Characteristic Function of Normal Distribution
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Contents
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
| \(\displaystyle \phi(t)\) | \(=\) | \(\displaystyle c \frac{1}{ \sqrt{2\pi \sigma^2} } \int_{x \in \R} e^{ - \left( \frac{x - k}{\sqrt{2}\sigma} \right)^2 } \d x\) | $\quad$ | $\quad$ |
Proof
The characteristic function is defined as
| \(\displaystyle \phi(t)\) | \(=\) | \(\displaystyle E[e^{itX}]\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \int_{x \in \R} e^{itx} \frac{1} { \sqrt {2 \pi \sigma^2} } e^{ - \frac { \left( x-\mu \right)^2} {2\sigma^2} } \d x\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac{1} { \sqrt {2 \pi \sigma^2} } \int_{x \in \R} e^{itx} e^{ - \frac { \left( x-\mu \right)^2} {2\sigma^2} } \d x\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac{1} { \sqrt {2 \pi \sigma^2} } \int_{x \in \R} e^{ itx - \frac { \left( x-\mu \right)^2} {2\sigma^2} } \d x\) | $\quad$ (1) | $\quad$ |
Begin by verifying that
| \(\displaystyle itx - \frac{ \left( x - \mu \right)^2 }{2\sigma^2}\) | \(=\) | \(\displaystyle - \frac{\left( x - k \right)^2 + 2\mu it\sigma^2 - t^2\sigma^4 }{2\sigma^2}\) | $\quad$ | $\quad$ |
We can then simplify the integral in (1):
| \(\displaystyle \int_{x \in \R} e^{ itx - \frac { \left( x-\mu \right)^2} {2\sigma^2} } \d x\) | \(=\) | \(\displaystyle \int_{x \in \R} e^{ - \frac {\left( x - k \right)^2 + 2\mu it\sigma^2 - t^2\sigma^4} {2\sigma^2} } \d x\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle e^{ \frac {2\mu it\sigma^2 - t^2\sigma^4 } {2\sigma^2} } \int_{x \in \R} e^{ - \frac{\left( x - k \right)^2 } {2\sigma^2} } \d x\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle e^{ \mu it - \frac{1}{2}t^2\sigma^2 } \int_{x \in \R} e^{ - \left( \frac{x - k}{\sqrt{2}\sigma} \right)^2 } \d x\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle c \int_{x \in \R} e^{ - \left( \frac{x - k}{\sqrt{2}\sigma} \right)^2 } \d x\) | $\quad$ | $\quad$ |
Lemma 1 $\blacksquare$
Lemma 2
Statement
| \(\displaystyle \lim_{\alpha \rightarrow \infty} \int_{\frac{-\alpha-\mu}{\sqrt{2}\sigma} - i\frac{t\sigma}{ \sqrt{2} } } ^{\frac{\alpha-\mu}{\sqrt{2}\sigma} - i\frac{t\sigma}{ \sqrt{2} } } e^{ -z^2 } \d z\) | \(=\) | \(\displaystyle \sqrt{2\pi \sigma^2}\) | $\quad$ | $\quad$ |
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
| \(\displaystyle \oint_{z \in \Gamma} e^{-z^2} \d z\) | \(=\) | \(\displaystyle \int_{c_1}^{c_2} e^{ -z^2 } \d z + \int_{c_2}^{c_3} e^{ -z^2 } \d z + \int_{c_3}^{c_4} e^{ -z^2 } \d z + \int_{c_4}^{c_1} e^{ -z^2 } \d z\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle 0\) | $\quad$ | $\quad$ |
We now evaluate each linear contour integral in the limit as $\alpha$ goes to infinity:
Between $c_3$ and $c_4$
| \(\displaystyle \lim_{\alpha \rightarrow \infty} \int_{c_3}^{c_4} e^{ -z^2 } \d z\) | \(=\) | \(\displaystyle \lim_{\alpha \rightarrow \infty} \int_{\frac{ \alpha-\mu }{ \sqrt{2}\sigma} }^{\frac{ -\alpha-\mu }{ \sqrt{2}\sigma} } e^{ -z^2 } \d z\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle -\lim_{\alpha \rightarrow \infty} \int_{\frac{ -\alpha-\mu }{ \sqrt{2}\sigma} }^{\frac{ \alpha-\mu }{ \sqrt{2}\sigma} } e^{ -z^2 } \d z\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle -\sqrt{2\pi \sigma^2}\) | $\quad$ Integral of Gaussian Distribution | $\quad$ |
Between $c_2$ and $c_3$
| \(\displaystyle \norm{ \lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} e^{ -z^2 } \d z }\) | \(= \norm{ \lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} e^{ -z^2 } \d z }\) | \(\displaystyle \) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(\le \lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} \norm{ e^{ -z^2 } } \d z\) | \(\displaystyle \) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(\le \lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} \max_{z \in [c_2,c_3]} \norm{ e^{ -z^2 } } \d z\) | \(\displaystyle \) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(\le \lim_{\alpha \rightarrow \infty} \int_{c_2}^{c_3} \norm{ e^{ - \left( \frac{\alpha-\mu}{\sqrt{2}\sigma} \right)^2 } } \d z\) | \(\displaystyle \) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(\le \lim_{\alpha \rightarrow \infty} \norm{ e^{ - \left( \frac{\alpha-\mu}{\sqrt{2}\sigma} \right)^2 } } \int_{c_2}^{c_3} \d z\) | \(\displaystyle \) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(= 0\) | \(\displaystyle \) | $\quad$ | $\quad$ |
Between $c_4$ and $c_1$
By similar argument, the integral between $c_4$ and $c_1$ is also 0.
Between $c_1$ and $c_2$
| \(\displaystyle \oint_{z \in \Gamma} e^{-z^2} \d z\) | \(=\) | \(\displaystyle \int_{c_1}^{c_2} e^{ -z^2 } \d z + \int_{c_2}^{c_3} e^{ -z^2 } \d z + \int_{c_3}^{c_4} e^{ -z^2 } \d z + \int_{c_4}^{c_1} e^{ -z^2 } \d z\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \int_{c_1}^{c_2} e^{ -z^2 } \d z + 0 - \sqrt{2\pi \sigma^2} + 0\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle 0\) | $\quad$ | $\quad$ |
and therefore
| \(\displaystyle \int_{c_1}^{c_2} e^{ -z^2 } \d z\) | \(=\) | \(\displaystyle \sqrt{2\pi \sigma^2}\) | $\quad$ | $\quad$ |
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
| \(\displaystyle \phi(t)\) | \(=\) | \(\displaystyle c \frac{1}{ \sqrt{2\pi \sigma^2} } \int_{\frac{-\alpha-\mu}{\sqrt{2}\sigma} - i\frac{t\sigma}{ \sqrt{2} } } ^{\frac{\alpha-\mu}{\sqrt{2}\sigma} - i\frac{t\sigma}{ \sqrt{2} } } e^{ -z^2} \d z\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle c \frac{1}{ \sqrt{2\pi \sigma^2} } \sqrt{2\pi \sigma^2}\) | $\quad$ by Lemma 2 | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle c\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle e^{ \mu it - \frac{1}{2}t^2\sigma^2 }\) | $\quad$ | $\quad$ |
$\blacksquare$
