Finding the Standard Deviation of a Poisson Given A Relationship Between Probabilities

Suppose a certain random variable  
  follows a Poisson distribution with unknown standard deviation  
(for a Poisson distribution the variance equal the mean).
The following relationship between probabilities is known:  
\[P(X=x)=\frac{e^{- \lambda} \lambda^x}{x!}\]
  we obtain
\[\frac{e^{- \lambda} \lambda^2}{2!}-\frac{e^{- \lambda} \lambda^1}{1!}=3 \frac{e^{- \lambda} \lambda^0}{0!}\]



\[\lambda = \frac{--2 \pm \sqrt{(-2)^2 - 4 \times 1 \times -6}}{2 \times 1} = \frac{2 \pm \sqrt{28}}{2} = 1 \pm \sqrt{7}\]
The standard deviation is
\[\sqrt{\lambda } = \sqrt{\frac{2 \pm \sqrt{28}}{2} = \sqrt{1 \pm \sqrt{7}}}\]
and since  
\[1 - \sqrt{7} \lt 0\]
  (and so has no real square root) the standard deviation is  
\[\sigma = \sqrt{ \lambda} = \sqrt{1 + \sqrt{7}}\]

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