Expectation 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:

$\expect X = \mu$

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 Moment in terms of Moment Generating Function:

$\expect X = \map {M'_X} 0$

We have:

\(\ds \map {M_X'} t\)

\(=\)

\(\ds \frac \d {\d t} \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}\)

\(\ds \)

\(=\)

\(\ds \map {\frac \d {\d t} } {\mu t + \frac 1 2 \sigma^2 t^2} \frac \d {\map \d {\mu t + \dfrac 1 2 \sigma^2 t^2} } \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}\)

Chain Rule for Derivatives

\(\ds \)

\(=\)

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

Derivative of Power, Derivative of Exponential Function

Setting $t = 0$:

\(\ds \map {M_X'} 0\)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \mu \exp 0\)

\(\ds \)

\(=\)

\(\ds \mu\)

Exponential of Zero

$\blacksquare$