Convergence of sum of expectations

Question

Let $X_1,X_2,...$ be independent random variables with finite expectation. If $\sum_{i=0}^\infty Var(X_i) < \infty$, Show that $\sum_{i=0}^\infty(X_i-E[X_i])$ converges almost surely.

$E[X_i]$ means a expectation of $X_i$

How to solve it? Use Weak Law of large numbers?

Answer 1

Since you aim to show convergence almost surely, it will be an application of the strong law of large numbers. Therefore just define $Y_i = X_i -\mathbb{E}[X_i]$ and it remains to verify, that the $Y_i$ are uncorrelated, have the same expectation (both clear) and $\mathbb{E}[Y_i^2] < \infty$ and have bounded variance (both given due to the finiteness of the series).

Answer 2

First note that by taking $Y_i = X_i - E[X_i]$, we assume that $E[X_i] = 0$.

Then, we use Kolmogorov's three-series theorem https://en.wikipedia.org/wiki/Kolmogorov%27s_three-series_theorem

Thus, it suffices to verify the three conditions. For any $A>0$,

(1) $$\sum^\infty_{n=1} P(|X_n > A|) \le \sum^\infty_{n=1} A^{-2}E[X_n^2] =\sum^\infty_{n=1} A^{-2}Var[X_n] \le \infty $$

(2) $$\sum^\infty_{n=1} E[X_n 1_{X_n\le A }] = - \sum^\infty_{n=1} E[X_n 1_{X_n> A }] ,$$

and $$\sum^\infty_{n=1} |E[X_n 1_{X_n> A }]| \le \sum^\infty_{n=1} E[|X_n 1_{|X_n|> A }|] \le A^{-1} \sum^\infty_{n=1} E[|X_n^2 | ] < \infty,$$ which implies that $\sum^\infty_{n=1} E[X_n 1_{X_n\le A }]$ converges.

(3) Finally, $$\sum^\infty_{n=1} |Var[X_n 1_{X_n\le A }] |\le \sum^\infty_{n=1} E[X_n^2 1_{X_n \le A }] + (E[|X_n| 1_{X_n \le A }] )^2 \le 2\sum^\infty_{n=1} E[X_n^2 ] < \infty,$$ where we used Holder's inequality for the second term in the last inequality.