# Characteristic Function of Normal Distribution

## Theorem

The characteristic function of the normal distribution with mean $\mu$ and variance $\sigma^2$ is

$\phi(t) = e^{ i t \mu - \frac{1}{2} t^2 \sigma^2 }$

## Lemma 1

### Statement

Let

$k = \mu + it\sigma^2$
$c = e^{ \mu it - \frac{1}{2}t^2\sigma^2 }$

Then

 $\displaystyle \phi(t)$ $=$ $\displaystyle c \frac{1}{ \sqrt{2\pi \sigma^2} } \int_{x \in \R} e^{ - \left( \frac{x - k}{\sqrt{2}\sigma} \right)^2 } \d x$ $\quad$ $\quad$

### Proof

The characteristic function is defined as

 $\displaystyle \phi(t)$ $=$ $\displaystyle E[e^{itX}]$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \int_{x \in \R} e^{itx} \frac{1} { \sqrt {2 \pi \sigma^2} } e^{ - \frac { \left( x-\mu \right)^2} {2\sigma^2} } \d x$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac{1} { \sqrt {2 \pi \sigma^2} } \int_{x \in \R} e^{itx} e^{ - \frac { \left( x-\mu \right)^2} {2\sigma^2} } \d x$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac{1} { \sqrt {2 \pi \sigma^2} } \int_{x \in \R} e^{ itx - \frac { \left( x-\mu \right)^2} {2\sigma^2} } \d x$ $\quad$ (1) $\quad$

Begin by verifying that

 $\displaystyle itx - \frac{ \left( x - \mu \right)^2 }{2\sigma^2}$ $=$ $\displaystyle - \frac{\left( x - k \right)^2 + 2\mu it\sigma^2 - t^2\sigma^4 }{2\sigma^2}$ $\quad$ $\quad$

We can then simplify the integral in (1):

 $\displaystyle \int_{x \in \R} e^{ itx - \frac { \left( x-\mu \right)^2} {2\sigma^2} } \d x$ $=$ $\displaystyle \int_{x \in \R} e^{ - \frac {\left( x - k \right)^2 + 2\mu it\sigma^2 - t^2\sigma^4} {2\sigma^2} } \d x$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle e^{ \frac {2\mu it\sigma^2 - t^2\sigma^4 } {2\sigma^2} } \int_{x \in \R} e^{ - \frac{\left( x - k \right)^2 } {2\sigma^2} } \d x$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle e^{ \mu it - \frac{1}{2}t^2\sigma^2 } \int_{x \in \R} e^{ - \left( \frac{x - k}{\sqrt{2}\sigma} \right)^2 } \d x$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle c \int_{x \in \R} e^{ - \left( \frac{x - k}{\sqrt{2}\sigma} \right)^2 } \d x$ $\quad$ $\quad$

Lemma 1 $\blacksquare$

## Lemma 2

### Statement

 $\displaystyle \lim_{\alpha \rightarrow \infty} \int_{\frac{-\alpha-\mu}{\sqrt{2}\sigma} - i\frac{t\sigma}{ \sqrt{2} } } ^{\frac{\alpha-\mu}{\sqrt{2}\sigma} - i\frac{t\sigma}{ \sqrt{2} } } e^{ -z^2 } \d z$ $=$ $\displaystyle \sqrt{2\pi \sigma^2}$ $\quad$ $\quad$

### Proof

Let $\Gamma \subset \C$ be the rectangular contour with corners

$c_1 = \frac{-\alpha-\mu}{\sqrt{2}\sigma} - i\frac{t\sigma}{ \sqrt{2} }$
$c_2 = \frac{\alpha-\mu}{\sqrt{2}\sigma} - i\frac{t\sigma}{ \sqrt{2} }$
$c_3 = \frac{\alpha-\mu}{\sqrt{2}\sigma}$
$c_4 = \frac{-\alpha-\mu}{\sqrt{2}\sigma}$

Since $e^{-z^2}$ is holomorphic everywhere in the region bounded by $\Gamma$, the Cauchy Integral Theorem states that

 $\displaystyle \oint_{z \in \Gamma} e^{-z^2} \d z$ $=$ $\displaystyle \int_{c_1}^{c_2} e^{ -z^2 } \d z + \int_{c_2}^{c_3} e^{ -z^2 } \d z + \int_{c_3}^{c_4} e^{ -z^2 } \d z + \int_{c_4}^{c_1} e^{ -z^2 } \d z$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle 0$ $\quad$ $\quad$

