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I have a poisson random Variable $X$ with parameter $\lambda$. I need to find the value of $P(X \leq \lambda)$.

I know that, this is given by

$$P(X \leq \lambda) = e^{-\lambda} \sum_{x=0}^{\lambda}{\frac{\lambda^x}{x!}}$$

How to calculate further? I found an answer which says this value is 0.5, but it calculates the value numerically using a R script. Is it possible to calculate this value directly? Thanks!

Angina Seng
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Nagabhushan S N
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    That answer only says that the value converges to $0.5$ as $\lambda$ tends to infinity. There is probably no simple, exact formula for this sum. – Jair Taylor Oct 28 '18 at 05:40
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    The CDF of a Poisson distribution can be expressed in terms of the incomplete Gamma function: $$P(X\le k)=\frac{1}{\Gamma(k+1)}\int_{\lambda}^\infty e^{-t}t^k,dt$$ – StubbornAtom Oct 28 '18 at 08:02

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That formula is valid when $\lambda$ is an integer. In general $$P(X\le\lambda)=e^{-\lambda}\sum_{n=0}^{\lfloor\lambda\rfloor}{\frac{\lambda^n}{n!}}$$ where $\lfloor\lambda\rfloor$ is the integer part of $\lambda$. This will not equal $\frac12$ in general, but for large $\lambda$ will be approximately $\frac12$.

Angina Seng
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