Existence of the expectation for some random variables on $\mathbb{N}^*$

Question

If $\Omega$= {1,2,3,...} and p({n})=$\frac{1}{n(n+1)}$ n=1,2,..

I need to find Ex , Ey

a)X(n)=n Y(n)=0 if n is odd and X(n)=0 Y(n)=n if n is even

I begin solve it by

$Ex=\sum_{x=0}^{\infty}(xp(x))$=1p(1)+2p(2)+3p(3)+.....

=1($\frac{1}{n}-\frac{1}{n+1}$)+2($\frac{1}{n}-\frac{1}{n+1}$)+ 3($\frac{1}{n}-\frac{1}{n+1}$)+......

please hint for this question??

Answer 1

You are using a bad definition for $EX$.

The expectation of a random variable $X$ defined on $(\mathbb{N}^*,\mathcal{P}(\mathbb{N}^*),p)$ is $$ EX= \sum_{n\geq 1} X(n) p(n). $$

Since, in the way you put it, $p$ is the probability of $\Omega$ and $X:\Omega \rightarrow \mathbb{N}$, so $p(\{n\})\neq P(X=n):=P_X({n})$ (in the latter case you could use the transfer theorem and have $ EX= \sum_{n\geq 0} n P_X(n) $). Notice that it is easy to see the problem, since $X$ can have values equal to $0$ and $p$ is not defined on $0$....

So here, $$ EX = \sum_{n\ \text{odd}} \frac{1}{n+1} = +\infty$$

And quite the same result for $EY$.