# What is the UMA 1-alpha confidence interval around lambda given sample iid Poisson(lambda)?

From Casella and Berger (2002), exercise 9.44a asks:

Let $X_1, ..., X_n$ be iid Poisson($\lambda$). Find a UMA 1-$\alpha$ confidence interval based on inverting the UMP level $\alpha$ test of $H_0:\lambda=\lambda_0$ versus $H_1:\lambda>\lambda_0$.

What I have so far:

By the Karlin-Rubin Theorem, $T>t_0$ where $T=\sum(X_i)\sim Poisson(n\lambda )$ is the UMP level $\alpha$ test. For level $a$, $t_0 = Poisson_{n\lambda, \alpha}$. Then, by the Gamma-Poisson relationship, I have that

$$\alpha \ge \left(P(T\ge t_0+1)=P(U \ge 2n\lambda) \right)$$

where $U\sim \chi^2(2t_0 + 2)$.

Inverting the test, I eventually get to

$$CI=\left\{\lambda:0<\lambda<{U\over2n}\right\}.$$

And I know the answer is

$$CI=\left\{\lambda:0<\lambda<{{\chi^2_{2\sum{x_i} + 2,\alpha}}\over2n}\right\}.$$

With simulations, I can show that $U = {\chi^2_{2\sum{x_i} + 2,\alpha}}$. But how do I prove this analytically?