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Let $(X_1,\ldots,X_n)$ denote a random sample from a Poisson distribution with parameter $\lambda$.

Maximum Likelihood Estimate of $\lambda$ is given by

$\hat{\lambda} = \bar{X} = \frac{1}{n} \sum\limits_{i=1}^n X_i$

a) Show that $\frac{\bar{X}-\mu}{\sqrt{\lambda/n}}\sim N(0,1)$ approx.

b) Using this normal approximation find the theoretical confidence interval with the $1-\alpha/2$ quantile from the $N(0,1)$ distribution.


Thoughts: I'm quite stuck on problem a). I don't quite se om the MLE which is equal the the sample mean is relevant here but I am probably (most definitely) missing something. My only idea is to use CLT in some way since

$$\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}=\frac{\bar{X}-\mu}{\sqrt{\lambda/n}}$$

as the variance of Poisson is just $\lambda$. But from here I don't know what to do.... Can someone help me?

And for problem b) I have 0 ideas...

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  • $\begingroup$ The result in part a) should be that $\frac{\sqrt{n}(\overline X-\lambda)}{\sqrt{\lambda}}\stackrel{a}\sim N(0,1)$, which follows from Lindeberg-Lévy CLT. Using this in part b), you can get a confidence interval for $\lambda$ from $\overline X-z_{\alpha/2}\sqrt{\frac{\overline X}{n}}<\lambda<\overline X+z_{\alpha/2}\sqrt{ \frac{\overline X}{n}}$ or consider a variance stabilising transform to find the C.I. $\endgroup$ – StubbornAtom Mar 14 at 19:36
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You have an answer in the comments, so just slightly formalizing it. For (a) you have $n$ i.i.d Poisson random variables, thus a finite second moment, hence by the CLT $$ \sqrt{n}\frac{(\bar{X} - \mathbb{E}X)}{\sqrt{ Var(X) }} \xrightarrow{D} N(0,1), $$ in your case $\mathbb{E}X = Var(X) = \lambda$. Thus for any finite $n$, $$ \sqrt{n}\frac{(\bar{X} - \lambda )}{\sqrt{ \lambda}} \approx N(0,1). $$ For (b), using (a) you know that $$ \mathbb{P}\left(z_{a/2} \le \sqrt{n}\frac{(\bar{X} - \lambda )}{\sqrt{ \lambda}} \le z_{1-a/2} \right) \approx 1 - a $$ re-arranging the inequality you have $$ \mathbb{P}\left( \bar{X} - z_{1-a/2}\sqrt{\lambda/n} \le \lambda \le \bar{X} + z_{1-a/2}\sqrt{\lambda/n} \right) \approx 1 - a, $$ replace the $\lambda$ with its estimator $\bar{X}$ in $\sqrt{\lambda/n}$ and you have the CI.

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