\[X\]
follows a Poisson distribution with unknown standard deviation \[\sigma\]
(for a Poisson distribution the variance equal the mean).The following relationship between probabilities is known:
\[P(X=2)-P(X=1)=3P(X=0)\]
.Using
\[P(X=x)=\frac{e^{- \lambda} \lambda^x}{x!}\]
we obtain\[\frac{e^{- \lambda} \lambda^2}{2!}-\frac{e^{- \lambda} \lambda^1}{1!}=3 \frac{e^{- \lambda} \lambda^0}{0!}\]
\[\frac{\lambda^2}{2}-\lambda=3\]
\[\lambda^2-2\lambda-6=0\]
Hence
\[\lambda = \frac{--2 \pm \sqrt{(-2)^2 - 4 \times 1 \times -6}}{2 \times 1} = \frac{2 \pm \sqrt{28}}{2} = 1 \pm \sqrt{7}\]
.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 \[\sigma = \sqrt{ \lambda} = \sqrt{1 + \sqrt{7}}\]
.