# Generating Functions for Dice

So last week I was writing an article for Betting Expert about laws of large numbers, and I was trying to produce some representations of distributions to illustrate the Weak LLN and the Central Limit Theorem. Because tossing a coin feels too simplistic, and also because the natural state space for this random variable, at least verbally, is not a subset of the reals, I decided to go for dice instead. So it’s clear what the distribution of the outcome of a single dice roll is, and with a bit of thought or a 6×6 grid, we can work out the distribution of the average of two dice rolls. But what about 100 rolls? Obviously, we need large samples to illustrate the laws of large numbers! In this post, we discuss how to calculate the distribution of the sample mean of n dice rolls.

First we observe that the total set of outcomes of n dice rolls is $6^n$. The sum of the outcomes must lie between n and 6n inclusive. The distribution of the sum and the distribution of the sample mean are equivalent up to dividing by n. The final observation is that because the total number of outcomes has a nice form, we shouldn’t expect it to make any difference to the method if we calculate the probability of a given sum, or the number of configurations giving rise to that sum.

Indeed, tying in nicely with the first year probability course, we are going to use generating functions, and there is no difference in practice between the probability generating function, and the combinatorial generating function, if the underlying mechanism is a uniform choice. Well, in practice, there is a small difference, namely a factor of 6 here. The motivation for using generating functions is clear: we are considering the distribution of a sum of independent random variables. This is pretty much exactly why we bother to set up the machinery for PGFs.

Anyway, since each of {1,2,…,6} is equally likely, the GF of a single dice roll is

$x+x^2+\ldots+x^6=x\cdot \frac{1-x^6}{1-x}.$

So, if we want the generating function of the sum of n independent dice rolls, we can obtain this by raising the above function to the power n. We obtain

$x^n(1-x^6)^n(1-x)^{-n}.$

Note the factor of $x^n$ at the beginning arises because the minimum value of the sum is n. So to work out the number of configurations giving rise to sum k, we need to evaluate the coefficient of $x^k$. We can deal with $(1-x^6)^n$ fairly straightforwardly, but some thought it required regarding whether it’s possible to do similar job on $(1-x)^{-n}$.

We have to engage briefly with what is meant by a binomial coefficient. Note that

$\binom{x}{k}=\frac{x(x-1)\ldots(x-k+1)}{1\cdot\ldots\cdot k}$

is a valid definition even when x is not a positive integer, as it is simply a degree k polynomial in x. This works if x is a general positive real, and indeed if x is a general negative real. At this stage, we do need to keep k a positive integer, but that’s not a problem for our applications.

So we need to engage with how the binomial theorem works for exponents that are not positive integers. The tricky part with the standard expression as

$(a+b)^n=\binom{n}{0}a^n+\ldots + \binom{n}{n}b^n,$

is that the attraction of this symmetry in a and b prompts us to work in more generality than is entirely necessary to state the result. Note if we instead write

$(1+x)^n=1+\binom{n}{1}x+\binom{n}{2}x^2+\ldots,$

we have unwittingly described this finite sum as an infinite series. It just happens that all the binomial coefficients apart from the first (n+1) are zero. The nice thing about this definition is that it might plausibly generalise to non-integer or negative values of n. And indeed it does. I don’t want to go into the details here, but it’s just a Taylor series really, and the binomial coefficients are set up with factorials in the right places to look like a Taylor series, so it all works out.

It is also worth remarking that it follows straight from the definition of a negative binomial coefficient, that

$\binom{-n}{j}=(-1)^j \binom{n+j-1}{j}.$

In any case, we can rewrite our expression for the generating function of the IID sum as

$x^n\left[\sum_{k=0}^n \binom{n}{k}(-1)^k x^{6k}\right]\left[\sum_{j\ge 0} \binom{-n}{j}(-1)^j x^j\right]$

By accounting for where we can gather exponents from each bracket, we can evaluate the coefficient of $x^m$ as

$\sum_{6k+j=m+n}\binom{n}{k}\binom{n+j-1}{j}(-1)^k.$

Ie, k in the sum takes values in $\{0,1,\ldots, \lfloor \frac{m+n}{6}\rfloor\}$. At least in theory, this now gives us an explicit way to calculate the distribution of the average of multiple dice rolls. We have to be wary, however, that many compilers will not be happy dealing with large binomial coefficients, as the large factorials grow extremely rapidly. An approximation using logs is likely to be more tractable for larger settings.

Anyway, I leave you with the fruits of my labours.

