Problem regarding calculation of expectation when we have a truncated random variable.

Question

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

Answer 1

Is this seems to be okay ? Thank you.