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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?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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?

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  • $\begingroup$ 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

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