Resnick -Probability Path - Exercise 7.40

Question

In this problem, Resnick states that if $\{X_n,n \geq 1\}$ is an independent sequence of random variables and $S_n := \sum_{i=1}^{n} X_n$, then $\frac{S_n}{n} \overset{a.s}{\to} 0$ if and only if $\frac{S_n}{n} \overset{P}{\to} 0$ and $\frac{S_{2^n}}{2^n} \overset{a.s}{\to} 0$

My question is the following: Why did Resnick include the assumption of independence?

Answer 1

Without independence, it is possible to find a sequence of random variables such that $S_n/n\to 0$ in probability and $S_{2^n}/2^n\to 0$ almost surely but we do not have $S_n/n\to 0$ almost surely.

Indeed, let $\left(Y_i\right)_{i\geqslant 1}$ be an i.i.d. sequence of non-negative random variables such that $$ \sum_{i\geqslant 1}2^i\mathbb P\left(Y_1>2^i\right)=\infty \tag{*} $$ and $$ \forall \varepsilon>0, \sum_{i\geqslant 1} \mathbb P\left(Y_1>\varepsilon 2^i\right)<\infty \tag{**}. $$ Finally, let $X_i=Y_i-Y_{i+1}$, so that $S_n=Y_1-Y_{n+1}$ for each $n$. Then

• $S_n/n\to 0$ in probability is equivalent to $Y_n/n\to 0$ in probability which holds because $\mathbb P\left(\lvert Y_n\rvert/n>\varepsilon\right)=\mathbb P\left( Y_1>n\varepsilon\right)\to 0$.
• We have $S_{2^n}/2^n\to 0$ almost surely because $Y_{2^n+1}/2^n\to 0$ almost surely, by (**) and the Borel-Cantelli lemma.
• But $S_n/n\to 0$ almost surely is equivalent to $Y_{n+1}/n\to 0$ which does not hold, by the second Borel-Cantelli lemma and (*).

Answer 2

I do not believe this is a necessary assumption. The non-immediate direction is to show that $\frac{S_n}{n} \overset{P}{\to} 0$ and $\frac{S_{2^n}}{2^n} \overset{a.s}{\to} 0$ implies $\frac{S_n}{n} \overset{a.s}{\to} 0$. But to do this, we do not need independence. Since $\frac{S_n}{n}$ is cauchy i.p -- by virtue of $\frac{S_n}{n}$ converging in probability -- it is also cauchy a.s (since $\psi_N = \sup_{m,n \geq N} |\frac{S_n}{n} - \frac{S_m}{m}|$ is a monotonically decreasing sequence). And $\frac{S_n}{n}$ being cauchy a.s + $\frac{S_{2^n}}{2^n} \overset{a.s}{\to} 0$ is all we need to conduct a straightforward argument to conclude $\frac{S_n}{n} \overset{a.s}{\to} 0$.

The other direction is an immediate consequence of the definition of almost sure convergence (and does not rely on the independence assumption)