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Let $X_1, \dots, X_n$ be a random sample from a Poisson population with parameter $\lambda$ and define $Y = \sum_i X_i$. Y is sufficient for $\lambda$ and $Y \text{~} \text{Poisson}(n \lambda)$.

Identity: If $X \text{~ gamma($\alpha, \beta$)}$ for an integer $\alpha$, then for any $x$, $P(X \le x) = P(Y \ge \alpha)$ where $Y \text{~ Poission}(x/\beta)$.

By the identity above, $P(Y \le y_0 | \lambda) = P(\chi^2_{2(y_0+1)} \gt 2n\lambda)$.

Can someone explain how the identity above shows this? My thoughts were to put it into this form:

$P(X \le x) = P(Y \ge \alpha) \Rightarrow 1-P(X \le x) = 1-P(Y \ge \alpha)$

$\Rightarrow P(X \gt x) = P(Y \lt \alpha) \Rightarrow P(X \ge x) = P(Y \lt \alpha)$

But $P(Y \lt \alpha) \ne P(Y \le \alpha)$.

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Oliver G
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  • You may want to see this. – Lee David Chung Lin Jun 28 '19 at 19:58
  • Can you elaborate? I'm having trouble finding the answer to this question in the linked question. – Oliver G Jun 30 '19 at 12:59
  • See the 2nd-to-last line the of question statement: $\displaystyle F(t) =1 - \sum_{i=0}^{\alpha-1}{\frac {(\lambda t)^i} {i!}}e^{-\lambda t}$. The left hand side is exactly your $P(X \leq x)$, and the right hand side is exactly your $P(Y \geq \alpha)$. Note that for the Gamma distribution, your scale parameter $\beta$ is the inverse of the underlying rate of Poisson process $\displaystyle\frac1{\beta} = \lambda$. Your question is exactly the same as his. – Lee David Chung Lin Jun 30 '19 at 22:58
  • https://math.stackexchange.com/q/2427795/321264, https://math.stackexchange.com/q/4107340/321264 – StubbornAtom Apr 19 '21 at 14:20

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