Find $E\left(\frac{2X}{(Y+1)^2}\right)$ For $Y\sim \operatorname{Pois}(1)$ and $E(X\mid Y)=\frac{Y+1}{2}$

Question

For $Y\sim \operatorname{Pois}(1)$ we know that $E(X\mid Y)=\frac{Y+1}{2}$. Find $E\left(\frac{2X}{(Y+1)^2}\right)$.

In the solution they said that: $$ E\left(\frac{2X}{(Y+1)^2}\right)=E\left(E\left(\frac{2X}{(Y+1)^2}\mid Y\right)\right)=E\left(\frac{2}{(Y+1)^2}E(X\mid Y)\right) $$ I don't understand why $E\left(E\left(\frac{2X}{(Y+1)^2}\mid Y\right)\right)$ is equal to $E\left(\frac{2}{(Y+1)^2}E(X\mid Y)\right)$. What general property did they use for that?

Is it true to say that for every function $g(y)$ we get $E(g(Y)X\mid Y)=g(Y)E(X\mid Y)$?

Answer 1

The conditional random variable $$\frac{2X}{(Y+1)^2} \mid Y$$ is random only with respect to $X$, because $Y$ is given. Therefore, $$\operatorname{E}\left[\frac{2X}{(Y+1)^2} \mid Y\right] = \frac{2}{(Y+1)^2} \operatorname{E}[X \mid Y],$$ just as we write $$\operatorname{E}[cX] = c \operatorname{E}[X]$$ for a scalar $c$. Now take the expectation with respect to $Y$ and we get the claimed equality.

Answer 2

Yes, as you guessed, it's true that

$$E(g(Y)X\mid Y)=g(Y)E(X\mid Y)$$

Informally: conditioning on $Y$ is like fixing the value of $Y$, then $g(Y)$ behaves like a constant, then it goes outside the expectation , analogous to $E[aX] = a E[X]$

Answer 3

If we assume $X \mid Y$ has density function $f_{X \mid Y}$,

$$\begin{align} \mathbb{E}\left[\dfrac{2X}{(Y+1)^2} \mid Y \right] &= \int_{-\infty}^{\infty}\dfrac{2x}{(Y+1)^2}f_{X \mid Y}(x \mid Y)\text{ d}x \\ &= \dfrac{2}{(Y+1)^2}\int_{-\infty}^{\infty}xf_{X \mid Y}(x \mid Y)\text{ d}x \\ &= \dfrac{2}{(Y+1)^2}\mathbb{E}\left[X \mid Y \right] \end{align}$$ The proof is similar in the case where $f_{X \mid Y}$ is a probability mass function, with sums instead of integrals.