Ask linking the Poisson and gamma families

Question

Let $X_1,..,X_n$ be a random sample from a Poisson population with parameter $\lambda$ and define $Y=\Sigma X_i$. Y is sufficient for $\lambda$ and $Y \sim Poisson(n\lambda)$. Now $Y=y_0$ is observed, one equation is

$$\Sigma_{k=0}^{y_0} e^{-n\lambda} \frac{(n\lambda)^k}{k!}=\frac{\alpha}{2}$$

Recall the identity linking the Poisson and gamma families: If X is a gamma($\alpha, \beta$) random variable, where $\alpha$ is an integer, then for any x, P(X $\le$x)=P(Y$\ge \alpha$), where $Y \sim Poisson(x/\beta)$.

We can write (remembering that $y_0$ is the observed value of Y): $$\frac{\alpha}{2}=\Sigma_{k=0}^{y_0} e^{-n\lambda} \frac{(n\lambda)^k}{k!}= P(Y \le y_0|\lambda)= P(\chi^2_{2(y_0+1)}>2n\lambda)$$.

I didn't get the above equality. My attempt is: $$\frac{\alpha}{2}=\Sigma_{k=0}^{y_0} e^{-n\lambda} \frac{(n\lambda)^k}{k!}= P(Y \le y_0|\lambda)=1-P(Y \ge y_0+1|\lambda)= 1-P(X \le x) =P(X>x)$$, where x is gamma($y_0+1,\frac{x}{n\lambda}$). Because I think $x/\beta = n\lambda$. I don't know how to do next. And I don't know my attempt is correct or not.

I also know the identity that $\chi^2_v=gamma(v/2,2)$

Background of this question:

When you say "one equation is" what do you mean? Is that a given? A definition of $\alpha$? Something else? Also, I think maybe you have two different meanings for $\alpha$ is this problem? — RobertTheTutor, Apr 18 '21 at 19:27
One way of showing this: https://math.stackexchange.com/q/2427795/321264. — StubbornAtom, Apr 18 '21 at 19:42
Does this answer your question? Proving the Gamma-Poisson relationship using induction — Jean Marie, Apr 18 '21 at 21:12
In fact, the text "Does this answer your question" is generated automatically. It comes from a remark of mine that this question has already been asked, therefore doesn't need a new answer, but the answer had already been done... — Jean Marie, Apr 18 '21 at 21:45

score 1 · Accepted Answer · answered Apr 18 '21 at 20:06

$\frac{x}{\beta} = n\lambda$. Then you have taken $\beta = \frac{x}{n\lambda}$.

Instead, note that x is a free variable that you want to set as the parameter in the probability.

Therefore if you set $\beta = 2$, then you obtain, $x = 2n\lambda$. The rest of the working looks right.

\begin{align*} \mathbb P(X>x) &= \mathbb P\left(X\sim \Gamma({y_0+1},2)>2n\lambda\right)\\ &= \mathbb P\left(X\sim \Gamma(\frac{2(y_0+1)}{2},2)>2n\lambda\right)\\ &= \mathbb P\left(X\sim \chi^2_{2(y_0+1)}>2n\lambda\right)\\ \end{align*}

Ask linking the Poisson and gamma families

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