty))$ for the tail of the Levy measure is often more useful in this setting.
• We can think of this decomposition as the sum of a drift, and a PPP with characteristic measure $\Pi+k\delta_\infty$. As we said above, we do not want to consider the case that X is a step process, so either d>0 or $\Pi((0,\infty))=\infty$ is enough to ensure this.

Analytic Methods

We give a snapshot of a couple of observations which make these nice to work with. Define the renewal measure U(dx) by:

$\int_{[0,\infty)}f(x)U(dx)=\mathbb{E}\left(\int_0^\infty f(X_t)dt\right).$

If we want to know the distribution function of this U, it will suffice to consider the indicator function $f(x)=1_{X_t\leq x}$ in the above.

The reason to exclude step processes specifically is to ensure that X has a continuous inverse:

$L_x=\sup\{t\geq 0:X_t\leq x\}$ so $U(x)=\mathbb{E}L_x$ is continuous.

In fact, this renewal measure characterises the subordinator uniquely, as we see by taking the Laplace transform:

$\mathcal{L}U(\lambda)=\int_{[0,\infty)}e^{-\lambda x}U(dx)=\mathbb{E}\int e^{-\lambda X_t}dt$

$=\int \mathbb{E}e^{-\lambda X_t}dt=\int\exp(-t\Phi(\lambda))dt=\frac{1}{\Phi(\lambda)}.$

The Arcsine Law

X is Markov, which induces a so-called regenerative property on the range of X, $\mathcal{R}$. Formally, given s, we do not always have $s\in\mathcal{R}$ (as the process might jump over s), but we can define $D_s=\inf\{t>s:t\in\mathcal{R}\}$. Then

$\{v\geq 0:v+D_s\in\mathcal{R}\}\stackrel{d}{=}\mathcal{R}.$

In fact, the converse holds as well. Any random set with this regenerative property is the range of some subordinator. Note that $D_s$ is some kind of dual to X, since it is increasing, and the regenerative property induces some Markovian properties.

In particular, we consider the last passage time $g_t=\sup\{s, in the case of a stable subordinator with $\Phi(\lambda)=\lambda^\alpha$. Here, $\mathcal{R}$ is self-similar with scaling exponent $\alpha$. The distribution of $\frac{g_t}{t}$ is thus independent of t. In this situation, we can derive the generalised arcsine rule for the distribution of $g_1$:

$\mathbb{R}(g_1\in ds)=\frac{\sin \alpha\pi}{\pi}s^{\alpha-1}(1-s)^{-\alpha}ds.$

The most natural application of this is to the hitting time process of Brownian Motion, which is stable with $\alpha=\frac12$. Then $g_1=S_1-B_1$, in the usual notation for the supremum process. Furthermore, we have equality in distribution of the processes (see previous posts on excursion theory and the short aside which follows):

$(S_t-B_t)_{t\geq 0}\stackrel{d}{=}(|B_t|)_{t\geq 0}.$

So $g_1$ gives the time of the last zero of BM before time 1, and the arcsine law shows that its distribution is given by:

$\mathbb{P}(g_1\leq t)=\frac{2}{\pi}\text{arcsin}\sqrt{t}.$

Dubins-Schwarz Theorem

In developing the stochastic integral, much of our motivation has come from considering integrals with respect to Brownian Motion. In this section, we develop some results which justify that Brownian Motion is the canonical stochastic process with non-zero quadratic variation (which is related, but not directly equivalent to the property of infinite total variation). In particular, we shall observe the Dubins-Schwarz theorem, which shows that martingales with unbounded (as time $\rightarrow\infty$) quadratic variation ARE Brownian Motion, up to a (stochastic) time change.

Recall Levy’s characterisation of a d-dimensional BM, which allows us to avoid considering independent normal increments. Given $X^1,\ldots,X^d\in\mathcal{M}_{c,loc}$:

$X=(X^1,\ldots,X^d)$ a BM iff $[X^i,X^j]_t=\delta_{ij}t$

Obviously, one direction has been shown as part of the construction and properties of quadratic variation. For the other direction,, because laws are precisely defined by characteristic functions, it suffices to show that

$\mathbb{E}\left[\exp(i\langle \theta,X_t-X_s\rangle)|\mathcal{F}_s\right]=\exp(-\frac12||\theta||^2(t-s))$

We set $Y_t:=\langle \theta,X_t\rangle$, and deduce $[Y]=t||\theta||^2$ and $Z=\mathcal{E}(iY)=\exp(iY_t+\frac12[Y]_t)\in\mathcal{M}_{c,loc}$, and furthermore is bounded on compact [0,t], hence is a true martingale. So $\mathbb{E}\left(\frac{Z_t}{Z_s}|\mathcal{F}_s\right)=1$ which is pretty much what was required.

