# Find the expected value of $(X^2+Y^2)/(XY)$ whwre $X,Y$ are independent geometric random variables [closed]

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

## closed as unclear what you're asking by Did, Claude Leibovici, Harish Chandra Rajpoot, user91500, user133281Oct 31 '15 at 18:54

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• Start by simplying the expression within the expectation and using the fact that the expectation is linear (en.wikipedia.org/wiki/Expected_value#Linearity). – Erik M Oct 29 '15 at 19:11
• How $\frac{X^2+Y^2}{XY}$ is defined when both $X$ and $Y$ equal zero? – Jack D'Aurizio Oct 29 '15 at 19:14
• $=E(X/Y+Y/X)=2E(X/Y)=2E(X)E(1/Y)=\frac 2 p\sum_{i\geq 1}\frac 1 ip(1-p)^{i-1}=\frac 2 p \frac p {1-p}\log \frac 1 p=\frac 2 {1-p} \log \frac 1 p$ – A.S. Oct 29 '15 at 19:20
• @JackD'Aurizio There are two distributions called geometric, the distribution of trials until success and that of failures before success. The former is supported over $\{1,...,\infty\}$ – Graham Kemp Oct 29 '15 at 19:56

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 $2q\ln(1/p)$. (I rewrote $-\ln p$ as $\ln(1/p)$ in order to avoid minus signs.)