I have been struggling to derive the formulae for CI of Poisson distribution parameters.
Suppose I have a sample $X=(X_1,\dots,X_n)\in\text{Poisson}(\lambda)$. Parameter $\lambda=\theta$ is what I want a CI for.
Given $n,\overline{X}$ as well as the confidence level $\gamma$, I consider the statistic: $$T(X)=n\overline{X}$$ which is a sum of $n$ independent Poisson random variable, thus a $\text{Poisson}(n\theta)$ r.v. Its CDF is $$F_{\theta}(t)=\sum_{k=0}^t\frac{(n\theta)^k}{e^{n\theta}k!}$$ Checking for monotonicity: $$\frac{\partial F_{\theta}(t)}{\partial\theta}=\sum_{k=0}^t\frac{nk(n\theta)^{k-1}e^{n\theta}-n\theta e^{n\theta}(n\theta)^k}{e^{n\theta}k!}=\sum_{k=0}^t\frac{(n\theta)^{k-1}n(k-n\theta)}{e^{n\theta}k!}$$ which is negative for any $k<n\overline{X}$, therefore $F_{\theta}(t)$ is decreasing from 0 to $n\overline{X}$.
The CI is then given by $$(\underline{\theta_n},\overline{\theta_n})=(\theta:\gamma_1<F_{\theta}(T+0),F_{\theta}(T)<\gamma_2)\quad\text{where }\gamma_2-\gamma_1=\gamma$$ where \begin{align*} &F_{\theta}(T)=\sum_{k=0}^{T}\frac{(n\theta)^k}{e^{n\theta}k!}\\ &F_{\theta}(T+0)=\sum_{k=0}^{T+1}\frac{(n\theta)^k}{e^{n\theta}k!}\end{align*} Using the relationship between Poisson and $\chi^2$, write: \begin{align*} &F_{\theta}(T)=1-H_{2(n\overline{X})}(2n\theta)=\gamma_2\\ &F_{\theta}(T+0)=1-H_{2(n\overline{X}+1)}(2n\theta)=\gamma_1 \end{align*} where H is the CDF of the $\chi^2$-distribution with 2T or 2T+2 degrees of freedom respectively.
I understand, I need to show that $n\underline{\theta}=\chi^2_{\frac{1-\gamma}{2}},n\overline{\theta}=\chi^2_{\frac{1+\gamma}{2}}$ for $\chi^2$-distributions with 2T and 2T+2 degrees of freedom respectively, but I don't know how to argue to obtain this result after the transition from Poisson to $\chi^2$.
My question is, how do I proceed to derive the formulae that allow to calculate the CI?
I have seen the threads with solved numeric examples and I did read the articles about multiple ways to evaluate this CI, but none of this addressed my problem above, hence the question.