$P(X=x, Y=y)= e^{-2}\binom{x}{y} \frac{3}{4}^{y}\frac{1}{4}^{x-y}\frac{2^{x}}{x!}$ ; Then $E(Y)$? Let the random variables $X$ and $Y$ have the joint probability mass function
$$P(X=x, Y=y)= e^{-2}\binom{x}{y} \dfrac{3}{4}^{y}\frac{1}{4}^{x-y}\frac{2^{x}}{x!} ; \ \ y=0,1,...x ; x = 0,1,2...$$
Then $E(Y)$?
I tried to simplify this but couldn't. I can see one part is of binomial other is of Poisson with parameter $2$ but they are not independent so that is not gonna help me. How should i simplify it ?
 A: If it's too hard to find the marginals, write
$$\begin{align}
\mathbb{E}[Y] &= \sum_{\text{all x}}\sum_{\text{all }y}y \cdot \mathbb{P}(X = x, Y = y)\\
&=\sum_{x=0}^{\infty}\sum_{y=0}^{x}y\cdot e^{-2}\binom{x}{y}\left(\dfrac{3}{4}\right)^y\left(\dfrac{1}{4}\right)^{x-y}\dfrac{2^x}{x!} \\
&= \sum_{x=0}^{\infty}e^{-2}\dfrac{2^x}{x!}\left[\sum_{y=0}^{x}y\binom{x}{y}\left(\dfrac{3}{4}\right)^y\left(\dfrac{1}{4}\right)^{x-y}\right]
\end{align}$$
The sum 
$$\sum_{y=0}^{x}y\binom{x}{y}\left(\dfrac{3}{4}\right)^y\left(\dfrac{1}{4}\right)^{x-y}$$
is the expected value of a binomial distribution based on $x$ trials and "success" probability $3/4$. Hence, 
$$\sum_{y=0}^{x}y\binom{x}{y}\left(\dfrac{3}{4}\right)^y\left(\dfrac{1}{4}\right)^{x-y} = x\left(\dfrac{3}{4}\right)$$
Thus,
$$\mathbb{E}[Y]=\sum_{x=0}^{\infty}e^{-2}\dfrac{2^x}{x!}x\dfrac{3}{4}=\dfrac{3}{4}\sum_{x=0}^{\infty}x\cdot\dfrac{e^{-2}2^x}{x!}$$
The sum 
$$\sum_{x=0}^{\infty}x\cdot\dfrac{e^{-2}2^x}{x!}$$
is the expected value of a Poisson distribution with parameter $2$, hence we have
$$\mathbb{E}[Y] = \dfrac{3}{4} \cdot 2 = \dfrac{3}{2}\text{.}$$
