# Variance of Poisson Distribution/Proof 3

## 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

From Moment Generating Function of Poisson Distribution, the moment generating function of $X$, $M_X$, is given by:

$\map {M_X} t = e^{\lambda \paren {e^t - 1} }$
$\var X = \expect {X^2} - \paren {\expect X}^2$
$\expect {X^2} = \map {M_X''} 0$

In Expectation of Poisson Distribution, it is shown that:

$\map {M_X'} t = \lambda e^t e^{\lambda \paren {e^t - 1} }$

Then:

 $\ds \map {M''_X} t$ $=$ $\ds \map {\frac \d {\d t} } {\lambda e^t e^{\lambda \paren {e^t - 1} } }$ $\ds$ $=$ $\ds \lambda \map {\frac \d {\d t} } {e^{\lambda \paren {e^t - 1} + t} }$ $\ds$ $=$ $\ds \lambda \map {\frac \d {\d t} } {\lambda \paren {e^t - 1} + t} \frac \d {\map \d {\lambda \paren {e^t - 1} + t} } \paren {e^{\lambda \paren {e^t - 1} + t} }$ Chain Rule for Derivatives $\ds$ $=$ $\ds \lambda \paren {\lambda e^t + 1} e^{\lambda \paren {e^t - 1} + t}$ Derivative of Power, Derivative of Exponential Function

Setting $t = 0$:

 $\ds \expect {X^2}$ $=$ $\ds \lambda \paren {\lambda e^0 + 1} e^{\lambda \paren {e^0 - 1} + 0}$ $\ds$ $=$ $\ds \lambda \paren {\lambda + 1}$ Exponential of Zero $\ds$ $=$ $\ds \lambda^2 + \lambda$
$\expect X = \lambda$

So:

$\var X = \lambda^2 + \lambda - \lambda^2 = \lambda$

$\blacksquare$