I am trying to solve the following probability problem:
Please calculate $ E(X) $ where the joint density function of x and y is:
$$f(x,y) = \frac{1}{y}e^{-(y+\frac{x}{y})} \text{ for } x,y > 0$$
It is well known that the expected value of a continuous random variable is:
$$ E(X) = \int_{-\infty}^\infty{xf_X(x)dx} $$
That means that we have to find $f_X(x)$ first by doing:
$$ f_X(x)=\int_{0}^\infty{\frac{1}{y}e^{-(y+\frac{x}{y})}}dy $$
Note: The zero lower bound is due to the given inequality $ y > 0$.
However, this integral turned out to be a handful (I tried substitution and integration by parts to no avail). Then, I tried this on WolframAlpha instead. However, it gave me an answer which I did not understand (link).
Am I even supposed to integrate this integral in the first place? I do not know much about the bessel function, which was present in the answer given by WolframAlpha. Furthermore, I am only undergoing an introductory probability course so it is not possible for the integral to be so complex (I was not taught bessel function anyway).
Could someone please advise me on how I could calculate the value of $E(X)$?
It is simple. First find $f_Y(y)=\int_{0}^{\infty}\frac{1}{y} e^{-(y+\frac{x}{y})} dx= e^{-y}$ Thus find expectation of $Y$ ,$E(Y)=\int_{0}^{\infty} ye^{-y} dy=1$ Now, find the distribution of $X|Y$ , which is $f_{Y|X}(y)=\frac{1}{y}e^{-\frac{x}{y}}$.Thus $X|Y \sim \text{Exponential}(y)$,with mean $y$.
Now, use law of conditional expectation, $E(X)=E(E(X|Y))=E(Y)=1$