We now evaluate each linear contour integral in the limit as $\alpha$ goes to infinity:

#### Between $c_3$ and $c_4$

 $\displaystyle \lim_{\alpha \rightarrow \infty} \int_{c_3}^{c_4} e^{ -z^2 } \d z$ $=$ $\displaystyle \lim_{\alpha \rightarrow \infty} \int_{\frac{ \alpha-\mu }{ \sqrt{2}\sigma} }^{\frac{ -\alpha-\mu }{ \sqrt{2}\sigma} } e^{ -z^2 } \d z$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle -\lim_{\alpha \rightarrow \infty} \int_{\frac{ -\alpha-\mu }{ \sqrt{2}\sigma} }^{\frac{ \alpha-\mu }{ \sqrt{2}\sigma} } e^{ -z^2 } \d z$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle -\sqrt{2\pi \sigma^2}$ $\quad$ Integral of Gaussian Distribution $\quad$

#### Between $c_2$ and $c_3$

 $\displaystyle \norm{ \lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} e^{ -z^2 } \d z }$ $= \norm{ \lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} e^{ -z^2 } \d z }$ $\displaystyle$ $\quad$ $\quad$ $\displaystyle$ $\le \lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} \norm{ e^{ -z^2 } } \d z$ $\displaystyle$ $\quad$ $\quad$ $\displaystyle$ $\le \lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} \max_{z \in [c_2,c_3]} \norm{ e^{ -z^2 } } \d z$ $\displaystyle$ $\quad$ $\quad$ $\displaystyle$ $\le \lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} \norm{ e^{ - \left( \frac{\alpha-\mu}{\sqrt{2}\sigma} \right)^2 } } \d z$ $\displaystyle$ $\quad$ $\quad$ $\displaystyle$ $\le \lim_{\alpha \rightarrow \infty} \norm{ e^{ - \left( \frac{\alpha-\mu}{\sqrt{2}\sigma} \right)^2 } } \int_{c_2}^{c_3} \d z$ $\displaystyle$ $\quad$ $\quad$ $\displaystyle$ $= 0$ $\displaystyle$ $\quad$ $\quad$

#### Between $c_4$ and $c_1$

By similar argument, the integral between $c_4$ and $c_1$ is also 0.

#### Between $c_1$ and $c_2$

 $\displaystyle \oint_{z \in \Gamma} e^{-z^2} \d z$ $=$ $\displaystyle \int_{c_1}^{c_2} e^{ -z^2 } \d z + \int_{c_2}^{c_3} e^{ -z^2 } \d z + \int_{c_3}^{c_4} e^{ -z^2 } \d z + \int_{c_4}^{c_1} e^{ -z^2 } \d z$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \int_{c_1}^{c_2} e^{ -z^2 } \d z + 0 - \sqrt{2\pi \sigma^2} + 0$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle 0$ $\quad$ $\quad$

and therefore

 $\displaystyle \int_{c_1}^{c_2} e^{ -z^2 } \d z$ $=$ $\displaystyle \sqrt{2\pi \sigma^2}$ $\quad$ $\quad$

Lemma 2 $\blacksquare$

## Proof

By Lemma 1

$\phi(t) = c \frac{1}{ \sqrt{2\pi \sigma^2} } \displaystyle \int_{x \in \R} e^{ - \left( \frac{ x - k } {\sqrt{2}\sigma } \right)^2} \d x$

Let $z = \left( \frac{ x - k } {\sqrt{2}\sigma } \right)$, then

 $\displaystyle \phi(t)$ $=$ $\displaystyle c \frac{1}{ \sqrt{2\pi \sigma^2} } \int_{\frac{-\alpha-\mu}{\sqrt{2}\sigma} - i\frac{t\sigma}{ \sqrt{2} } } ^{\frac{\alpha-\mu}{\sqrt{2}\sigma} - i\frac{t\sigma}{ \sqrt{2} } } e^{ -z^2} \d z$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle c \frac{1}{ \sqrt{2\pi \sigma^2} } \sqrt{2\pi \sigma^2}$ $\quad$ by Lemma 2 $\quad$ $\displaystyle$ $=$ $\displaystyle c$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle e^{ \mu it - \frac{1}{2}t^2\sigma^2 }$ $\quad$ $\quad$

$\blacksquare$