Conditional Expectation for nth Moment of Univariate Discrete Distribution

Question

Suppose I have a discrete random variable $X$ with a known distribution, and $X\ge0$. Now suppose I define a new random variable $Y=X|_{X>0}$. That is, $Y$ is equal to $X$ given that $X>0$. Now suppose I am interested in obtaining the $n^{th}$ moment of $Y$. Is it true that:

$$E(Y^n)=\frac{E(X^n)}{1-P(X=0)}?$$

Answer 1

Let $X=\begin{cases} 1 &, \text{with probability} \frac12 \\ -1 & \text{,with probability}\frac12 \end{cases}$

Let $n=1$. We have $E[X]=0$.

but $E[X|X>0]=1 \neq 0$.

Edit:

If $X \ge 0$, by law of total expectation, $$E[X^n]=E[X^n|X>0]P(X>0)+E[X^n|X=0]P(X=0)$$

hence

$$E[X^n]=E[X^n|X>0]P(X>0)$$

$$E[X^n|X>0]=\frac{E[X^n]}{1-P(X=0)}$$