Thread: What is the variance of the division of 2 random variables?

1. What is the variance of the division of 2 random variables?

More specifically, if X and Y are Poisson-distributed, what is the variance of X/Y (assuming that X and Y are independent)?

Thanks!

2. Originally Posted by floats
More specifically, if X and Y are Poisson-distributed, what is the variance of X/Y (assuming that X and Y are independent)?

Thanks!
As far as I know, Poisson random variables can be 0 (with probability $e^{-\lambda}$), so why would the ratio $\frac{X}{Y}$ make sense?

3. I see, that is a good point.

What if we consider X/(Y+1) instead? What would the variance of that be?

4. Originally Posted by floats
What if we consider X/(Y+1) instead? What would the variance of that be?
To get the variance, you need the expectation, and the expectation of the square. Since the variables are independent, you need $E[X]E\left[\frac{1}{Y+1}\right]$ and $E[X^2]E\left[\frac{1}{(1+Y)^2}\right]$. I guess you know how to get $E[X]$ and $E[X^2]$ ( $\lambda$ and $\lambda+\lambda^2$ since I know the variance is $\lambda$). So let's consider the other ones.

$E\left[\frac{1}{Y+1}\right]=e^{-\lambda}\sum_{n=0}^\infty \frac{1}{n+1}\frac{\lambda^n}{n!}=e^{-\lambda}\sum_{n=0}^\infty \frac{\lambda^n}{(n+1)!}$ $=\frac{1}{\lambda}e^{-\lambda}\sum_{n=0}^\infty \frac{\lambda^{n+1}}{(n+1)!}$ ; then let $m=n+1$ in the sum to see that it equals $e^{\lambda}-1$.

$E\left[\frac{1}{(Y+1)^2}\right]=e^{-\lambda}\sum_{n=0}^\infty\frac{1}{(n+1)^2}\frac{\l ambda^n}{n!}$ is more tricky. I'd let $f(\lambda)=\sum_{n=0}^\infty \frac{1}{(n+1)^2}\frac{\lambda^{n+1}}{n!}$ and compute $f'(\lambda)=\sum_{n=0}^\infty \frac{1}{n+1}\frac{\lambda^n}{n!}$, which we have computed above ; now integrate to get $f(\lambda)$, and multiply by $e^{-\lambda}\frac 1\lambda$ to get $E[\frac{1}{(Y+1)^2}]$.

5. I got the $E[\frac{1}{Y+1}]$ part but was struggling with the $E[\frac{1}{(Y+1)^2}]$ part.

Thanks for the help!