# Characteristic Function of Normal Distribution

## Theorem

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

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

## Proof

### Lemma 1

Let:

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

Then:

$\map \phi t = c \dfrac 1 {\sqrt {2 \pi \sigma^2} } \ds \int_{x \mathop \in \R} e^{-\paren {\frac {x - k} {\sqrt 2 \sigma} }^2} \rd x$

### Proof

The characteristic function is defined as

 $\ds \map \phi t$ $=$ $\ds \expect {e^{i t X} }$ where $\expect {\, \cdot \,}$ denotes expectation $\ds$ $=$ $\ds \int_{x \mathop \in \R} e^{i t x} \frac 1 {\sqrt {2 \pi \sigma^2} } e^{-\frac {\paren {x - \mu}^2} {2 \sigma^2} } \rd x$ $\ds$ $=$ $\ds \frac 1 {\sqrt {2 \pi \sigma^2} } \int_{x \mathop \in \R} e^{i t x} e^{-\frac {\paren {x - \mu}^2} {2 \sigma^2} } \rd x$ $\text {(1)}: \quad$ $\ds$ $=$ $\ds \frac 1 {\sqrt {2 \pi \sigma^2} } \int_{x \mathop \in \R} e^{i t x - \frac {\paren {x - \mu}^2} {2 \sigma^2} } \rd x$

Begin by verifying that:

$i t x - \dfrac {\paren {x - \mu}^2} {2 \sigma^2} = -\dfrac {\paren {x - k}^2 + 2 \mu i t \sigma^2 - t^2 \sigma^4} {2 \sigma^2}$

We can then simplify the integral in $(1)$:

 $\ds \int_{x \mathop \in \R} e^{i t x - \frac {\paren {x - \mu}^2} {2\sigma^2} } \rd x$ $=$ $\ds \int_{x \mathop \in \R} e^{-\frac {\paren {x - \mu}^2 + 2 \mu i t \sigma^2 - t^2 \sigma^4} {2 \sigma^2} } \rd x$ $\ds$ $=$ $\ds e^{\frac {2 \mu i t \sigma^2 - t^2 \sigma^4} {2 \sigma^2} } \int_{x \mathop \in \R} e^{-\frac {\paren {x - k}^2} {2 \sigma^2} } \rd x$ $\ds$ $=$ $\ds e^{\mu i t - \frac 1 2 t^2 \sigma^2} \int_{x \mathop \in \R} e^{-\paren {\frac {x - k} {\sqrt 2 \sigma} }^2} \rd x$ $\ds$ $=$ $\ds c \int_{x \mathop \in \R} e^{-\paren {\frac {x - k} {\sqrt 2 \sigma} }^2} \rd x$

$\Box$

### Lemma 2

$\ds \lim_{\alpha \mathop \to \infty} \int_{\frac{-\alpha - \mu} {\sqrt 2 \sigma} - i \frac {t \sigma} {\sqrt 2} }^{\frac {\alpha - \mu} {\sqrt 2 \sigma} - i \frac {t \sigma} {\sqrt 2} } e^{-z^2} \rd z = \sqrt {2 \pi \sigma^2}$

### Proof

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

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

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

 $\ds \oint_{z \mathop \in \Gamma} e^{-z^2} \rd z$ $=$ $\ds \int_{c_1}^{c_2} e^{-z^2} \rd z + \int_{c_2}^{c_3} e^{-z^2} \rd z + \int_{c_3}^{c_4} e^{-z^2} \rd z + \int_{c_4}^{c_1} e^{-z^2} \rd z$ $\ds$ $=$ $\ds 0$

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

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

 $\ds \lim_{\alpha \mathop \to \infty} \int_{c_3}^{c_4} e^{-z^2} \rd z$ $=$ $\ds \lim_{\alpha \mathop \to \infty} \int_{\frac {\alpha - \mu} {\sqrt 2 \sigma} }^{\frac {-\alpha - \mu} {\sqrt 2 \sigma} } e^{-z^2} \rd z$ $\ds$ $=$ $\ds -\lim_{\alpha \mathop \to \infty} \int_{\frac {-\alpha - \mu} {\sqrt 2 \sigma} }^{\frac {\alpha - \mu} {\sqrt 2 \sigma} } e^{ -z^2 } \rd z$ $\ds$ $=$ $\ds -\sqrt {2 \pi \sigma^2}$ Integral of Gaussian Distribution

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

 $\ds \norm {\lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} e^{-z^2} \rd z}$ $=$ $\ds \norm {\lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} e^{-z^2} \rd z}$ $\ds$ $\le$ $\ds \lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} \norm {e^{-z^2} } \rd z$ $\ds$ $\le$ $\ds \lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} \max_{z \mathop \in \closedint {c_2} {c_3} } \norm {e^{-z^2} } \rd z$ $\ds$ $\le$ $\ds \lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} \norm {e^{-\paren {\frac {\alpha - \mu} {\sqrt 2 \sigma} }^2} } \rd z$ $\ds$ $\le$ $\ds \lim_{\alpha \rightarrow \infty} \norm {e^{-\paren {\frac {\alpha - \mu} {\sqrt 2 \sigma} }^2} } \int_{c_2}^{c_3} \rd z$ $\ds$ $=$ $\ds 0$

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

By a similar argument as for between $c_2$ and $c_3$, the integral between $c_4$ and $c_1$ is also $0$.

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

 $\ds \oint_{z \mathop \in \Gamma} e^{-z^2} \rd z$ $=$ $\ds \int_{c_1}^{c_2} e^{-z^2} \rd z + \int_{c_2}^{c_3} e^{-z^2} \rd z + \int_{c_3}^{c_4} e^{-z^2} \rd z + \int_{c_4}^{c_1} e^{-z^2} \rd z$ $\ds$ $=$ $\ds \int_{c_1}^{c_2} e^{-z^2} \rd z + 0 - \sqrt {2 \pi \sigma^2} + 0$ $\ds$ $=$ $\ds 0$

and therefore:

 $\ds \int_{c_1}^{c_2} e^{-z^2} \rd z$ $=$ $\ds \sqrt {2 \pi \sigma^2}$

$\Box$

By Lemma 1:

$\map \phi t = c \dfrac 1 {\sqrt {2 \pi \sigma^2} } \ds \int_{x \mathop \in \R} e^{-\paren {\frac {x - k} {\sqrt 2 \sigma} }^2} \rd x$

Let $z = \paren {\dfrac {x - k} {\sqrt 2 \sigma} }$.

Then:

 $\ds \map \phi t$ $=$ $\ds c \frac 1 {\sqrt {2 \pi \sigma^2} } \int_{\frac {-\alpha - \mu} {\sqrt 2 \sigma} - i \frac {t \sigma} {\sqrt 2} }^{\frac {\alpha - \mu} {\sqrt 2 \sigma} - i \frac {t \sigma} {\sqrt 2} } e^{-z^2} \rd z$ $\ds$ $=$ $\ds c \frac 1 {\sqrt {2 \pi \sigma^2} } \sqrt {2 \pi \sigma^2}$ by Lemma 2 $\ds$ $=$ $\ds c$ $\ds$ $=$ $\ds e^{\mu i t - \frac 1 2 t^2 \sigma^2}$

$\blacksquare$