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.
 A: 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

*

*(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
$$\Bigg(0.8-1.64\sqrt{0.8/10};0.8+1.64\sqrt{0.8/10}\Bigg)$$
$$\Bigg(0.3348;1.2652\Bigg)$$


*(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 $\Sigma_i X_i =4$

Graphically:
Left tail: $5.1\%$

Right tail: $1.6\%$

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