4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $$\left(\frac{\lambda_1}{\sum_j \lambda_j}\right)\left(\frac{1}{\lambda_1} - \frac{1}{\sum_j \lambda_j}\right)$$ $\endgroup$ – jld May 31 '17 at 15:01
  • 2
    $\begingroup$ 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. $\endgroup$ – wolfies May 31 '17 at 16:48
  • $\begingroup$ @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 $\endgroup$ – jld May 31 '17 at 17:14
  • $\begingroup$ So you'll have 0/0. It's still undefined. $\endgroup$ – soakley May 31 '17 at 18:33
  • $\begingroup$ So my "binomial-idea" is not valid? $\endgroup$ – infstat May 31 '17 at 18:49
0
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How is this an answer to the question $\endgroup$ – Do not reinstate Monica Jun 5 at 13:21
  • $\begingroup$ Well, maybe its more a comment. $\endgroup$ – kjetil b halvorsen Jun 5 at 15:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.