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I've been given a Poisson distribution for a question sheet and I'm trying to find the mean of the distribution. The solution reads:

$$ \langle x \rangle=\sum_{n=0}^\infty n \frac{a^n}{n!}e^{-a}=e^{-a}\sum_{n=1}^\infty \frac{a^n}{(n-1)!}=e^{-a}a\sum_{n=1}^\infty \frac{a^{n-1}}{(n-1)!}= e^{-a}ae^{a}=a$$

Where $a$ is a real constant. I'm struggling to understand why the lower boundary of the sum has changed during the second step. In the question we are also given the series expansion of the exponential function as:

$$e^x = \sum_{n=0}^\infty \frac{x^n}{n!} $$

Any help would be very much appreciated thanks a lot in advance!

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    $\begingroup$ It is because in the first term, $n=0$ hence it dissapears ($0 \times \text{something}=0) . Also note that the "something" is well defined since $0 \neq 0! =1$. $\endgroup$ Jun 14, 2017 at 11:56

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It is because \begin{align*}\sum_{n=0}^{\infty} n \frac{a^n}{n!} e^{-a} & = e^{-a} \sum_{n=0}^{\infty} n \frac{a^n}{n!} \\ & = e^{-a} \left( 0 \cdot \frac{a^0}{0!} + \sum_{n=1}^{\infty} n \frac{a^n}{n!}\right) \\ & = e^{-a} \sum_{n=1}^{\infty} n \frac{a^n}{n!} \\ & = e^{-a} \sum_{n=1}^{\infty} \frac{a^n}{(n-1)!} \end{align*}

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