Variance of Poisson Distribution/Proof 2

Proof 2
From Variance of Discrete Random Variable from PGF, we have:
 * $\var X = \map {\Pi''_X} 1 + \mu - \mu^2$

where $\mu = \expect X$ is the expectation of $X$.

From the Probability Generating Function of Poisson Distribution, we have:
 * $\map {\Pi_X} s = e^{-\lambda \paren {1 - s} }$

From Expectation of Poisson Distribution, we have:
 * $\mu = \lambda$

From Derivatives of PGF of Poisson Distribution, we have:
 * $\map {\Pi''_X} s = \lambda^2 e^{-\lambda \paren {1 - s} }$

Putting $s = 1$ using the formula $\map {\Pi''_X} 1 + \mu - \mu^2$:
 * $\var X = \lambda^2 e^{-\lambda \paren {1 - 1} } + \lambda - \lambda^2$

and hence the result.