Convergence of sum of expectations 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?
 A: 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).
A: Law of large numbers is not applicable because the $X_i$'s are not identically distributed.
Step 1: Suppose on the contrary that $$P(\sum_{i=0}^\infty(X_i-E[X_i]) = \infty) > 0$$
Step 2: Then
$$E[\sum_{i=0}^\infty(X_i-E[X_i])] = \infty$$
Step 3: By Fubini's Theorem, we have
$$\sum_{i=0}^\infty E[(X_i-E[X_i])] = \infty$$
Step 4:
$$\to \sum_{i=0}^\infty 0 = \infty$$
Step 5:
$$\to 0 = \infty$$
QED
Some notes:

*

*The variance is a red herring much like in the law of large numbers.


*It is indeed possible that $P(\sum_k X_k = \infty)>0$ while $\sum_k Var(X_k) < \infty$. However, if $E[X_k] = 0$, then $P(\sum_k X_k = \infty)=0$
A: This requires a few theorems. First of all, for a series of independent random variables all types of convergence (almost sure convergence, convergence in prbability, convergence in distribution, convergence in $L^{p}$ in the special case when the p-th absolute moments exist)  are equivalent. Secondly convergence of the given series in $L^{2}$, i.e. in mean square is an elementary fact about convergence of a series in the norm of a Hilbert space when the terms are orthogonal.  From these two fact we get almost sure convergence.
Alternative proof: the partial sums of the series form a martingale. This martingale is $L^{2}$ bounded by hypothesis. By Martingale Convergence Theorem the series converges almost surely.
A: 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.
