Characteristic Function of Gaussian Distribution/Lemma 1

Lemma for Characteristic Function of Gaussian Distribution

Let $\map \phi t$ denote the characteristic function of the Gaussian distribution with mean $\mu$ and variance $\sigma^2$.

Let:

$k = \mu + i t \sigma^2$

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

Then:

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

$\ds \map \phi t$

$=$

$\ds \expect {e^{i t X} }$

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 $

$=$

$\ds c \int_{x \mathop \in \R} e^{-\paren {\frac {x - k} {\sqrt 2 \sigma} }^2} \rd x$

$\blacksquare$