Variance of Poisson Distribution/Proof 2

Theorem

Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.

Then the variance of $X$ is given by:

$\var X = \lambda$

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.

$\blacksquare$

Sources

• 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 4.3$: Moments: Exercise $5$