Find $\text{Cov}(X/Y,Y)$ when $X$ and $Y$ are i.i.d. geometric with parameter $p$

Question

Suppose $X$ and $Y$ are i.i.d. $\text{Geometric}(p)$ with $0 < p < 1.$ I want to find $\operatorname{Cov}(X/Y,Y)$.

I was thinking of $\operatorname{Cov}(X/Y,Y) = E(X) -E(X/Y)E(Y)$, but I don't know how to compute $E(X/Y)$ since the distribution of $X/Y$ goes through all rational numbers and it's hard (impossible(?)) to give a closed formula since $a/b = 2a/2b=\cdots$

Answer 1

Since $X$ and $Y$ are independent, $E(X/Y) = E(X)E(1/Y)$, so what is $E(1/Y)$?

$X$ is a Geometric random variable find the expectation of $1/X$

$$E(1/Y) = \frac{-p\ln p}{1-p}$$

Thus,

$$\begin{align} \operatorname{Cov}(X/Y,Y) &= E(X) - E(X)E(1/Y)E(Y) \\ &= 1/p - (1/p)\left( \frac{-p\ln p}{1-p} \right) (1/p) \\ &=(1/p)\left( 1 + \frac{p\ln p}{1-p} \right) \end{align}$$ Properties of note, this converges $\infty$ as $p\rightarrow 0$, and to $0$ as $p\rightarrow 1$.