Gaussian Distribution - 1. Gaussian Random Variable

Fourier transform of the pdf $f_X(x)$ of a Gaussian r.v. $X \sim \mathcal{N}(0, \sigma_X^2)$

We first compute the Fourier transform of the Gaussian integral

$$ g(x) = e^{-\pi x^2}. $$

We have

$$ \mathcal{F}g(s) = \int_{-\infty}^\infty e^{-2\pi isx}e^{-\pi x^2}dx. $$

Differentiate both sides with respect to $s$:

$$ \frac{d}{ds} \mathcal{F} g(s) = \int_{-\infty}^\infty e^{-2\pi isx}(-2\pi isx)e^{-\pi x^2}dx.$$

This can be evaluated by an integration by parts: setting 

$$ \begin{cases} u=e^{-2\pi isx}, & dv = -2\pi ixe^{-\pi x^2} \\ \frac{du}{dx} = (-2\pi is)e^{-2\pi isx}, & v = ie^{-\pi x^2} \end{cases} $$

Thus

$$ \begin{align*} \mathcal{F}g(s) &= \left[ ie^{-2\pi isx}e^{-\pi x^2} \right]_{-\infty}^\infty - \int_{-\infty}^\infty ie^{-\pi x^2}(-2\pi is)e^{-2\pi isx}dx \\ &= \left[ ie^{-\pi (x^2+2isx)} \right]_{-\infty}^\infty - 2\pi s \int_{-\infty}^\infty e^{-2\pi isx}e^{-\pi x^2}dx \\ &= 0 - 2\pi s \mathcal{F} g(s) \end{align*} $$

$$ \frac{d}{ds}\mathcal{F}g(s) = -2\pi s \mathcal{F}g(s) \Longrightarrow \mathcal{F}g(s) = \mathcal{F}g(0)e^{-\pi s^2} $$

But

$$ \mathcal{F}g(0) = \int_{-\infty}^\infty e^{-\pi x^2}dx = 1 $$

Hence

$$ \mathcal{F} g(s) = e^{-\pi s^2}. $$

Using this Fourier transform pair, we now find the Fourier transform of $f_X(x),$ denoted by $\hat{f_X}(\theta)$:

$$ \begin{align*} \hat{f_X}(\theta) &= \mathcal{F}f_X(\theta) = \frac{1}{\sqrt{2\pi \sigma_X^2}} \left( \mathcal{F} e^{-\frac{x^2}{2\sigma_X^2}} \right) (\theta) \\ &= \frac{1}{\sqrt{2\pi \sigma_X^2}} \left( \mathcal{F} g \left( \frac{x}{\sqrt{2\pi \sigma_X^2}} \right) \right) (\theta) \\ &= \frac{1}{\sqrt{2\pi \sigma_X^2}} \sqrt{2\pi \sigma_X^2} \mathcal{F}g(\sqrt{2\pi \sigma_X^2} \theta) \\ &= e^{-\pi (\sqrt{2\pi \sigma_X^2} \theta)^2} = e^{-2\pi^2 \sigma_X^2 \theta^2}\end{align*} $$

$$ \therefore f_X(x) = \frac{1}{\sqrt{2\pi \sigma_X^2}}e^{-\frac{x^2}{2\sigma_X^2}} \hspace{1cm} \longleftrightarrow \hspace{1cm} \hat{f_X}(\theta) = e^{-2\pi^2 \sigma_X^2 \theta^2} $$

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Characteristic function $\varphi_X(t)$ of a zero-mean Gaussian r.v. $X \sim \mathcal{N}(0, \sigma_X^2)$

The characteristic function $\varphi_X(\theta)$ is defined as the expected value of $e^{itX}$, i.e.

$$ \varphi_X(t) = E[e^{itX}]. $$

Using the foregoing derivation of the Fourier transform of the univariate Gaussian r.v. $X$, we have

$$ \begin{align*} \varphi_X(t) &= \int_{-\infty}^\infty e^{itx} f_X(x)dx \\ &= \left( \mathcal{F} f_X \right) \left( -\frac{t}{2\pi} \right) \\ &= \hat{f_X} \left( -\frac{t}{2\pi} \right) \\ &= e^{-2 \pi^2 \sigma_X^2 \left( -\frac{t}{2\pi} \right)^2 } \\ &= e^{-\sigma_X^2 t^2 /2} \end{align*} $$

$$ \therefore f_X(x) = \frac{1}{\sqrt{2\pi \sigma_X^2}} e^{- \frac{x^2}{2\sigma_X^2}} \hspace{1cm} \longleftrightarrow \hspace{1cm} \varphi_X(t) = e^{-\frac{ \sigma_X^2 t^2}{2}} $$

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