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  
\[8=2 \lambda \rightarrow \lambda =4\]
.

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