Let $X_1, X_2, \ldots$ be independent r.v.'s with $0 \leq X_n \leq 1$ and $\sum_n E(X_n) = \infty$. Show $\sum_n X_n = \infty$ with probability 1?

Question

Let $X_1, X_2, \ldots$ be independent random variables with $0 \leq X_n \leq 1$ and $\sum_n E(X_n) = \infty$. I'd like to show that $\sum_n X_n = \infty$ with probability 1. This seems like a Borel Cantelli Problem to be but I am having a hard time defining the sets to work with. Is there another easier approach here? Thanks!

Answer 1

We can use Kolmogorov's three-series theorem.

Let $X_1,X_2,\ldots$ be independent random variables. Let $A>0$ and let $Y_i=X_i1_{\{|X_i|\le A\}}$. In order that $\sum_{n=1}^\infty X_n$ converges almost surely, it is necessary and sufficient that $$ \sum_{n=1}^\infty\Pr\{|X_n|>A\}<\infty,\ \sum_{n=1}^\infty\operatorname EY_n\ \text{converges, and}\ \sum_{n=1}^\infty\operatorname{Var}Y_n<\infty. $$

Let us choose any $A\ge1$. Then we have that $$ \sum_n\operatorname E[X_n1_{\{|X_n|\le A\}}]=\sum_n\operatorname EX_n=\infty. $$ Hence, the series $\sum_nX_n$ diverges almost surely. Since $X_n\ge0$, we have that $\sum_nX_n=\infty$ almost surely.

Answer 2

Independence and boundedness of the variables can be exploited efficiently with the following computation (which is an easy version of the Chernoff bound): for any fixed $\alpha>0$ we have $$\mathbf{P}(X_1+\dots+X_n\le \alpha)=\mathbf{P}(e^{-(X_1+\dots+X_n)}\ge e^{-\alpha})\le\frac{\mathbf{E}[e^{-(X_1+\dots+X_n)}]}{e^{-\alpha}} =e^\alpha\prod_{k=1}^n\mathbf{E}[e^{-X_k}].$$ Now for any $0\le s\le 1$ we have the simple inequality $$1-e^{-s}=se^\xi\ge se^{-1}$$ (we applied Lagrange's theorem to the function $x\mapsto e^x$ on the interval $[-s,0]$, which gives us some $\xi\in [-s,0]$ such that the first equality holds; then we used the fact that $\xi\ge -1$), thus $$\mathbf{E}[e^{-X_k}]\le \mathbf{E}[1-e^{-1}X_k]=1-e^{-1}\mathbf{E}[X_k]$$ and finally $$\prod_{k=1}^n\mathbf{E}[e^{-X_k}]\le\prod_{k=1}^n\left(1-e^{-1}\mathbf{E}[X_k]\right)\le\prod_{k=1}^n\exp\left(-e^{-1}\mathbf{E}[X_k]\right) =\exp\left(-e^{-1}\sum_{k=1}^n\mathbf{E}[X_k]\right)\to 0$$ as $n\to\infty$. So $$\mathbf{P}(\sum X_k\le \alpha)=\lim_{n\to\infty}\mathbf{P}(\sum_{k=1}^n X_k\le\alpha)=0$$ and since $\alpha$ was arbitrary we get the thesis.