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What is $\Pr\left( \sum_{i=1}^n X_i \le c \,\middle|\, \lambda = \lambda_0 \right) $ when $X_i \sim \text{Poisson}(\lambda)$?

Is it right to say $X_1+\dots+X_n\sim\text{Poisson}(n\lambda_0)$ then $\Pr\left( \sum_{i=1}^n X_i \le c \,\middle|\, \lambda = \lambda_0 \right)=\sum_{k=1}^cn^k\lambda_0^k\frac{e^{-n\lambda_0}}{k!}$?

Cure
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  • The $X_i$ need to be [conditionally] independent [given $\lambda = \lambda_0$]. – angryavian Oct 20 '17 at 05:52
  • @angryavian Could you explain further? I took $\lambda=\lambda_0$ to mean $X_i=\sum \lambda_0^k\frac{e^{-\lambda_0}}{k!}$. – Cure Oct 20 '17 at 05:56
  • If $X_i \sim \text{Poisson}(\lambda_0)$ are **independent**, then $X_1 + \cdots + X_n \sim \text{Poisson}(n \lambda_0)$. Also your comment does not make sense; did you try to write the CDF of $X_i$? – angryavian Oct 20 '17 at 06:00
  • @angryavian Yes,isn't it the CDF? I was considering the accepted answer for that question: https://math.stackexchange.com/questions/221078/poisson-distribution-of-sum-of-two-random-independent-variables-x-y If $\mu=\lambda=\lambda_0$ then $\text{Pr}(X_1+X_2=c)=\frac{(2\lambda_0)^c}{c!}e^{-2\lambda_0}$. In my case $\text{Pr}(X_1+\dots+X_n=c|\lambda=\lambda_0)=\frac{(n\lambda_0)^c}{c!}e^{-n\lambda_0}$ then $\text{Pr}(X_1+\dots+X_n\leq c|\lambda=\lambda_0)=\sum_{k=1}^c\frac{(n\lambda_0)^k}{k!}e^{-n\lambda_0}$. – Cure Oct 20 '17 at 06:33
  • The point of my comments is to emphasize that you need an **independence** assumption (which you have not explicitly mentioned) in order to use that result. But if you do have independence, then what you have written in your original question is fine. – angryavian Oct 20 '17 at 06:44
  • What is this conditioning on the parameter $\lambda$? Is $\lambda$ a RV too in this example? – Nap D. Lover Oct 20 '17 at 11:06

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