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# Large Deviations 1 – Motivation and Cramer’s Theorem

I’ve been doing a lot of thinking about Large Deviations recently, in particular how to apply the theory to random graphs and related models. I’ve just writing an article about some of the more interesting aspects, so thought it was probably worth turning it into a few posts.

Motivation

Given $X_1,X_2,\ldots$ i.i.d. real-valued random variables with finite expectation, and $S_n:=X_1+\ldots+X_n$, the Weak Law of Large Numbers asserts that the empirical mean $\frac{S_n}{n}$ converges in distribution to $\mathbb{E}X_1$. So $\mathbb{P}(S_n\geq n(\mathbb{E}X_1+\epsilon))\rightarrow 0$. In fact, if $\mathbb{E}X_1^2<\infty$, we have the Central Limit Theorem, and a consequence is that $\mathbb{P}(S_n\geq n\mathbb{E}X_1+n^\alpha)\rightarrow 0$ whenever $\alpha>\frac12$.

In a concrete example, if we toss a coin some suitably large number of times, the probability that the proportion of heads will be substantially greater or smaller than $\frac12$ tends to zero. So the probability that at least $\frac34$ of the results are heads tends to zero. But how fast? Consider first four tosses, then eight. A quick addition of the relevant terms in the binomial distribution gives:

$\mathbb{P}\left(\text{At least }\tfrac34\text{ out of four tosses are heads}\right)=\frac{1}{16}+\frac{4}{16}=\frac{5}{16},$

$\mathbb{P}\left(\text{At least }\tfrac34\text{ out of twelve tosses are heads}\right)=\frac{1}{2^{12}}+\frac{12}{2^{12}}+\frac{66}{2^{12}}+\frac{220}{2^{12}}=\frac{299}{2^{12}}.$

There are two observations to be made. The first is that the second is substantially smaller than the first – the decay appears to be relatively fast. The second observation is that $\frac{220}{2^{12}}$ is substantially larger than the rest of the sum. So by far the most likely way for at least $\tfrac34$ out of twelve tosses to be heads is if exactly $\tfrac34$ are heads. Cramer’s theorem applies to a general i.i.d. sequence of RVs, provided the tail is not too heavy. It show that the probability of any such large deviation event decays exponentially with n, and identifies the exponent.

Theorem (Cramer): Let $(X_i)$ be i.i.d. real-valued random variables which satisfy $\mathbb{E}e^{tX_1}<\infty$ for every $t\in\mathbb{R}$. Then for any $a>\mathbb{E}X_1$,

$\lim_{n\rightarrow \infty}\frac{1}{n}\log\mathbb{P}(S_n\geq an)=-I(a),$

$\text{where}\quad I(z):=\sup_{t\in\mathbb{R}}\left[zt-\log\mathbb{E}e^{tX_1}\right].$

Remarks

• So, informally, $\mathbb{P}(S_n\geq an)\sim e^{-nI(a)}$.
• I(z) is called the Fenchel-Legendre transform (or convex conjugate) of $\log\mathbb{E}e^{tX_1}$.
• Considering t=0 confirms that $I(z)\in[0,\infty]$.
• In their extremely useful book, Dembo and Zeitouni present this theorem in greater generality, allowing $X_i$ to be supported on $\mathbb{R}^d$, considering a more general set of large deviation events, and relaxing the requirement for finite mean, and thus also the finite moment generating function condition. All of this will still be a special case of the Gartner-Ellis theorem, which will be examined in a subsequent post, so we make do with this form of Cramer’s result for now.

The proof of Cramer’s theorem splits into an upper bound and a lower bound. The former is relatively straightforward, applying Markov’s inequality to $e^{tS_n}$, then optimising over the choice of t. This idea is referred to by various sources as the exponential Chebyshev inequality or a Chernoff bound. The lower bound is more challenging. We reweight the distribution function F(x) of $X_1$ by a factor $e^{tx}$, then choose t so that the large deviation event is in fact now within the treatment of the CLT, from which suitable bounds are obtained.

To avoid overcomplicating this initial presentation, some details have been omitted. It is not clear, for example, whether I(x) should be finite whenever x is in the support of $X_1$. (It certainly must be infinite outside – consider the probability that 150% or -40% of coin tosses come up heads!) In order to call this a Large Deviation Principle, we also want some extra regularity on I(x), not least to ensure it is unique. This will be discussed in the next posts.

# Large Deviations and the CLT

Taking a course on Large Deviations has forced me to think a bit more carefully about what happens when you have large collections of IID random variables. I guess the first thing think to think about is ‘What is a Large Deviation‘? In particular, how large or deviant does it have to be?

