# Variance of quotient of Poisson random variable and sum of the Poisson sample

Let

$$Y_1\sim \operatorname{Poisson}(\lambda_1)\\Y_2\sim \operatorname{Poisson}(\lambda_2),$$ where $$Y_1$$ and $$Y_2$$ are independent, and $$\lambda_1, \lambda_2>0$$.

What is the variance of $$\frac{Y_1} {Y_1+Y_2}?$$

Or more general, what is the variance of $$\frac{X_1} {X_1+ \cdots +X_n},$$where $$X_i\sim \operatorname{Poisson}(\lambda_i)$$, $$i=1,...,n$$?

EDIT:

This idea came to mind: It is well-known that if $$Y_1\sim \operatorname{Poisson}(\lambda_1)$$ and $$Y_2\sim \operatorname{Poisson}(\lambda_2)$$, where $$Y_1$$ and $$Y_2$$ are independent, then $$Y_1|Y_1+Y_2\sim \operatorname{Binomial}(y_1+y_2,p)$$. Since the variance of a Binomially distributed random variable is $$np(1-p)$$, I just need to find the MLE of $$p$$.

Could this be a solution to my problem?

• you could approximate it with a Taylor series, as described here: stat.cmu.edu/~hseltman/files/ratio.pdf. The first order approximation to the variance of the ratio is $$\left(\frac{\lambda_1}{\sum_j \lambda_j}\right)\left(\frac{1}{\lambda_1} - \frac{1}{\sum_j \lambda_j}\right)$$
– jld
May 31, 2017 at 15:01
• May I pose the consideration that the variance you seek is not defined? The sum of Poissons in the denominator will be Poisson ... and notably include 0 with discrete mass --> such that 1/denominator will yield infinite expressions. May 31, 2017 at 16:48
• @wolfies because of the dependence between them $\sum_j X_j = 0 \implies X_1 = 0$ so we'll never have a non-zero number divided by 0
– jld
May 31, 2017 at 17:14
• So you'll have 0/0. It's still undefined. May 31, 2017 at 18:33
• So my "binomial-idea" is not valid? May 31, 2017 at 18:49