Real-world application where strong LLN is needed (weak LLN is not enough)

Question

Do you know of any real world (algorithm, physics, ...) application of the law of large numbers where we need the strong LLN and the weak LLN by itself is not enough to prove that the application is working as expected ?

Answer 1

This gives details on my comment about systems with random walks and thresholds that we do not want to cross. It shows that if the average stepsize converges to 1 with probability 1, then we can start far enough to the right of the threshold to ensure the probability of ever crossing it is as small as we like. But if the average stepsize converges to 1 in probability (but not with probability 1) there are examples where we (unfortunately) cannot avoid crossing the threshold, no matter how far away from it we start.

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Fix $x >0$ as a non-random starting location for the random walk. Let $\{X_i\}_{i=1}^{\infty}$ be a sequence of random variables that represent step sizes. The $\{X_i\}$ variables are possibly dependent and can have different distributions. Define $S_n$ as the location on step $n$:
$$ S_n = x + \sum_{i=1}^n X_i \quad \forall n \in \{1, 2, 3, ...\} $$ Let $C_{x}$ be the event that we eventually cross the 0 threshold, assuming starting location $x$:
$$ C_{x} = \cup_{n=1}^{\infty} \left\{x +\sum_{i=1}^n X_i \leq 0\right\}$$

Claim 1:

If $\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^n X_i =1$ with probability 1, then $\lim_{x\rightarrow\infty} P[C_x] = 0$.

Proof: Define $M = \inf_{n\in \{1, 2, 3, ...\}}\left\{\sum_{i=1}^n X_i \right\}$. Then $M\leq X_1$ and it can be shown that $P[M=-\infty]=0$ and $M$ has a valid CDF function $P[M\leq y]$ for all $y \in \mathbb{R}$. For $x \in \mathbb{R}$ we have $$ C_x = \cup_{n=1}^{\infty}\left\{x + \sum_{i=1}^n X_i \leq 0\right\} \subseteq \{x + M \leq 0\} $$ Thus $$P[C_x] \leq P[x+M\leq 0] = P[M\leq -x] $$ Hence $$ \lim_{x\rightarrow\infty} P[C_x] \leq \lim_{x\rightarrow\infty} P[M\leq -x] = 0 \quad \Box$$

Claim 2:

There are processes $\{X_i\}_{i=1}^{\infty}$ that satisfy $\frac{1}{n}\sum_{i=1}^n X_i \rightarrow 1$ in probability (but not with probability 1) such that $P[C_x]=1$ for all $x>0$.

We construct such a process: Define $\{Y_n\}_{n=1}^{\infty}$ as independent random variables that satisfy $$ Y_n = \left\{\begin{array}{ll} 1 &\mbox{with prob $(1-1/n)$}\\ -n & \mbox{with prob $1/n$} \end{array}\right.$$ Then $Y_n\rightarrow 1$ in probability (but not with probability 1) and by Borel-Cantelli we have $Y_n=-n$ infinitely often. Define $\{X_i\}_{i=1}^{\infty}$ as follows: Define $X_1=Y_1$ and $$ X_n = nY_n - (n-1)Y_{n-1} \quad \forall n \in \{2, 3, 4, ...\}$$ It follows that $$ \frac{1}{n}\sum_{i=1}^n X_i = Y_n \quad \forall n \in \{1, 2, 3, ...\}$$ Thus, $\frac{1}{n}\sum_{i=1}^n X_i \rightarrow 1$ in probability (but not with probability 1).

Fix $x>0$. Notice that if $Y_n=-n$ for some particular positive integer $n$, then $x + \sum_{i=1}^n X_i = x-n^2$. If $Y_n=-n$ for infinitely many positive integers $n$, then $x + \sum_{i=1}^n X_i \leq 0$ for infinitely many $n$. But we already know $\{Y_n=-n\}$ occurs infinitely often with probability 1. Thus $P[C_x]=1$.