The joint density function of $X$ and $Y$ is given by $$f(x,y)=\frac{1}{y}e^{-(y+\frac{x}{y})},\quad x>0,y>0.$$ Find $E[X]$, $E[Y]$ and $Cov\left(X,Y\right)$.
Calculating $E[Y]$ was easy for me. \begin{align} f_{Y}(y)&=\int_{0}^{\infty} \frac{1}{y}e^{-(y+\frac{x}{y})}dx\\ &=e^{-y}, \quad y>0 \end{align} Therefore $Y$ is an exponential random variable with parameter $1$ so $$E[Y]=1$$
Now, I am stuck in calculating the density function of X. I tried to calculate it from the usual integral but I couldn't. I looked at my calculus book to revise the integrals chapter but I haven't stumble accross something like it. Any help would be much appreciated.
Thanks to
@sudeep5221for the suggestion. We do not need to explicitly calculate the density function of $X$.\begin{align} E[X]&=\int_{0}^{\infty}\int_{0}^{\infty}xf(x,y)\ dx\ dy\\ &=\int_{0}^{\infty}e^{-y}\int_{0}^{\infty}\frac{x}{y}e^{-x/y}\ dx\ dy \end{align} It is noticeable that $\int_{0}^{\infty}\frac{x}{y}e^{-x/y}\ dx$ is the mean of an exponential random variable with parameter $1/y$, and thus equals to $y$. $$ E[X]=\int_{0}^{\infty}\int_{0}^{\infty}ye^{-y}dy=1 $$
To end, calculate $Cov(X,Y)$. $$Cov(X,Y)= E[XY]-E[X]E[Y]$$ Also \begin{align} E[XY]&=\int_{0}^{\infty}\int_{0}^{\infty}xyf(x,y)\ dx\ dy\\ &=\int_{0}^{\infty}ye^{-y}\left(\int_{0}^{\infty}\frac{x}{y}e^{-x/y}dx\right)\ dy\\ &= \int_{0}^{\infty}y^{2} e^{-y}dy \end{align} We can now integrate by parts where $U=y^2$ and $dV= e^{-y}dy$. \begin{align} E[XY]&=\int_{0}^{\infty}y^2e^{-y} dy= \left(-y^2 e^{-y}|_{0}^{\infty}\right)+\int_{0}^{\infty}2ye^{-y}dy\\ &=2E[Y]\\ &=2 \end{align} As a result, \begin{align}Cov(X,Y)&=2-(1)(1)\\ Cov(X,Y)&=1 \end{align}