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Fishy Problem

Problem

Given $X$ follows a Poisson distribution, with mean $\lambda$, prove that $P(X = \lambda) = P(X = \lambda - 1)$ for all positive integer values of $\lambda$.

For example, if $X \sim Po(5)$ then $P(X = 5) = P(X = 4) = 0.175$ (3 s.f.).


Solution

If $X \sim Po(\lambda)$ then $P(X = r) = \dfrac{e^{-\lambda} \lambda^r}{r!}$.

$\begin{align}\therefore P(X = \lambda) &= \dfrac{e^{-\lambda} \lambda^\lambda}{\lambda !}\\&= \dfrac{e^{-\lambda} \lambda \lambda^{\lambda - 1}}{\lambda (\lambda - 1)!}\\&= \dfrac{e^{-\lambda} \lambda^{\lambda - 1}}{(\lambda - 1)!}\\&= P(X = \lambda - 1)\end{align}$

What if $\lambda$ is non-integer?

Problem ID: 324 (30 May 2007)     Difficulty: 3 Star

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