Calculate $E(e^{X+Y})$ Let $(X,Y)$ be a two-dimensional abs. cont. vector. I am given:
$$f_{X,Y}(x,y)=\begin{cases} 
      1 & 0<x<1, \quad x<y<x+1 \\
      0 & \text{otherwise}
   \end{cases}$$
 and
$$f_X(x)=\begin{cases}
   1 & 0<x<1 \\
   0 & otherwise
\end{cases}$$
Now I need to calculate $E(e^{X+Y})$. I wanted to do the following:
$$E(e^{X+Y})=E(e^Xe^Y)=E(e^X)E(e^Y)$$
But realized that this requires that $X$ and $Y$ are independent, which I do not know.  So here I am kinda stuck, would appreciate if someone could help me fowards. 
 A: We have that:
\begin{equation}
E[e^{X+Y}] = \int_{X\times Y}e^{x+y}1_{x\in[0,1],y\in[x,x+1]}\mathrm{d}x\mathrm{d}y
\end{equation}
We can apply Fubini's theorem to this to get that:
\begin{align*}
E[e^{X+Y}] &= \int_{x = 0}^1 \int_{y = x}^{x+1} e^{x+y}\mathrm{d}y\mathrm{d}x \\
& = \int_{x = 0}^1 e^x\int_{y = x}^{x+1}e^y\mathrm{d}y\mathrm{d}X \\
& = \int_0^1 e^x \left(e^{x+1}-e^x\right)\mathrm{d}x \\
& = \int_0^1 e^{2x+1}-e^{2x}\mathrm{d}x \\
& = \left(\frac{1}{2}e^{2x+1}-\frac{1}{2}e^{2x}\right|_{x = 0}^1 = \frac{1}{2}(e^3-e^2)-\frac{1}{2}(e-1) = \frac{1}{2}(e^3-e^2-e+1)
\end{align*}
Now, we have easily that:
\begin{equation}
E[e^X] = \int_0^1 e^x\mathrm{d}x = e-1
\end{equation}
To find the marginal density of $Y$, note that geometrically $f_{X,Y}$'s support is a parallelogram with corners $(0,0),(1,0),(1,1),(1,2)$.  The marginal density $f_Y(y)$ is the "height" (with respect to the $y$-axis) of this parallelogram, which is clearly $y$ if $y\in(0,1)$, and $2-y$ (a little less clearly, but not too hard to motivate) if $y\in(1,2)$.
We can now find the expectation as:
\begin{equation}
E[e^Y] = \int_0^1 e^y\mathrm{d}y+\int_1^2 e^y(2-y)\mathrm{d}{y} = \int_0^2 e^y\mathrm{d}y+\int_1^2 e^y-ye^y\mathrm{d}y = e^2-1 -e
\end{equation}
So:
\begin{equation}
E[e^Y]E[e^X] = (e-1)(e^2-e-1) = e^3-e^2-e-e^2+e+1 = e^3-2e^2+1\neq \frac{1}{2}(e^3-e^2-e+1) = E[e^{X+Y}]\end{equation}
This confirms that computing $E[e^Y]E[e^X]$ would have been insufficient.
