Utilize series of the form $\sum\limits_{n\ge1}\frac{1}{n^p}$ to construct independent, nonnegative random variables $X_n$ such that $\sum\limits_{n\ge1}X_n$ converges a.s. but $\sum\limits_{n\ge1}EX_n$ diverges.
I am quite stumped on this one. I know $X_n=n\cdot\mathbb{1}_{(0,\frac{1}{n})}$ is a typical example of random variables such that
\begin{align*}
\sum_{n\ge1}EX_n=\sum_{n\ge1}n\cdot P\big(\big(0,\frac{1}{n}\big)\big)=\sum_{n\ge1}(1)=\infty
\end{align*}
However these random variables are not independent and I am not sure that $\sum\limits_{n\ge1}X_n$ converges a.s. If we let $A_n$ be disjoint intervals of length $\frac{1}{n}$ and set $X_n=n\cdot\mathbb{1}_{A_n}$, then the $X_n$ are independent this time and $\sum_{n\ge1}EX_n=\infty$ again, as above. But I am not sure that $\sum\limits_{n\ge1}X_n$ converges a.s., if they do, is there an nice way to see this? Any help with this or any other example of $X_n$'s that will satisfy the required properties would be greatly appreciated.
Best Answer
Let $X_n:= n^\alpha \mathbf{1}_{A_n}$, where $(A_n)_{n\geqslant 1}$ is a sequence of independent sets and $A_n$ has probability $p_n$, with $\alpha$ and $p_n$ specified later. If $\sum_{n\geqslant 1}p_n$ converges, so does $\sum_{n\geqslant 1}X_n$, by using Borel-Cantelli lemma. Note that $EX_n=n^\alpha p_n$ hence we can choose $p_n=n^{-2}$ and $\alpha =2$ for example.
For the construction of the sequence of sets, one can work on an infinite product of the unit interval endowed with the Lebesgue measure.