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The PDF of a Poisson distribution is $P(X=k)=\frac{\lambda^k}{k!}e^\lambda$. As a PDF, it should sum to one; i.e. $\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}e^\lambda=1$. I'm having trouble proving this though.

From the Taylor series for $e$, I can see that $\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}<\sum_{k=0}^{\infty}\frac{1}{k!}=e$ so it must be less than $e^{\lambda+1}$, but that's the tightest bound I've been able to get.

Any help?

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You're missing a minus sign in the exponent. The series sums to $\mathrm e^\lambda$, and the factor in the distribution is $\mathrm e^{-\lambda}$. –  joriki Jan 27 '13 at 20:08

1 Answer 1

up vote 4 down vote accepted

The Maclaurin series for $e^x$ is $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$.

It converges to $e^x$ for all $x$.

Remark: For a "nice" function $f(x)$, the coefficient of $x^n$ in the Maclaurin series expansion of $f(x)$ is $\frac{f^{(n)}(0)}{n!}$, where $f^{(n)}(x)$ is the $n$-th derivative of $f(x)$. In the case $f(x)=e^x$, all the derivatives are $e^x$, so the coefficient of $x^n$ is $\frac{1}{n!}$.

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