in a poisson-distribution, why is the probability of e-1 the same as e?

Question

I'm not sure if I'm doing it wrong, but I checked the Wiki page, and there seems to be 2 maximum points regardless of input, which the probability of getting the expected value minus 1 the same as getting the expected value. Why is this?

Only if the expectation is an integer. – Did Nov 13 '14 at 19:56 — Did, Nov 13 '14 at 19:56

score 0 · Answer 1 · answered Nov 13 '14 at 19:33

0

We have $$P(X=n) = e^{-\lambda}\frac{\lambda^n}{n!}$$ for some $\lambda>0$.

For $\lambda=1$ this implies $P(X=0)=P(X=1)=\frac{1}{e}$

More generally, if $\lambda=n$, we get $P(X=n)=P(X=n-1)=e^{-n}\frac{n^{n-1}}{(n-1)!}$

answered Nov 13 '14 at 19:33

Peter

84,454

Note, that $\frac{n^n}{n!}=\frac{n^{n-1}n}{(n-1)!n}$ – Peter Nov 13 '14 at 19:36

in a poisson-distribution, why is the probability of e-1 the same as e?

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