# A criterion of convergence almost surely.

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.

• 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! – Michael Apr 23 at 3:52
• 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. – Michael Apr 23 at 5:33
• Duplicate of this question (... which hasn't an answer; it's just to link the two questions) – saz Apr 23 at 17:45

## 1 Answer

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 $$S_n = \sum_{i=1}^n X_i$$.

### Claim:

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 $$E[|A-B|]\geq E[|A|]P[|B|\leq x] - x \geq E[|A|](1/2) - x = \infty$$