Suppose random variables $\{X_k\}_{k\in \Bbb{N}}$ are $i.i.d.$ and set $S_n=X_1+...+X_n$, show that if $S_n/n\rightarrow 0$ in probability and $$S_{2^n}/2^n\rightarrow 0 \ \ a.s.$$ then $S_n/n\rightarrow 0$ almost surely.

I know convergence in probability implies there exists a subsequence converges a.s., but I don't see how to use the condition of $S_{2^n}/2^n\rightarrow 0 \ \ a.s.$, any help will be appreciated.

  • $\begingroup$ If $E[|X_1|]<\infty$ then $E[X_1]$ is finite and you can use the standard (strong) LLN to prove almost-sure convergence. If $E[|X_1|] =\infty$ then $S_n/n$ does not converge almost surely. So presumably you can use the given conditions to prove that $E[|X_1|]<\infty$. I do not immediately see how, though! $\endgroup$ – Michael Apr 23 at 3:52
  • $\begingroup$ We know if $E[|X_1|]=\infty$ then $\sum_{n=1}^{\infty} P[|X_1|>n]=\infty$ and so $\sum_{n=1}^{\infty} P[|X_n|>n] = \infty$ and so (with prob 1) there are infinitely many indices $n$ such that $|X_n|>n$, including in intervals $k \in \{2^n, ..., 2^{n+1}\}$, any such index $k$, when replaced by $X_k = O(1)$, will change the $S_{2^{n+1}}/2^{n+1}$ sum significantly. The problem is that there may be many such indices $k \in \{2^n, ..., 2^{n+1}\}$, and their $X_k$ values could potentially cancel each other out. $\endgroup$ – Michael Apr 23 at 5:33
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    $\begingroup$ Duplicate of this question (... which hasn't an answer; it's just to link the two questions) $\endgroup$ – saz Apr 23 at 17:45

The only argument I can see is a stochastic coupling argument. Let me know if someone finds a more direct proof.

I remove a condition that $S_n/n\rightarrow 0$ in probability that I find to be unnecessary.

Suppose $\{X_i\}_{i=1}^{\infty}$ are i.i.d. random variables and define $S_n = \sum_{i=1}^n X_i$.


If $\frac{S_{2^n}}{2^n}\rightarrow 0$ with probability 1 then $\frac{S_n}{n}\rightarrow 0$ with probability 1.

Proof for case 1.

Suppose $E[|X_1|]<\infty$ (case 1). Then $E[X_1]$ is well defined and finite, and the usual strong law of large numbers implies that $S_n/n \rightarrow E[X_1]$ with probability 1. This implies $S_{2^n}/2^n\rightarrow E[X_1]$ with probability 1. Since we already know $S_{2^n}/2^n\rightarrow 0$ with probability 1, it follows that $E[X_1]=0$. So $S_n/n\rightarrow 0$ with probability 1 and we are done.

Proof for case 2.

Suppose $E[|X_1|]=\infty$ (case 2). We want to show this case is impossible. Let $\{X_i\}_{i=1}^{\infty}$, $\{\tilde{X}_i\}_{i=1}^{\infty}$ be independent and i.i.d. random variables with the same distribution as $X_1$. Define $Y_i = X_i - \tilde{X}_i$. Then $\{Y_i\}_{i=1}^{\infty}$ are i.i.d. with a symmetric distribution about $0$, in the sense that $Y_i$ and $-Y_i$ have the same distribution. It can be shown that $E[|Y_1|]=E[|X_1-\tilde{X}_1|]=\infty$ (see footnote below). Define $$D_n = \sum_{i=1}^n Y_i$$ It is clear that $D_{2^n}/2^n\rightarrow 0$ with probability 1.

Now we have \begin{align} \infty&=E[|Y_1|] \\ &= \int_{0}^{\infty} P[|Y_1|\geq t]dt \\ &\leq \sum_{i=0}^{\infty}P[|Y_1|\geq i] \\ &= 1 + \sum_{i=1}^{\infty} P[|Y_1|\geq i]\\ &= 1 + \sum_{i=1}^{\infty} P[|Y_i|\geq i] \end{align} By independence of the $\{Y_i\}$ and Borel-Cantelli it means that $$ P[\{|Y_i|\geq i\} \quad i.o.] =1$$

Now define frame $k$ for $k \in \{1, 2, ...\}$ as the set of indices $F_k=\{2^{k-1}+1, ..., 2^{k}\}$. From the i.i.d. sequence $\{Y_i\}$ define a new sequence $\{W_i\}$ as follows: Define $W_1=Y_1$. For each frame $k \in \{1, 2, ...\}$, identify the smallest index $i \in F_k$ for which $|Y_i|\geq i$ and call this index $i^*[k]$. If there is no such index, simply define $i^*[k]=2^k$. Then define for all indices $i \in F_k$: $$ W_i = \left\{ \begin{array}{ll} Y_i &\mbox{ if $i \neq i^*[k]$} \\ -Y_i & \mbox{ if $i = i^*[k]$} \end{array} \right.$$ Define $\tilde{D}_n = \sum_{i=1}^n W_i$. We observe:

1) With probability 1, there are infinitely many frames $k \in \{1, 2, 3, ...\}$ for which $|Y_{i^*[k]}|\geq i^*[k]\geq 2^{k-1}$.

2) In any frame $k$ such that $|Y_{i^*[k]}|\geq i^*[k]\geq 2^{k-1}$ we have $$ \tilde{D}_{2^k} = \tilde{D}_{2^{k-1}} + (D_{2^k}-D_{2^{k-1}}) + \delta_k$$ where $$|\delta_k| = |Y_{i^*[k]}-(-Y_{i^*[k]})| = 2|Y_{i^*[k]}|\geq 2i^*[k] \geq 2^k$$ Hence $$ 1 \leq \frac{|\delta_k|}{2^k} = \frac{|(\tilde{D}_{2^k} -\tilde{D}_{2^{k-1}}) - (D_{2^k} -D_{2^{k-1}})|}{2^k}$$ If there are infinitely many such frames $k$, it is impossible for both $D_{2^n}/2^n$ and $\tilde{D}_{2^n}/2^n$ to converge to $0$. In view of point 1, the probability that both $D_{2^n}/2^n$ and $\tilde{D}_{2^n}/2^n$ converge to $0$ is zero.

3) (Coupling) The sequence $\{Y_i\}$ is stochastically equivalent to the sequence $\{W_i\}$ (that is, for all finite $n$, the joint distribution for $(Y_1, ..., Y_n)$ is the same as that for $(W_1, ..., W_n)$). Thus, $\{D_n\}$ and $\{\tilde{D}_n\}$ are stochastically equivalent. Since $D_{2^n}/2^n\rightarrow 0$ with probability 1, it holds that $\tilde{D}_{2^n}/2^n\rightarrow 0$ with probabilty 1. In view of point 2, this gives a contradiction.

Footnote: If $A, B$ are independent random variables with $E[|A|]=\infty$ then $E[|A-B|]=\infty$ by the following argument: Let $x>0$ be such that $P[|B|\leq x]\geq 1/2$. Then $$ |A-B| \geq |A-B|1_{\{|B|\leq x\}} \geq (|A|-x)1_{\{|B|\leq x\}} \geq |A|1_{\{|B|\leq x\}}-x$$ hence $$ E[|A-B|]\geq E[|A|]P[|B|\leq x] - x \geq E[|A|](1/2) - x = \infty$$


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