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

Let $Y_1\sim \operatorname{Poisson}(\lambda_1)$ and $Y_2\sim \operatorname{Poisson}(\lambda_2)$, 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 $\frac{X_1} {X_1+...+X_n}$, 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 '17 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. – wolfies May 31 '17 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 '17 at 17:14
• So you'll have 0/0. It's still undefined. – soakley May 31 '17 at 18:33
• So my "binomial-idea" is not valid? – infstat May 31 '17 at 18:49

As pointed out in comments, with a certain positive (maybe small, but always positive) probability, the expression $$Y_1/(Y_1+Y_2)$$ becomes $$0/0$$, so the random variable $$Y_1/(Y_1+Y_2)$$ is not unambiguously defined. You must clarify your question, how do you define your random variable in that case?