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For 2 independent Poisson processes $X,Y$, with parameters $\lambda_X, \lambda_Y$ respectively, how can I prove that $X+Y$ is a Poisson process with parameter $\lambda_X+\lambda_Y$?

To do this, I suppose you would use the rule $$ P(X+Y = k) = \sum_{n=0}^k P(X = n) \cdot P(Y = k-n) \\ = \sum_{n=0}^k\frac{\lambda_X^n e^{\lambda_X}}{n!} \cdot \frac{\lambda_Y^n e^{\lambda_Y}}{(k-n)!} $$

However, I am unsure of how to continue from here. Can anyone prompt me in the right direction?

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That's a good start

$$\begin{align}\sum_{n=0}^k\frac{\lambda_X^n e^{-\lambda_X}}{n!} \cdot \frac{\lambda_Y^{k-n} e^{-\lambda_Y}}{(k-n)!} &= e^{-\lambda_X-\lambda_Y} \sum_{n=0}^{k}{\lambda_X^n\over n!}{\lambda_Y^{k-n}\over (k-n)!}\\&= {1\over k!}e^{-\lambda_X-\lambda_Y}\sum_{n=0}^{k} {k\choose n}\lambda_X^n\lambda_Y^{k-n}\\&= {e^{-\lambda_X-\lambda_Y}(\lambda_X+\lambda_Y)^k\over k!} \end{align}$$

(Note you forgot minus signs in front of the $\lambda$ in the exponentials)

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  • $\begingroup$ Could you explain the downvote so I can post better answers in the future ? $\endgroup$ – Astyx May 21 '17 at 10:36
  • $\begingroup$ It's not $\binom{n}{k}$... $\endgroup$ – Mesmerized student May 21 '17 at 16:21
  • $\begingroup$ @Mesmerizedstudent Indeed, a typo from my part, thanks ! $\endgroup$ – Astyx May 21 '17 at 16:22

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