Consider $Y_1\sim \operatorname {Poisson}(\lambda\psi)$ and $Y_2\sim \operatorname {Poisson}(\lambda)$, where $Y_1$ and $Y_2$ are independent, and where we are interested in testing $H_0: \psi =1$. The joint likelihood function is given by $$f_{Y_{1},Y_{2}}(y_{1},y_{2})=\frac{e^{-\lambda\psi}(\lambda\psi)^{y_{1}}}{y_{1}!}\times\frac{e^{-\lambda}\lambda^{y_{2}}}{y_{2}!}=\frac{e^{-\lambda(1+\psi)}\lambda^{y_{1}+y_{2}}\psi^{y_{1}}}{y_{1}!y_{2}!},$$ and the log-likelihood function is given by $$l(\lambda, \psi)=-\lambda(1+\psi)+(y_1+y_2)\log(\lambda)+y_1\log(\psi)-\log(y_1!y_2!).$$ By calculating the partial derivatives of $l(\lambda, \psi)$ with respect to $\lambda$ and $\psi$, respectively, setting them equal to zero and solving for $\lambda$ and $\psi$, respectively, the MLEs are given by $\hat{\lambda}=y_2$ and $\hat{\psi}=y_1/y_2$.

Now as stated, I want to test $H_0: \psi=1$, and I want to do so by constructing a Wald test: $$W=\frac{(\hat{\psi}-\psi_0)^2} {Var(\hat{\psi})}$$ So I already got the MLE of $\psi$, $\hat{\psi}=y_1/y_2$, and $\psi_0=1$ in my case. The problem is $Var(\hat{\psi})$ since it might not be defined.

My idea is to reformulate the hypothesis as $H_0: \frac \psi {1+\psi}=0.5$. I have that $\frac{\hat{\psi}} {1+\hat{\psi}}=\frac{y_1} {y_1+y_2}$ which means that I need the variance of this estimator to construct the Wald test statistic. Does anyone know what the variance of a Poisson random variable divided with the sum of the sample of random variables is?

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    $\begingroup$ Since $y_2$ has a non-zero probability of being 0, the variance is not finite (neither is the mean!). $\endgroup$
    – Glen_b
    May 29, 2017 at 14:12
  • $\begingroup$ How about the fact that testing $\psi=1$ is the same as testing $y_1=y_2$? And $\hat{\lambda}=y_2$, so I get $W=(y_2-y_1)^2/Var(y_2)$. Could that be a way to get around the problem that @Glen_b Points out? $\endgroup$
    – infstat
    May 29, 2017 at 14:58
  • $\begingroup$ Clearly $y_1\neq y_2$; they're random variables. But testing equality of their Poisson parameters would be a much more promising direction to head I think. $\endgroup$
    – Glen_b
    May 29, 2017 at 21:35
  • $\begingroup$ The "problem" is that testing $\lambda=\psi$ is not at all what I'm interested in doing. I want to test wether the Poisson intensity has changed from one time period to another, i.e. if $\psi\ne1$ $\endgroup$
    – infstat
    May 30, 2017 at 6:21
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    $\begingroup$ @Glen_b The resolution of this conundrum is that when $y_2=0,$ the MLE is not given by $y_1/y_2:$ in that case $\hat\psi=1.$ Its variance is finite. However, the use of the Wald test is questionable and would require careful analysis. The problems stem from the unfortunate and ill-defined parameterization used here ($\psi$ is not unique when $\lambda=0$). The usual way to analyze this situation is to rephrase $H_0$ as $\psi\lambda-\lambda=0$ and to test the MLE of this parameter, equal to $y_2-y_1,$ which has a Skellam distribution. $\endgroup$
    – whuber
    Nov 6, 2020 at 16:15


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