The joint density of random variables X and Y is given by $$f(x,y)= \begin{cases} \frac{2e^{-2x}}{x} , & 0\le x \lt \infty \ , \ 0\le y \le x \\ 0\quad , & \text{otherwise} \end{cases}$$ Compute $\text{Cov}(X ,Y)$.
So in order to get the covariance, I need to find the marginal density function for $x$ and $y$ first, then compute $E(X)$ and $E(Y)$.
I am able to get the marginal density function for $x$ but I am stucked with the marginal density function for $y$: $$ f_Y(y)=\int_{-\infty}^{\infty}f(x,y)\:dx=\int_{0}^{\infty}\frac{2e^{-2x}}{x}\:dx $$ The integral does not converge at $0$.
Or should I use $\int_{y}^{\infty}\frac{2e^{-2x}}{x}\:dx$ instead of $\int_{0}^{\infty}\frac{2e^{-2x}}{x}\:dx$ , since $0\le y \le x\lt \infty$ ?
But I am also not sure how to deal with $\int_{y}^{\infty}\frac{2e^{-2x}}{x}\:dx$...
Any help is highly appreciated! Thanks.
Note the domain is the infinite triangle region under the line $y=x$ and above the x-axis. So when you find the marginal function, your integration limit for $x$ is from $y$ to $\infty$
$$f_Y(y)=\int_y^\infty f(x,y) dx$$
This is exponential integral, see here: https://en.wikipedia.org/wiki/Exponential_integral
You cannot proceed more from here to find an analytical result for $f_Y(y)$. But you still can calculate the $E(Y)$ for example, $$E(Y)=\int_0^\infty yf_Y(y)dy=\int_0^\infty \int_y^\infty yf(x,y) dxdy$$
Interchange the integration orders:
$$E(Y)=\int_0^\infty \int_0^x yf(x,y) dydx=\int_0^\infty\frac{2e^{-2x}}{x} \left(\int_0^x y dy\right)dx $$