## Motivation:

A couple weeks ago I was working on a problem that involved the expected value of a ratio of two random variables:

$$\mathbb{E}\big[\frac{X_n}{Z_n}\big] \approx \frac{\mu_{X_n}}{\mu_{Z_n}} - \frac{\mathrm{Cov}(X_n,Z_n)}{\mu_{Z_n}^2} + \frac{\mathrm{Var(Z_n)}\mu_{X_n}}{\mu_{Z_n}^3}$$

where $Z_n$ was a sum of $n$ i.i.d. random variables with a symmetric distribution centred at zero.

Everything about this approximation worked fine in computer simulations where $n$ was large but mathematically there appeared to be a problem since:

$$\mathbb{E}\big[Z_n\big] = 0$$

Given that (2) didn’t appear to be an issue in simulation, I went through the code several times to check whether there was an error but found none. After thinking about the problem for a bit longer it occurred to me to formalise the problem and analyse:

$$P(\sum_{n=1}^N a_n = 0)$$

where $a_n$ are i.i.d. random variables with a uniform distribution centred at zero so $\mathbb{E}[a_i]=0$. My intuition suggested that under relatively weak assumptions:

$$\lim_{N \to \infty} P(\sum_{n=1}^N a_n = 0) = 0$$

We may think of this as a measure-theoretic phenomenon in high-dimensional spaces where $N \in \mathbb{N}$ is our dimension and $\vec{a} \in \mathbb{R}^N$ is a random vector.

## Analysis of a special case:

Given that (3) is a very general problem, I decided to start by analysing the special case of $a_i \sim \mathcal{U}(\{-1,1\})$ where:

$$\forall n \in \mathbb{N}, P(a_n=1)=P(a_n=-1)=\frac{1}{2}$$

$$S_0 = \{ (a_n)_{n=1}^N \in \{-1,1\} : \sum_n a_n = 0\}$$

Knowing that $S_0$ is non-empty only if we have parity of positive and negative terms, we may deduce that:

$$S_0 \neq \emptyset \iff N \in 2\mathbb{N}$$

For the above reason, I focused my analysis on the following sequence:

$$u_N = P(\sum_{n=1}^{2N} a_n = 0)= \frac{2N \choose N}{2^{2N}} = \frac{(2N)!}{2^{2N}(N!)^2}$$

## Proof that $u_N$ is decreasing:

We can demonstrate that $u_N$ is strictly decreasing by considering the ratio:

$$\frac{u_{N+1}}{u_N}=\frac{\frac{(2N+2)!}{2^{2N+2}((N+1)!)^2}}{\frac{(2N)!}{2^{2N}(N!)^2}}=\frac{(2N+2)(2N+1)}{4(N+1)^2}=\frac{2N+1}{2N+2} < 1$$

Now, with (9) we have what is necessary to show that:

$$\lim_{n \to \infty} u_N = 0$$

## Analysis of the limit $\lim\limits_{N \to \infty} u_N$:

Using (9) we may derive a recursive definition of $u_N$:

$$u_{N+1}=\frac{2N+1}{2N+2} \cdot u_N$$

and given that $u_0=1$ we have:

$$u_{N}=\prod_{n=0}^{N-1} \frac{2n+1}{2n+2}= \frac{1}{3} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot …$$

At this point we can make the useful observation:

$$\lim_{N \to \infty} u_N = 0 \implies \lim_{N \to \infty} - \ln u_N = \infty$$

## Proof that $\lim\limits_{N \to \infty} u_N=0$:

By combining (12) and (13) we find that:

$$-\ln u_N = -\ln \prod_{n=0}^{N-1} \frac{2n+1}{2n+2}= \sum_{n=0}^{N-1} \ln \frac{2n+2}{2n+1}= \sum_{n=0}^{N-1} \ln \big(1+\frac{1}{2n+1}\big)$$

We note that when $n\in \mathbb{N}$ is large:

$$\ln \big(1+\frac{1}{n}\big) \approx \frac{1}{n}$$

Now, from (15) it follows that:

$$\sum_{n=1}^\infty \frac{1}{2n+1} = \infty \implies \sum_{n=0}^{\infty} \ln \big(1+\frac{1}{2n+1}\big) = \infty$$

Using (15) we may conclude that (10) is indeed true. In some sense, when $n$ is large we can expect to observe the expected value with vanishing probability.

## Discussion:

A natural question that follows is whether the above method may be used to handle other cases. Let’s consider $a_i \sim \mathcal{U}(\{-1,0,1\})$ where:

$$\forall n \in \mathbb{N}, P(a_n=1)=P(a_n=0)=P(a_n=-1)=\frac{1}{3}$$

so we may define:

$$u_N = P(\sum_{n=1}^{4N} a_n = 0)= \frac{(4N)!}{3^{4N}} \sum_{k=1}^N \frac{1}{(2k)!^2 (4N-4k)!}$$

I actually tried to analyse the combinatorics of this sequence but quickly realised that even if I managed to show that this sequence converged to zero, it wasn’t clear how this method would manage to handle the most general setting, the case of all integer dimensions $N \in \mathbb{N}$, and it didn’t appear to be very effective in terms of the number of calculations per case.

In order to make progress, I decided to model this problem from a different perspective.