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

• Mar 22nd 2010, 05:43 PM
floats
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!
• Mar 23rd 2010, 01:01 AM
Laurent
Quote:

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 $\displaystyle e^{-\lambda}$), so why would the ratio $\displaystyle \frac{X}{Y}$ make sense?
• Mar 23rd 2010, 06:35 AM
floats
I see, that is a good point.

What if we consider X/(Y+1) instead? What would the variance of that be?
• Mar 23rd 2010, 11:20 AM
Laurent
Quote:

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 $\displaystyle E[X]E\left[\frac{1}{Y+1}\right]$ and $\displaystyle E[X^2]E\left[\frac{1}{(1+Y)^2}\right]$. I guess you know how to get $\displaystyle E[X]$ and $\displaystyle E[X^2]$ ($\displaystyle \lambda$ and $\displaystyle \lambda+\lambda^2$ since I know the variance is $\displaystyle \lambda$). So let's consider the other ones.

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

$\displaystyle E\left[\frac{1}{(Y+1)^2}\right]=e^{-\lambda}\sum_{n=0}^\infty\frac{1}{(n+1)^2}\frac{\l ambda^n}{n!}$ is more tricky. I'd let $\displaystyle f(\lambda)=\sum_{n=0}^\infty \frac{1}{(n+1)^2}\frac{\lambda^{n+1}}{n!}$ and compute $\displaystyle f'(\lambda)=\sum_{n=0}^\infty \frac{1}{n+1}\frac{\lambda^n}{n!}$, which we have computed above ; now integrate to get $\displaystyle f(\lambda)$, and multiply by $\displaystyle e^{-\lambda}\frac 1\lambda$ to get $\displaystyle E[\frac{1}{(Y+1)^2}]$.
• Mar 23rd 2010, 12:23 PM
floats
I got the $\displaystyle E[\frac{1}{Y+1}]$ part but was struggling with the $\displaystyle E[\frac{1}{(Y+1)^2}]$ part.

Thanks for the help!