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Let $X$ be a Poisson random variable with mean $\lambda$. How do I show that $P[X \geq \lambda] = 1/2$? Also, I was wondering what distributions have this property that the density is concentrated equally around the mean. I see that it's true for Gaussian and Uniform distributions.

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But it's not true. If $\lambda = 1$, then $P[X \ge \lambda] = 1 - P[X=0] \neq 1/2$. – Rahul Oct 24 '10 at 6:01
Not sure if $P(X \geq \lambda) = 1/2$. Take $\lambda = 1$. Then $P(X \geq 1) = 1-e^{-1}$. – Dinesh Oct 24 '10 at 6:04
up vote 2 down vote accepted

P[X ≥ μ] = 1/2 if and only if the mean (μ) and median "coincide". This is true for Gaussian and uniform distribution, but not always true for Poisson.

There is exactly one μ between every unit interval [n, n+1) such that P[X ≥ μ] = 1/2. Some of them are: ln 2, 1.6784…, 2.674…, 3.672…, ...

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Actually this fact was used in bounding the max load when N = Poi(n) balls are thrown into n bins. The proof goes like this: – Naga Oct 25 '10 at 13:13
Let $M$ denote the max load in any bin when $n$ balls are thrown at random (and independently) into $n$ bins. Let $M'$ denote the max load in a bin when $N = Poi(n)$ balls thrown into $n$ bins. The proof goes like this: $P(M > \epsilon) <= P(M' > \epsilon | N >= n) = P(M' > \epsilon, N >= n)/P(N >= n) <= P(M' > \epsilon) / P(N>=n$. Now since $P(N >= n) = 1/2$ (as $n = E[N]$), we get the final bound as $2P(M'>\epsilon)$. – Naga Oct 25 '10 at 13:20
@Naga: Obviously, that proof is not correct for small $n$. The larger $n$ is, though, the closer $P(N>n)$ gets to 1/2. Are you maybe only interested in asymptotics? – Jens Oct 26 '10 at 15:18

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