# Confidence interval for Poisson variables

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.

• Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here. For typesetting equations please use MathJax. – dantopa Feb 27 '19 at 5:33
• I am having trouble understanding your unusual equation for $\bar X.$ The title of the question mentions confidence intervals, but the question itself seems not to make a connection with CIs. Please clarify. – BruceET Feb 27 '19 at 9:19
• Yes, I just edited the question. I wasn't clear enough. I´m trying to find the sequences such that $c_n$ converges in distribution to a standard Gaussian random variable. – Ben C. Feb 27 '19 at 15:39

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 $$\lambda.$$

• Sorry, I wasn't clear enough. I´m trying to find the sequences such that $c_n$ converges in distribution to a standard Gaussian random variable. – Ben C. Feb 27 '19 at 15:40
• Thanks for revising the equation about $\bar X.$ Now you need to think about the connection between the sequences and the 'confidence interval' in the title. Can you show what you have done toward finding the sequences? – BruceET Feb 27 '19 at 17:11