Now, Dubins-Schwarz states

Theorem: Given $M\in\mathcal{M}_{c,loc}, M_0=0, [M]_\infty=\infty$ almost surely, if we set $\tau_s:=\inf\{t:[M]_t>s\}$, then $B_s:=M_{\tau_s}$ is a $(\mathcal{F}_{\tau_s})$-BM, with $M_t=B_{[M]_t}$.

This final result is clear if $[M]_t$ is almost surely strictly increasing in t: just take $s=[M]_t$ in the definition.

We know B is cadlag: we first show B as defined is almost surely continuous. It remains to show $B_{s-}=B_s\,\forall s>0\iff M_{\tau_{s-}}=M_{\tau_s}$, noting that $\tau_{s-}=\inf\{t\geq 0:[M]_t=s\}$ (by continuity) is a stopping time also.

The only interesting case is if $\tau_{s-}<\tau_s$, for which need to show [M] is constant. This is intuitively obvious, but formally, we must appeal to $(M^2-[M])^{\tau_s}$ which is UI, since $\mathbb{E}[M^{\tau_s}]_\infty<\infty$. Now may apply OST to obtain $\mathbb{E}[M_{\tau_s}^2-M_{\tau_{s-}}^2|\mathcal{F}_{\tau_{s-}}]=\mathbb{E}[(M_{\tau_s}-M_{\tau_{s-}})^2|\mathcal{F}_{\tau_{s-}}]=0$ which implies M is almost surely constant on $[\tau_{s-},\tau_s]$. We need to lift this to the case where it holds for all s simultaneously almost surely. Note that cadlag almost surely plus almost surely continuous at each point does not implies almost surely continuous everywhere (eg consider H(U(0,1)) with H the Heaviside function and U a uniform distribution). Instead, we record intervals of constancy of both $M_t,[M]_t$. That is, we set

$T_r=\inf\{t>r:M_t\neq M_r\},\quad S_r=\inf\{t>r:[M]_t\neq [M]_r\}$

Then these are cadlag, and by above $T_r=S_r\,\forall r\in\mathbb{Q}^+$ a.s. therefore $T_r=S_r\,\forall r$ almost surely. Thus M, [M] are constant on the same intervals.

We also check B is adapted to $\mathcal{G}_t=\mathcal{F}_{\tau_t}$. STP $X_T1_{\{T<\infty\}}$ is $\mathcal{F}_T$-measurable for X cadlag adapted. Approximating T discretely from above gives the result, exploiting that the result is clear if T has countable support. Now, obtain $M^{\tau_s}\in\mathcal{M}_c^2$, so $M_{t\wedge \tau_s}$ UI by Doob, so by OST, get $\mathbb{E}[M_{\tau_s}|\mathcal{F}_{\tau_s}]=M_{\tau_r}$, to get B a martingale. The finally:

$\mathbb{E}[B_s^2-s|\mathcal{G}_r]=\mathbb{E}[(M^2-[M])_{\tau_s}|\mathcal{F}_{\tau_s}]=M_{\tau_r}^2-[M]_{\tau_r}=B_r^2-r$

And so we can apply Levy’s characterisation to finish the result.

Martingales

Definition

$X_n$ is a stochastic process, integrable and adapted to filtration $(\Omega,\mathcal{F},(\mathcal{F}_n),\mathbb{P})$. Then $X$ is a martingale if $\mathbb{E}[X_n|\mathcal{F}_m]=X_m$ almost surely whenever $n\geq m$.

It is natural to think about a martingale as defined in the context of a process evolving in time. Then this definition is very reasonable:

• Integrable: the entire construction is about taking expectations. So these need to exist.
• Adapted: $X_n$ can’t be affected by what happens after time n. It must be defined by what has happened up to time n.
• Expectation condition: if you look at the process at a given time m, the best estimate for what $X$ will be in the future is in fact what it is now. As you might expect, it is sufficient that $\mathbb{E}[X_{n+1}|\mathcal{F}_n]=X_n$ almost surely for each n.

Motivation

There are many situations where the expected change in a variable over a time period is zero, whatever the value of the variable is at the start. For example, gambling. For illustration, assume we are speculating on the outcomes of tossing a coin repeatedly. You might have a complicated strategy, for example ‘double your stake when you lose’ (the so-called martingale strategy), or anything else. But ultimately, you can’t see into the future. So before every coin toss, you have to decide your stake, based on what’s happened up until now, and you will win or lose with equal probability, so your expected gain is 0, regardless of how you make your stake choice. Thus under any strategy determined without looking into the future (called previsible), the process recording your winnings is a martingale. Continue reading