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