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

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


Proof

Lemma 1

Let:

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

Then:

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


Proof

The characteristic function is defined as

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


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 $(1)$:

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

$\Box$


Lemma 2

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


Proof

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

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


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

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

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


Between $c_3$ and $c_4$

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


Between $c_2$ and $c_3$

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


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 $c_2$

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

and therefore:

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

$\Box$


By Lemma 1:

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


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

Then:

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

$\blacksquare$