Mean of a Poisson Distribution From Two Relative Probabilities
Suppose we have good reason to suspect that a particular random variable\[X\]
follows a Poisson distribution, so that \[Z \sim Po( \lambda )\]
, where \[\lambda\]
is the relevant parameter, typically a frequency rate. Then we can find \[\lambda\]
given the relative probabilities of observations of \[X\]
.For a Poisson distribution,
\[P(X=x)=\frac{\lambda^x e^{-\lambda}}{x!}\]
.Suppose that
\[P(X=7)=2P(X=8)\]
.Then
\[\frac{\lambda^7 e^{-\lambda}}{7!}=2 \frac{\lambda^8 e^{-\lambda}}{8!} \]
.Divide by
\[e^{- \lambda}\]
to give \[\frac{\lambda^7}{7!}=2 \frac{\lambda^8 }{8!} \]
.Multiply,y both sides by
\[8!\]
and divide by \[\lambda^7\]
. Then \[8=2 \lambda \rightarrow \lambda =4\]
.