Let $X$ and $Y$ be two independent Geometric(p) random variables. Find $E[(X^2+Y^2)/(XY)]$.
I am really struggling with this question because I want to apply the LOTUS equation but am unsure how to do it for geometric variables. Any help would be appreciated.
Geometric distribution understood as: number of trials until first success
We want the expectation of $\frac{Y}{X}+\frac{X}{Y}$, which by linearity is the expectation of $Y/X$ plus the expectation of $X/Y$. By independence, we have $E(Y/X)=E(Y)E(1/X)$. We know $E(Y)$ so all we need is $E(1/X)$.
For the expectation of $1/X$, we want the sum $$p\sum_1^\infty \frac{1}{k}q^{k-1},\tag{1}$$ where $q=1-p$.
Recall that the Maclaurin series of $-\ln(1-x)$ is $$x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots.$$ Now looking at (1) we see that $$E(1/X)=-\frac{p}{q}\ln p.$$ Now put the pieces together. You should get something like $\frac{2}{q}\ln(1/p)$. (I rewrote $-\ln p$ as $\ln(1/p)$ in order to avoid minus signs.)