# Poisson Distribution Problem

I need to derive an approximate $100(1 − α)$ percent confidence interval for $\lambda$ given data $X_1, X_2, \ldots, X_n$ from a Poisson distribution with parameter $\lambda$, I should use the asymptotic normality of the MLE $\widehat{\lambda}$, what do I need is a generic solution for this.

Thanks!

We know that the asymptotic distribution of the mle $\widehat{\theta}$ is described by

$$\sqrt{n} \left( \widehat{\theta}-\theta_0 \right) \xrightarrow{D} N \left(0, \frac{1}{I(\theta_0)} \right)$$

where $\theta_0$ is the true parameter and $I(\theta_0)$ Fisher's information, which is a scalar in this case, evaluated at the true parameter. You do not observe the true parameter but you can estimate it. We know that the mle is consistent and thus if the information is continuous $I(\widehat{\theta}) \xrightarrow{P} I(\theta_0 )$.

That's all you need to derive your asymptotic confidence interval. Let us know if you encounter any difficulties in these steps.

EDIT: Rather than Fisher's information, one may simply use the constistency of the mle and the Central Limit Theorem. The mle of the Poisson distribution is the sample mean which in iid situations has a limiting normal distribution. The variance of this limiting distribution might be computed directly, exploiting the fact that the sum of Poisson random variables is itself Poisson.

• Thanks for the answer! There's another way to do this? I don't think my teacher teach us about Fisher's information! But i'm gonna give it a try! Thanks a lot. – Francisco Magalhaes Jun 2 '15 at 13:18
• @user1674 There is another way in which you use the Central Limit Theorem, see my updated answer. The results are identical. – JohnK Jun 2 '15 at 13:23
• Thank you very much! I think i should use the Central Limit Theorem. Can you explain a little bit more how to use it? I didn't understand the part you say "The mle of the Poisson distribution is the sample mean which in iid situations has a limiting normal distribution. The variance of this limiting distribution might be computed directly, exploiting the fact that the sum of Poisson random variables is itself Poisson." – Francisco Magalhaes Jun 2 '15 at 13:41
• @user1674 We are not supposed to give the answer directly to homework questions, such as this one. I will just say that once you find the mle, you can use the fact that $$\bar{X} \xrightarrow{D} N\left(\mu ,\frac{\sigma^2}{n} \right)$$. – JohnK Jun 2 '15 at 13:43
• How can I use that formula to calculate de confidence interval. I've just calculated the λ^. – Francisco Magalhaes Jun 2 '15 at 23:03