moment generating function of a pmf A pmf is shown below along with its mgf. I was wondering how I would calculate moments for the marginal distributions. Would I let t1 = 0 to find the marginal distribution of y? Would I then differentiate that expression and evaluate it at 0 to find E(Y)? 
Also, how would I find E(XY) and E(X|y)? For E(XY), would the easiest way to find it be to do a double summation of x and y for xy * the pmf? And for E(X|y), I am not sure at all how to go about finding that... Thanks for the help!

 A: Let's first compute the moment generating function from the p.m.f.:
$$
  \begin{eqnarray}
  M(t_1, t_2) &=& \mathbb{E}( \exp( t_1 X + t_2 Y) ) =
     \sum_{x=0}^\infty \sum_{y=x}^\infty \mathrm{e}^{-2} \frac{\exp(t_1 x)}{x!} \frac{\exp(t_2 y)}{(y-x)!} \\ &=& \sum_{x=0}^\infty \sum_{y=0}^\infty \mathrm{e}^{-2} \frac{\exp(t_1 x)}{x!} \frac{\exp(t_2 (y+x))}{(y)!} = M_{\mathrm{Po}(1)}(t_1+t_2) M_{\mathrm{Po}(1)}(t_2)
  \end{eqnarray}
$$
where $M_{\mathrm{Po}(1)}(t) = \exp\left( \mathrm{e}^t - 1 \right)$. The factorization of $M(t_1, t_2)$ means that the random variable in question corresponds to vector $(X, Y) = (Z_1, Z_1+Z_2)$, where $Z_1$ and $Z_2$ are i.i.d. Poisson random variate with unit mean.
Thus $\mathbb{E}(X Y) = \mathbb{E}\left( Z_1(Z_1+Z_2) \right) = \mathbb{E}\left( Z_1^2 \right) + \mathbb{E}\left( Z_1 \right) \mathbb{E}\left( Z_2 \right) = (1^2 + 1) + 1 \times 1 = 3$. Using moment generating function
$$ \begin{eqnarray}
  \mathbb{E}(X Y) &=& \left. \frac{\mathrm{d}^2}{\mathrm{d} t_1 \mathrm{d} t_2} M(t_1, t_2) \right\vert_{t_1=0, t_2=0} = \left. \left( M(t_1,t_2) \mathrm{e}^{t_1+t_2}  \left(
  1 + \mathrm{e}^{t_1} + \mathrm{e}^{t_1+t_2} \right)\right)\right\vert_{t_1=0, t_2=0} \\&=& 1 \times 1 \times (1+1+1 ) = 3
  \end{eqnarray}
$$
Now, let's turn to $\mathbb{E}\left( X \vert Y=y \right)$. Doing computation directly:
$$
  \mathbb{E}(X \vert Y=y) = \frac{ \sum_{x=0}^\infty x \frac{1}{\mathrm{e}^2} \frac{1}{x!} \frac{1}{(y-x)!}}{\sum_{x=0}^\infty \frac{1}{\mathrm{e}^2} \frac{1}{x!} \frac{1}{(y-x)!}} 
  = \frac{ \sum_{x=0}^y x \frac{1}{\mathrm{e}^2} \frac{1}{x!} \frac{1}{(y-x)!}}{\sum_{x=0}^y \frac{1}{\mathrm{e}^2} \frac{1}{x!} \frac{1}{(y-x)!}}
$$
Now the changing variables $x \mapsto y-x$ in the sum in the numerator:
$$
  \mathbb{E}(X \vert Y=y) = y - \mathbb{E}(X \vert Y=y) \qquad \implies \qquad 
    \mathbb{E}(X \vert Y=y) = \frac{y}{2}  
$$
In order to get the same result using mgf, one should remark, that for positive discrete r.v. the $M_{\mathrm{Po(1)}}(\log t)$ is the probability generating function of Poisson random variable with unit mean, indeed:
$$
   M(\log t) = \exp( t -1 ) = \sum_{x=0}^\infty \frac{\mathrm{e}^{-1}}{x!} t^x
$$
It would then follow that 
$$
  \begin{eqnarray}
  \mathbb{E}(X \vert Y=y) &=& \frac{
   [t_2]^y \left. \frac{\mathrm{d}}{\mathrm{d} t_1} M\left( t_1, \log t_2 \right) \right\vert_{t_1=0}}{ [t_2]^y   M\left(0, \log t_2 \right)  } = 
  \frac{[t_2]^y \left. \frac{\mathrm{d}}{\mathrm{d} t_1} \exp\left( \mathrm{e}^{t_1} t_2 + t_2 - 2 \right) \right\vert_{t_1=0}}{ [t_2]^y   M\left(0, \log t_2 \right)  }  \\
  &=& \frac{[t_2]^y \left. t_2 \exp\left( t_2 \mathrm{e}^{t_1} + t_2 + t_1 - 2 \right) \right\vert_{t_1=0}}{ [t_2]^y   \exp(2(t_2-1))  } = 
  \frac{[t_2]^y \left( t_2 \exp\left( 2 (t_2 - 1) \right)   \right)}{
    [t_2]^y   \exp(2(t_2-1))
    } \\
  &=& \frac{ 2^{y-1} \mathrm{e}^{-2} /(y-1)! }{ 2^{y} \mathrm{e}^{-2} /(y)!} = \frac{y}{2}
  \end{eqnarray}
$$
Added As suggested by Didier Piau, the conditional expectation also follows by symmetry, using representation of $(X, Y)$ in terms of $Z_1$ and $Z_2$. Indeed:
$$
   \mathbb{E}(X \vert Y=y) = \mathbb{E}(Z_1 \vert Y=y) \stackrel{\text{symmetry}}{=} \frac{1}{2} \left( \mathbb{E}(Z_1 \vert Y=y) + \mathbb{E}(Z_2 \vert Y=y)) \right) = \frac{1}{2} \mathbb{E}(Z_1 +Z_2 \vert Y=y)= \frac{y}{2}
$$
Where the symmetry refers to the fact that, due to $Z_1$ and $Z_2$ being i.i.d., the following equation holds:
$$\mathbb{E}(Z_1 \vert Y=y) = \mathbb{E}(Z_1 \vert Z_1+Z_2 = y) = \mathbb{E}(Z_2 \vert Z_1+Z_2 = y) =  \mathbb{E}(Z_2 \vert Y=y).$$
