Variance of Poisson Distribution

{[refactor|Usual}}

Theorem
$\newcommand{\var}{\operatorname {var}\,}$ Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.

Then the variance of $X$ is given by:
 * $\var \left({X}\right) = \lambda$

Proof 1
From the definition of Variance as Expectation of Square minus Square of Expectation:
 * $\var \left({X}\right) = E \left({X^2}\right) - \left({E \left({X}\right)}\right)^2$

From Expectation of Function of Discrete Random Variable:
 * $\displaystyle E \left({X^2}\right) = \sum_{x \in \Omega_X} x^2 \Pr \left({X = x}\right)$

So:

Then:

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

where $\mu = E \left({x}\right)$ is the expectation of $X$.

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

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

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

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

and hence the result.

Comment
The interesting thing about the Poisson distribution is that its expectation and its variance are both equal to its parameter $\lambda$.