Law of large numbers along moving window The law of large numbers says that if $X_1, X_2, \ldots$ is an i.i.d. sequence, with $\mathbb{E}|X_1| < \infty$, then 
$$
    \frac{1}{n}(X_1 + \cdots + X_n) \stackrel{a.s.}{\longrightarrow} \mathbb{E}X_1 .
$$
If we let $(a_n)$ and $(b_n)$ be two deterministic sequences increasing to infinity, is it true that
$$
    \frac{1}{b_n} (X_{a_n} + \cdots + X_{a_n+b_n}) \stackrel{a.s.}{\longrightarrow} \mathbb{E}X_1?
$$
So, averaging along sliding, increasing windows?  
This should be a simple consequence of LLN, but I can't figure out a simple proof.  
 A: I'll take the liberty of changing your sum slightly, so that we divide 
by the number of terms. Define $M(n)={1\over n}\left(X_1+\cdots +X_{n}\right)$
 and $A(n)={1\over b_n}\left(X_{a_n+1}+\cdots +X_{a_n+b_n}\right).$
I will also write $\mu$ for $\mathbb{E}(X_1).$
Then 
$$A(n)-\mu={a_n+b_n\over b_n}\left(M(a_n+b_n)-\mu\right)
-{a_n\over b_n}\left(M({a_n})-\mu\right).\tag1$$
From the law of large numbers and (1), we see that if the ratio $a_n/b_n$ is bounded above, then 
$A(n)\to\mu$ almost surely.

On the other hand, the result is not universally true. Let $X_i$ take the values +1 and -1 
with equal probability so $\mu=0$. From large deviation theory, there exists a finite constant $c>0$ so 
that $P(M(n)>1)\geq \exp(-cn)$ for large $n$.
Now take $b_n=\lceil\ln(n)/c\rceil$ so that $P(M(b_n)>1)\geq 1/n.$
Choose $a_n=n^2$ so that $a_{n+1}>a_n+b_n$ for large $n$, and hence the averages $A(n)$ appearing 
in the moving windows are independent. 
Since $A(n)$ has the same distribution as $M({b_n})$, and $$\sum_n P(A(n)>1)=\sum_n P(M(b_n)>1)\geq \sum_n 1/n=\infty,$$ 
the Borel-Cantelli lemma shows that $A(n)$ does not converge almost surely.   
