# Normal approximation of MLE of Poisson distribution and confidence interval

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...

• 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. – StubbornAtom Mar 14 at 19:36

## 1 Answer

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.