Compute $E(Y\,|\,X)$ as a function of $X$

Question

Let $Y$ be a Poisson $\lambda$ random variable, and define $X=I_{[Y>0]}$. Compute $E(Y\,|\,X)$ as a function of $X$ and find $E(|Y-X|)$.

So far, I found the mass function of $X$, $$f_X(x)=\begin{cases} P(X=0)&=e^{-\lambda}\\ P(X=1)&= 1-e^{-\lambda} \end{cases}$$ I don't see how to find the mass function of $Y$ given $X$.

To calculate $E(|Y-X|)$, I think I need to separate into two cases. Since $X$ only equal to $0$ and $1$, $$\begin{align*} E(|Y-X|)&= \begin{cases} 0&\;\;\text{if $Y=0,1$}\\ E(Y-1)=\lambda-1&\;\;\text{if $Y\geq 2$} \end{cases} \end{align*}$$

Answer 1

If $X = 1$ then we know that $Y>0$ so we need the expected value $E(Y|Y>0).$ So we just only take into account the possibilities that $Y=1,2,3\ldots$ and adjust the normalization factor accordingly: $$ E(Y|X=1)= E(Y|Y>0) = \frac{\sum_{k=1}^\infty k \frac{\lambda^k}{k!}e^{-\lambda}}{\sum_{k=1}^\infty \frac{\lambda^k}{k!}e^{-\lambda}}$$ This can be cleaned up and computed in closed form by writing the numerator as a derivative.

If $X=0$ then $Y=0,$ so the expected value of $Y$ is just zero: $$E(Y|X=0) = E(Y|Y=0) = 0.$$

For $E(|X-Y|)$ there are no more cases (this is an unconditional expectation). If $Y=0$ then $X=0$ and $|X-Y| =0.$ If $Y>0$ then $X=1$ and $|X-Y| = Y-1.$ So we have $$ E(|X-Y|)=0P(Y=0)+\sum_{k=1}^\infty (k-1)P(Y=k) = \sum_{k=1}^\infty (k-1)\frac{\lambda^k}{k!}e^{-\lambda} $$

which can again be done in closed form using differentiation techniques.