# Example of nonnegative random variables $X_n$ such that $\sum\limits_{n\ge1}X_n$ converges a.s. but $\sum\limits_{n\ge1}EX_n$ diverges.

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.

• "$A_n$ be disjoint intervals of length $\frac{1}{n}$ and set $X_n=n\cdot\mathbb{1}_{A_n}$, then the $X_n$ are independent" is not what you want (disjoint is not the same as independent). You will also have an issue with $\sum\limits_{n\ge1}\frac{1}{n} = \infty$ which is why the questions suggests $\sum\limits_{n\ge1}\frac{1}{n^p}$ Jul 20 '20 at 18:00

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.

• Thanks for your reply! How does $\sum\limits_{n\ge1}p_n<\infty\implies\sum\limits_{n\ge1}X_n<\infty$ by BC? I am trying to work that out but am getting tripped up. Jul 20 '20 at 18:14
• $\limsup A_n$ has probability $0$, hence for almost every $\omega$, there is some $n_0(\omega)$ for which $\omega\notin A_n$ whenever $n\geq n_0(\omega)$. Jul 20 '20 at 18:18
• Awesome, thank you very much! Jul 20 '20 at 18:24

Consider on some probability space a sequence of independent random variables $$(X_n)_{n\in\mathbb N}$$ such that \begin{align*} X_n= \begin{cases} n&\text{with probability \dfrac{1}{n^2},}\\ 0&\text{with probability 1-\dfrac{1}{n^2};} \end{cases} \end{align*} for a general construction of the underlying probability space, see Theorem 20.4 in Billingsley (1995, p. 265).

For each $$m\in\mathbb N$$, let \begin{align*} E_m\equiv\{X_n=0\text{ for every n\geq m}\}. \end{align*} By independence, one has \begin{align*} \mathbb P(E_m)=\prod_{n=m}^{\infty}\left(1-\frac{1}{n^2}\right)=\frac{m-1}{m}. \end{align*} Define $$E\equiv\bigcup_{m=1}^{\infty} E_m$$. Since $$E_1\subseteq E_2\subseteq E_3\subseteq\cdots$$, it follows that \begin{align*} \mathbb P(E)=\lim_{m\to\infty}\mathbb P(E_m)=1. \end{align*} But on $$E$$, only finitely many members of $$(X_n)_{n\in\mathbb N}$$ can be positive, so \begin{align*} \sum_{n=1}^{\infty}X_n<\infty. \end{align*}

At the same time, \begin{align*} \sum_{n=1}^{\infty}\mathbb E(X_n)=\sum_{n=1}^{\infty}\left[(n)\left(\frac{1}{n^2}\right)+(0)\left(1-\frac{1}{n^2}\right)\right]=\sum_{n=1}^{\infty}\frac{1}{n}=\infty. \end{align*}

My first answer was wrong, as pointed out by @RobertIsrael. I just realized this amended answer is essentially the same as the much simpler one by @DavideGiraudo (once you exploit the Borel–Cantelli lemma).

• In your example, $\mathbb E[X_n] = 1/n^4$, not $1/n$. Jul 20 '20 at 18:11
• @RobertIsrael Whoa, this mistake is as embarrassing as it is obvious. Thanks for pointing it out. Jul 20 '20 at 18:12
• Awesome, thanks for the reply. Jul 20 '20 at 19:46