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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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