Confidence interval for Poisson variables

Question

Let $X_{i},...,X_{n}$ be i.i.d. Poisson random variables with parameter $\lambda>0$

I have:

$$\bar{X}={(1/n)\sum_{i=1}^n X_i}.$$

Find two sequences $(a_n)_{n>=1}$ and $(b_n)_{n>=1}$ such that $a_n(\bar{X}-b_n)$ converges in distribution to a standard Gaussian random variable $Z$~$N(0,1)$

I'm trying to find the values for the sequence $(a_n)_{n>=1}$ and $(b_n)_{n>=1}$ such that:

$$c_n=a_n*({(1/n)\sum_{i=1}^n X_i} + b_n)$$ converges in distribution to a standard Gaussian random variable.

Answer 1

Without explicitly working your exact problem, I will illustrate a few methods of finding a 95% CI for Poisson $\lambda.$

There is much literature on this general topic, with perhaps as many as a dozen reasonable types if CIs used in various applications. For large $\lambda$ (say above 30 or so) most of the methods give roughly the same numerical results. I suppose others may want to show additional types of Poisson CIs.

Wald asymptotic interval. In order to test $H_0: \lambda = \lambda_0$ against $H_a: \lambda \ne \lambda_0,$ based on $T$ Poisson events, one can use the test statistic $$Z = \frac{T - \lambda_0}{\sqrt{\lambda_0}} \stackrel{aprx}{\sim} \mathsf{Norm}(0,1).$$

Inverting this test in a naive way, one obtains the 95% CI $$T \pm 1.96\sqrt{T}.$$ The idea is that values of $\lambda_0$ that would not lead to rejection for a given $T$ at the 5% level of significance form a 95% CI for $\lambda.$

Two approximations involved here, both of which become reasonably good for large $T.$ One is that $\frac{T - \lambda_0}{\sqrt{\lambda_0}}$ is approximately standard normal; the other is that $\sqrt{\lambda_0}$ in the denominator is well-approximated by $T.$

"Agresti-Style" CI. If we solve the quadratic inequality for a somewhat more accurate inversion of the test, and conflate 1.96 with 2, we get the CI $$T +2 \pm 1.96\sqrt{T+1}.$$ This interval is somewhat in the spirit of the well-known "plus-Four" or "Agresti-Coull" Ci for the binomial success probability. [This type of interval uses only the asymptotic normality of a Poisson distribution with increaseing $\lambda.]$

Bayesian flat-prior interval estimate used as a CI. This interval uses the posterior distribution $\mathsf{Gamma}(.5 + t, 1)$ and cuts 2.5% of the probability from each tail. [Philosophical interpretations of Bayesian posterior probability intervals and frequentist confidence intervals differ, but many frequentists are willing to use numerical results of Bayesian methods in frequentist estimation.]

Examples: Suppose $T = 30$ Poisson events are observed in a domain (e.g., time period, area, or volume) and we want to estimate the Poisson rate $\lambda$ for that domain. [I choose $T = 20$ because it is small enough to give noticeably different results for the three intervals; for $T=50,$ the three results are much alike.]

The three types of CIs are computed as follows:

Wald: $30 \pm 1.96\sqrt{20}$ or $(11.24, 28.77).$

Agresti: $22 \pm 1.96\sqrt{21}$ or $(13.02, 30.98).$ A grid search for the exact inversion of the test gives the interval $(12.95, 30.89).$

lam = seq(5, 50, by=.0001)
z = (20 - lam)/sqrt(lam)
ci = lam[z > -1.96 & z < 1.96]
min(ci); max(ci)
[1] 12.9475
[1] 30.8941

Bayesian: $(12.61, 30.29)$

qgamma(c(.025, .975), 20.5, 1)
[1] 12.60726 30.28029

If $n$ observations are taken from the domain and their average is $T/n,$ simply divide the interval endpoints by $n.$

Actual coverage probability. Because of the discrete nature of Poisson counts and the use of approximations, alleged "95%" confidence intervals do not have exactly 95% coverage for all values of $\lambda.$ The "Agresti" style has very nearly the intended coverage for $\lambda$ sufficiently large. For this style of 95% confidence intervals, the graph below shows actual coverage for 2000 values of $\lambda \in [4,75].$

By contrast, the Wald-style CIs have less than the intended coverage probability for many values of $\lambda.$