Characteristic Function of Normal Distribution

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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

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

Proof

The characteristic function is defined as

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

Begin by verifying that

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

We can then simplify the integral in (1):

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

Lemma 1 $\blacksquare$

Lemma 2

Statement

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

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

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

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

Between $c_3$ and $c_4$

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

Between $c_2$ and $c_3$

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

Between $c_4$ and $c_1$

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

Between $c_1$ and $c_2$

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

and therefore

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

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

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


$\blacksquare$