Confidence interval for theta for sum of poisson random variables

Question

Assuming you are summing 10 poisson variables, and the sum of their results is 60, how would you approximate a 95% CI for theta?

I would start by figuring out the sample mean, but then I'm kinda stuck. Thanks!

Answer 1

The sample mean would give you a point estimate for the rate parameter. What you need in order to calculate a confidence interval is a variance of the sampling distribution.

Formally, if $\boldsymbol x = (x_1, \ldots, x_n)$ is an IID sample drawn from a Poisson distribution $$X \sim \operatorname{Poisson}(\theta), \quad \Pr[X = x] = e^{-\theta} \frac{\theta^x}{x!} \, \quad x = 0, 1, 2, \ldots,$$ and we wish to find a 95% CI for $\theta$, we can choose the point estimate based on the method of maximum likelihood: $$\hat \theta_{\text{MLE}} = \bar x,$$ which is a result I leave as an exercise. Then we would calculate the variance of this estimator: $$\operatorname{Var}[\hat\theta_{\text{MLE}}] = \operatorname{Var}\left[\frac{1}{n}\sum_{i=1}^n X_i\right] \overset{\text{iid}}{=} \frac{\theta}{n},$$ hence the estimated variance of the estimator is $$\widehat{\operatorname{Var}}[\hat \theta_{\text{MLE}}] = \frac{\bar x}{n},$$ and an asymptotic 2-sided $100(1-\alpha)\%$ Wald-type confidence interval for $\theta$ would then be $$\bar x \pm z_{\alpha/2}^* \sqrt{\frac{\bar x}{n}},$$ where $z_{\alpha/2}^*$ is the upper $\alpha/2$ quantile of the standard normal distribution. Of course, this is not the only confidence interval one can construct; this one, as pointed out above, is approximate and based on the asymptotic distribution of $\hat \theta$.

The exact same question was asked here on stats.SE: How to calculate a confidence level for a Poisson distribution? In particular, it is worthwhile to read the paper that was referenced in one of the answers, at https://www.ine.pt/revstat/pdf/rs120203.pdf