Confidence interval for Poisson distribution using CLT.

Question

In my lecture notes it is used that if $x_1,...,x_n$ is an i.i.d. sample from a Poisson distribution with parameter $\lambda$ then a confidence interval for $\lambda$ is given as $$\left[ \bar{x}+\frac{q^2}{n}-\frac{q}{\sqrt{n}}\sqrt{\bar{x}-\frac{q^2}{4n}},\bar{x}+\frac{q^2}{n}+\frac{q}{\sqrt{n}}\sqrt{\bar{x}+\frac{q^2}{4n}}\right]$$ where $\bar{x}=\sum_{i=1}^nx_i$ and $q$ is the $1-\alpha/2$ quantile of a standard normal distribution.

So I tried to to show that myself but got stuck. Here is what I got: by CLT for large $n$ we have $$\frac{\bar{x} - \lambda}{\sqrt{\lambda /n}}$$is approximately standard normal. Then using that $P\left(q_{\alpha/2}\le \frac{\bar{x} - \lambda}{\sqrt{\lambda /n}}\le q\right)$ is a approximately $1-\alpha$ and the MLE for $\lambda$ is $\hat{\lambda}=1/\bar{x}$ I get a interval of the form $$\bar{x}\pm q\sqrt{\hat{\lambda}/n}.$$ However I can't seem to arrive at the interval given in the lecture notes and some help would be appreciated.

Answer 1

The point is that we want the interval to be independent of the estimated value.

We can rewrite the event $\left\{q_{\alpha/2}\leqslant \frac{\bar{x} - \lambda}{\sqrt{\lambda /n}}\leqslant q\right\}$ as $\left\{q_{\alpha/2}\sqrt \lambda\leqslant \sqrt n(\bar{x} - \lambda)\leqslant q\sqrt{\lambda}\right\}$.

Solving the inequations $q_{\alpha/2}x\leqslant \sqrt{n}(\overline{x}-x^2)$ and $\sqrt{n}(\overline{x}-x^2)\leqslant qx$ will give that $x$ belongs to some interval; then replace $x$ by $\sqrt \lambda$ to get an interval for $\lambda$.