Hi My question is as follows, Let $X_n$ be a discrete random variable such that $P(X_n=1)=\frac12$ or $P(X_n=-1)=\frac12$.

I need to calculate the following expectation.

$$ \sum_{n=0}^\infty E\!\left[\frac{X_n}{n}1_{|X_n|\leq\frac{n}{2}}\right]$$.

My work as follows,

$$ \sum_{n=0}^\infty E\!\left[\frac{X_n}{n}1_{|X_n|\leq\frac{n}{2}}\right]= \sum_{n=0}^\infty \frac{X_n}{n} \times P(|X_n|\leq\frac{n}{2}) $$

$$= \sum_{n=0}^\infty \frac{X_n=1}{n} \times P(|X_n=1|\leq\frac{n}{2}) + \frac{X_n=-1}{n} \times P(|X_n=-1|\leq\frac{n}{2}) $$ when i expand this , i got this

$$ = 1 \times\frac12 + (-1)\times \frac12 + (\frac12)\times 1 + (-\frac12)\times 1 + ...... $$

So the final answer is zero. I want to know that i did this correctly .. Can anyone help ?

Thank you

  • 1
    $\begingroup$ I am confused by your notation. $$\mathbb 1(|X_n| \le n/2)$$ is always $1$ if $n \ge 2$, and is $0$ if $n \in \{0,1\}$, so why would you write it this way if only the first two terms of your sum do not obey the inequality? Your subsequent computation doesn't make sense in light of this. $\endgroup$
    – heropup
    May 16, 2018 at 17:44
  • $\begingroup$ @heropup Yeah i got you. This is a part of a question where i need to prove the Kolmogorov's three-series theorem . I didnt see that until you showed it to me. Thank you. Btw there is nothing wrong in my method although it doesn't make any sense. isn't it ? $\endgroup$ May 16, 2018 at 18:00
  • $\begingroup$ In addition to the issue heropup mentions, note that you cannot pull a random variable outside of an expectation. Your work suggests "$E[X_n 1_{|X_n|\leq n/2}]=X_nE[1_{|X_n|\leq n/2}]$," yet this is not correct because the left-hand-side is a real number while the right-hand-side is a random variable. Actually, the remaining work after that also does not make sense, it looks like you are making up your own probability rules. $\endgroup$
    – Michael
    May 16, 2018 at 18:24
  • $\begingroup$ Hi @Michael . In this question , I am trying to figure out whether i did it correctly or incorrectly and if it wrong then what would be the correct approach. So you are telling that my approach is incorrect. Isn't it ? $\endgroup$ May 16, 2018 at 18:54

1 Answer 1


This is my asnwer after i go through the suggested modifications

Is this seems to be okay ? Thank you.

  • $\begingroup$ The second expression "$\sum_{n=0}^{\infty} E[X_n/n]$ for $n\geq 2$" does not make sense, it should be $\sum_{n=2}^{\infty}E[X_n/n]$. Else it is fine, indeed $E[X_n/n] = (1/n)E[X_n]=0$ for all $n \geq 2$. [Of course there is a "divide by zero" issue for $n=0$, overall the problem itself is not stated well since it creates that divide-by-0 issue, and it seems to have no reason to use indicator functions.] $\endgroup$
    – Michael
    May 16, 2018 at 20:50
  • $\begingroup$ Hi Yeah it should start from n=2. My actual problem is prove $$ \sum_{n=0}^\infty X_n \exp(-n)$$. series converges with probability 1. I tried to use the Kolmogorov's Three-Series Theorem to prove that. I tried to generalize this result so that i can use it for the original problem. According to that theorem i need to create a truncated random variable. $\endgroup$ May 16, 2018 at 21:29
  • $\begingroup$ so in that question my truncation will be as follows. $$ Y_n=\sum_{n=0}^\infty \frac {X_n}{ e^{n}} 1_{|X_n|\leq\frac{e^n}{2}}$$. $\endgroup$ May 16, 2018 at 21:38
  • $\begingroup$ If $X_n \in \{-1, 1\}$ for all $n$, then $\sum_{n=0}^{\infty} X_n e^{-n}$ converges (surely) by the ratio test and/or by simply noting the series is absolutely convergent (i.e., the sum of absolute values $|X_n e^{-n}|$ converges): en.wikipedia.org/wiki/Convergence_tests $\endgroup$
    – Michael
    May 17, 2018 at 0:14
  • $\begingroup$ Hi @Michael Thank you for the reply. If i use the Kolmogorov's Three-Series Theorem, the way i approached this question seems to be correct. isnt it ? $\endgroup$ May 17, 2018 at 0:23

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