Variance of Gaussian Distribution/Proof 2

Theorem

Let $X \sim N \paren {\mu, \sigma^2}$ for some $\mu \in \R, \sigma \in \R_{> 0}$, where $N$ is the Gaussian distribution.

Then:

$\var X = \sigma^2$

Proof

By Moment Generating Function of Gaussian Distribution, the moment generating function of $X$ is given by:

$\map {M_X} t = \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$

From Variance as Expectation of Square minus Square of Expectation:

$\var X = \expect {X^2} - \paren {\expect X}^2$

From Moment Generating Function of Gaussian Distribution: Second Moment:

$\map { {M_X}''} t = \paren {\sigma^2 + \paren {\mu + \sigma^2 t}^2 } \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$

From Moment in terms of Moment Generating Function, we also have:

$\expect {X^2} = \map { {M_X}''} 0$

Setting $t = 0$, we obtain the second moment:

\(\ds \map {M''_X} 0\)

\(=\)

\(\ds \expect {X^2}\)

\(\ds \)

\(=\)

\(\ds \paren {\sigma^2 + \paren {\mu + \sigma^2 0}^2 } \map \exp {\mu 0 + \dfrac 1 2 \sigma^2 0^2}\)

\(\ds \)

\(=\)

\(\ds \paren {\sigma^2 + \mu^2 } \exp 0\)

\(\ds \)

\(=\)

\(\ds \sigma^2 + \mu^2\)

Exponential of Zero

So:

\(\ds \var X\)

\(=\)

\(\ds \sigma^2 + \mu^2 - \paren {\expect X}^2\)

\(\ds \)

\(=\)

\(\ds \sigma^2 + \mu^2 - \mu^2\)

Expectation of Gaussian Distribution

\(\ds \)

\(=\)

\(\ds \sigma^2\)

$\blacksquare$