$X$ and $Y$ have the following joint density:
$$f_{X, Y}(x,y) = \begin{cases} {\frac{1}{2}x^3e^{-xy-x}} & \text{$x \gt 0, y \gt 0$} \\{0} & \text{otherwise}\end{cases}$$
(a) Give the value of $Cov(X, Y)$.
(b) Give the value of $E(Y|X= 2)$
(a)
I asked this a few days ago but I made significant changes to my solution based on feedback I got. I think I have most of it right but am having difficulty calculating $f_Y(y)$. Wolfram Alpha and other online calculators do not give an output.
I have the following: $$Cov(X,Y)=E(XY)-E(X)E(Y)$$
$$E(XY)=\int_{0}^{\infty}\int_{0}^{\infty} xy\frac{1}{2}x^3e^{-xy-x} dydx = 1$$
$$f_Y(y)=\int_0^{\infty} \frac{1}{2}x^3e^{-xy-x} dx = ?$$
$$f_X(x)=\int_0^{\infty} \frac{1}{2}x^3e^{-xy-x} dy = \frac{1}{2}e^{-x}x^2$$
$$E(Y)=\int_0^{\infty} y\cdot f_Y(y)dy=?$$
$$E(X) = \int_0^{\infty} x\cdot f_X(x) dx = 3$$
(b)
$$f_{Y|X}(y|x)=\frac{f_{X,Y}(x,y)}{f_X(x)}=\frac{x\cdot e^{-xy-x}}{e^{-x}}$$
Then $$E(Y|X=2)=\int_0^{\infty} y\cdot \frac{2\cdot e^{-2y-2}}{e^{-2}}dy=\frac{1}{2}$$
Edit for part (a)
$$E(Y)=\int_0^{\infty} \int_0^{\infty} y\cdot\frac{1}{2}x^3 e^{-xy-x}dydx =0.5$$
Giving $$Cov(X,Y)=1-(3)\cdot(0.5)=-0.5$$
Is that right?