Exact Confidence Interval for Poisson using Gamma-Poisson Relationship

I'm reading Casella-Berger's Statistical Inference and trying to follow along in example 9.2.15, which constructs an exact confidence interval for a Poisson rate. In this example, the authors solve the sum

$$\frac{\alpha}{2}=\sum_{k=0}^{y_0} \frac{e^{-n\lambda}(n\lambda)^k}{k!}$$ by using a link between the gamma and Poisson distributions. Namely, they use the fact $$P(Y\geq \alpha)=P(X\leq x)$$ for $Y\sim Poisson(\frac{x}{\beta})$ and $X\sim Gamma(\alpha, \beta)$. Additionally, it helps to note $Gamma(p/2,\beta=2)$ is a $\chi^2$ random variable with $p$ degrees of freedom.

They then conclude $$\sum_{k=0}^{y_0} \frac{e^{-n\lambda}(n\lambda)^k}{k!}=P(Y\leq y_0)=P(\chi^2_{2(y_0+1)}>2n\lambda).$$ My question is why this is true. Using the link between the Poisson and Gamma mentioned above, I keep finding the degrees of freedom of the $\chi^2$ to be $2y_0$ and not $2(y_0+1)$. I feel like this might just be a simple calculation mistake, but I'm not seeing it.

Given \begin{equation} X \sim {{\chi}_{2n}}^2 \end{equation} and \begin{equation} Y \sim P(\lambda) \end{equation} Let's prove (the complement of your expression) \begin{equation} P(X < 2\lambda) = P(Y \ge n) \end{equation} Notice that $X \sim \chi_{2n}^{2}=\Gamma(n,2)$. In addition, \begin{equation} P(X<2\lambda)=\frac{1}{\Gamma(n)2^n} \displaystyle\int_{0}^{2\lambda}x^{n-1}e^{-\frac{x}{2}}dx \end{equation} and \begin{equation} P(Y\geq n)=1-P(Y \leq n-1)=1-\displaystyle\sum_{i=0}^{n-1}\frac{e^{-\lambda}\lambda^i}{i!} \end{equation} Do the derivative on both sides with respect to $\lambda$ \begin{equation} \frac{\partial}{\partial \lambda}P(X<2\lambda) = \displaystyle\frac{2}{\Gamma(n)2^n}(2\lambda)^{n-1}e^{-\lambda}=\frac{\lambda^{n-1}e^{-\lambda}}{(n-1)!} \end{equation} and \begin{equation} \frac{\partial}{\partial \lambda} P(Y \ge n) = \displaystyle\sum_{i=0}^{n-1}\frac{e^{-\lambda}\lambda^i}{i!}-\displaystyle\sum_{i=1}^{n-1}\frac{e^{-\lambda}\lambda^{i-1}}{(i-1)!}=\frac{\lambda^{n-1}e^{-\lambda}}{(n-1)!} \end{equation} Notice that the derivatives are the same, hence \begin{equation} P(X<2\lambda) = P(Y \ge n) + C \end{equation} where $C$ is a constant. Plug $\lambda = 0$, you'll get $C =0$. This means that ( by inverting both inequalities) \begin{equation} P(X>2\lambda) = P(Y < n) \end{equation} or \begin{equation} P(X>2\lambda) = P(Y \leq n-1) \end{equation} You've got an $n \lambda$ instead of my $\lambda$ and a $y_0+1$ instead of my $n$
• Perhaps this should be edited to have $G(n,2)$ instead of $\Gamma(n,2)$ that way the usual notation for the function $\Gamma(\cdot)$ isn't confused with the Gamma distribution. Just a suggestion. Sep 13 '18 at 19:39