# An exact and an approximate confidence interval for a Poisson distribution

$$X_{1}, ..., X_{10}$$ ~ $$Pois(\theta)$$

Observations: $$x_{1} = x_{3} = x_{6} = x_{8} = x_{9} = 0$$; $$x_{2} = x_{5} = x_{10} = 1$$; $$x_{4} = 2$$; $$x_{7} = 3$$.

I want to determine an exact (numerical) and an approximate (numerical) confidence interval for $$\theta$$ of confidence level 0.9:

• exact confidence interval: I don't know how to do this. On the internet I can only find things about an approximate confidence interval.

• approximate confidence interval:

$$X = (X_{1}, ..., X_{10})$$

MLE: $$\hat{\theta} = \bar{X} = 0.8$$

And: $$Var(X) = \theta$$. So an estimate for the variance is also $$\bar{X} = 0.8$$.

Then according to the answer: An approximate confidence interval is $$\bar{X}$$ $$\pm$$ $$\sqrt{\bar{X}/n}$$ $$\xi_{1 - \alpha/2}$$.

But this means that: T = $$\sqrt{n}$$ $$\frac{\theta - E[X]}{\sqrt{Var(X)}}$$ ~ N(0,1). But this does not hold for X ~ Pois($$\theta$$), right? Because a Poisson distribution does not have to be symmetric.

For $$n$$ greater enough ($$n=10$$ is borderline but actually enough to get your CI with a Gaussian distribution) you can apply CLT in the following way

$$\frac{\overline{X}_{10}-\theta}{\sqrt{\theta}}\sqrt{10}\sim \Phi$$

In fact this is a pivotal quantity with Standard Gaussian distribution so you have two ways to calculate an appropriate approximate CI

1. (suggested procedure) Estimate the standard deviation of $$\overline{X}_{10}$$ with $$\sqrt{\frac{\overline{X}_{10}}{10}}$$ finding the following CI

$$\Bigg(\overline{X}_{10}-1.64\sqrt{\frac{\overline{X}_{10}}{10}};\overline{X}_{10}+1.64\sqrt{\frac{\overline{X}_{10}}{10}}\Bigg)$$

That is

$$\Bigg(0.8-1.64\sqrt{0.8/10};0.8+1.64\sqrt{0.8/10}\Bigg)$$

$$\Bigg(0.3348;1.2652\Bigg)$$

1. (less common procedure) Soving the following double inequality

$$-1.64<\frac{0.8-\theta}{\sqrt{\theta}}\sqrt{10}<1.64$$

$$0.8+\frac{1.64^2}{20}\pm \sqrt{\Bigg(\frac{1.64^2}{20}\Bigg)^2+0.8\frac{1.64^2}{10}}$$

Exact Confidence interval

It is easy to use the following estimator (that is Complete and sufficient as the sample mean)

$$T=\Sigma_i X_i\sim Po(10\theta)$$

now let's set the two probability

$$0.05=\sum_{t=0}^{8}\frac{e^{-10\theta}(10\theta)^t}{t!}$$

$$0.05=\sum_{t=8}^{\infty}\frac{e^{-10\theta}(10\theta)^t}{t!}$$

In order to solve these two equation w.r.t $$\theta$$ with some attempts (you can start with the approximate bounds found before) you (quite) easy will find that an exact CI for $$\theta$$ at 90% is the following

$$\Big(0.3980;1.4435\Big)$$

Further explanation for the example "in the lecture": finding CI for bernulli with the Statistical Method. They suppose $$n=20$$ and $$\Sigma_i X_i =4$$

Graphically:

Left tail: $$5.1\%$$

Right tail: $$1.6\%$$

Confidence interval %: $$100-1.6-5.1=93.3\%$$

• The exact convidence interval: I do understand that T ~ Pois($10\theta$). But then. Those summations. Why do they hold? First: I think you mean that either one of them is equal to 0.95 instead of 0.05. And then: Why do you sum up to 8? Maybe it has something to do with a test with a critical region? – Laura van Leuven Jan 4 at 14:01
• @LauravanLeuven : it's a standard method of finding confidence interval named "Statistical Method". You can find it, i.e. in "Mood Graybill Boes", Chapter VIII, par 4.2: Statistical Methods (of finding Confidence intervals). It's only a matter of doing nasty calculations...You have to calculate the two tails, left and right, both with probability 5% given you observed $\Sigma_i X_i=8$ – tommik Jan 4 at 14:23
• I understand it know! There was also an example of this in the lecture, which I didn't understand completely. I didn't realize that the bounderies are a function of theta. So thanks! :) – Laura van Leuven Jan 4 at 15:39
• @LauravanLeuven : I added also a further explanation for the example you found in the lecture I mentioned. Hope now it's clear :) – tommik Jan 4 at 15:58

Actually, according to the central limit theorem, it does (in terms of limit distribution). You can check that the poisson distribution checks every requirement : $$X_1,..X_n$$ are independant, iid, $$\bar{X}, \sigma$$ are finite and $$\sigma\neq0$$