Of primary interest is the tail of the distribution function of $S_n=X_1+\ldots+X_n$, where the $X_i$ are independent and identically distributed as $X$. As we can always negate everything later if necessary, we typically consider the probability of events of the type:

$\mathbb{P}(S_n\geq \theta(n))$

where $\theta(n)$ is some function which almost certainly increases fairly fast with $n$. More pertinently, if we are looking for some limit which corresponds to an actual random variable, we perhaps want to look at lots of related $\theta(n)$s simultaneously. More concretely, we should fix $\theta$ and consider the probabilities

$\mathbb{P}(\frac{S_n}{\theta(n)}\geq \alpha).$ (*)

Throughout, we lose no generality by assuming that $\mathbb{E}X=0$. Of course, it is possible that this expectation does not exist, but that is certainly a question for another post!

Now let’s consider the implications of our choice of $\theta(n)$. If this increases with $n$ too slowly, and the likely deviation of $S_n$ is greater than $\theta(n)$, then the event might not be a large deviation at all. In fact, the difference between this event and the event ($S_n$ is above 0, that is, its mean) becomes negligible, and so the probability at (*) might be 1/2 or whatever, regardless of the value of $\alpha$. So object $\lim \frac{S_n}{\theta(n)}$ whatever that means, certainly cannot be a proper random variable, as if we were to have convergence in distribution, this would imply that the limit RV consisted of point mass at each of $\{+\infty, -\infty\}$.

On the other hand, if $\theta(n)$ increases rapidly with $n$, then the probabilities at (*) might become very small indeed when $\alpha>0$. For example, we might expect:

$\lim_{n\rightarrow\infty}\mathbb{P}(\frac{S_n}{\theta(n)}\geq \alpha)=\begin{cases}0& \alpha>0\\1&\alpha<0.\end{cases}$

and more information to be required when $\alpha=0$. This is what we mean by a large deviation event. Although we always have to define everything concretely in terms of some finite sum $S_n$, we are always thinking about the behaviour in the limit. A large deviation principle exists in an enormous range of cases to show that these probabilities in fact decay exponentially. Again, that is the subject for another post, or indeed the lecture course I’m attending.

Instead, I want to return to the Central Limit Theorem. I first encountered this result in popular science books in a vague “the histogram of boys’ heights looks like a bell” kind of way, then, once a normal random variable had been to some extent defined, it returned in A-level statistics courses in a slightly more fleshed out form. As an undergraduate, you see it in several forms, including as a corollary following from Levy’s convergence theorem.

In all applications though, it is generally used as a method of calculating good approximations. It is not uncommon to see it presented as:

$\mathbb{P}(a\sigma\sqrt{n}+\mu n\leq S_n\leq b\sigma\sqrt{n}+\mu n)\approx \frac{1}{\sqrt{2\pi}}\int_a^b e^{-x^2/2}dx.$

Although in many cases that is the right way to think use it, it isn’t the most interesting aspect of the theorem itself. CLT says that the correct scaling of $\theta(n)$ so that the deviation probabilities lie between the two cases outline above is the same (that is, $\theta(n)=O(\sqrt{n})$ in some sense) for an enormous class of distributions, and in particular, most distributions that one might encounter in practice (ie finite mean, finite variance). There is even greater universality, as furthermore the limit distribution at this interface has the same form (some appropriate normal distribution) whenever $X$ is in this class of distributions. I think that goes a long way to explaining why we should care about the theorem. It also immediately prompts several questions:

• What happens for less regular distributions? It is now more clear what the right question to ask in this setting might be. What is the appropriate scaling for $\theta(n)$ in this case, if such a scaling exists? Is there a similar universality property for suitable classes of distributions?
• What is special about the normal distribution? The theorem itself shows us that it appears as a universal scaling limit in distribution, but we might reasonably ask what properties such a distribution should have, as perhaps this will offer a clue to a version of CLT or LLNs for less regular distributions.
• We can see that the Weak Law of Large Numbers follows immediately from CLT. In fact we can say more, perhaps a Slightly Less Weak LLN, that

$\frac{S_n-\mu n}{\sigma \theta(n)}\stackrel{d}{\rightarrow}0$

• whenever $\sqrt{n}<<\theta(n)$. But of course, we also have a Strong Law of Large Numbers, which asserts that the empirical mean converges almost surely. What is the threshhold for almost sure convergence, because there is no a priori reason why it should be $\theta(n)=n$?

To be continued next time.