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

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.

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2 Answers 2

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Let's start with the approximate CI:

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

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

Leading to the following CI

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

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 $\Sigma_i X_i =4$

enter image description here

Graphically:

Left tail: $5.1\%$

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Right tail: $1.6\%$

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

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  • $\begingroup$ 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? $\endgroup$
    – Laura van Leuven
    Jan 4, 2021 at 14:01
  • $\begingroup$ @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 $\Sigma_i X_i=8$ $\endgroup$
    – tommik
    Jan 4, 2021 at 14:23
  • $\begingroup$ 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! :) $\endgroup$
    – Laura van Leuven
    Jan 4, 2021 at 15:39
  • $\begingroup$ @LauravanLeuven : I added also a further explanation for the example you found in the lecture I mentioned. Hope now it's clear :) $\endgroup$
    – tommik
    Jan 4, 2021 at 15:58
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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 $

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