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world:characteristic_function_gaussian


2017/10/07 23:39 ·

## Statement

We give a short proof of the expression for the characteristic function of a centered standard gaussian random variable. Let $X\sim N(0,1)$. The aim is to show that $\varphi(u)=\PE[\rme^{\rmi u X}]=\rme^{-u^2/2}$ for all $u \in \rset$.

$\bproof$ First write $$\varphi(u)=\int \rme^{\rmi u x} \frac{\rme^{-x^2/2}}{\sqrt{2\pi}} \rmd x=\rme^{-u^2/2} \int \underbrace{\frac{\rme^{-(x-\rmi u)^2/2}} {\sqrt{2\pi}}}_{\psi(u,x)} \rmd x$$ Denote $A(u)=\int \psi(u,x)\rmd x$. Since $A(0)=1$, we only need to show that $A$ is constant. But, note that $\frac{\partial \psi(u,x)}{\partial u} =-\rmi \frac{\partial \psi(u,x)}{\partial x}$ so that $$A'(u)\stackrel{(\star)}{=}\int \frac{\partial \psi(u,x)}{\partial u} \rmd x=-\int \rmi\frac{\partial \psi(u,x)}{\partial x}\rmd x=\left[-\rmi \psi(u,\cdot)\right]_{-\infty}^{+\infty}=-\rmi \left[\rme^{-(x-\rmi u)^2/2}/\sqrt{2\pi}\right]_{x \to -\infty}^{x \to \infty}=0$$ Note that $(\star)$ is true since for all $u\in \rset$ and all bounded neighborhood $V$ of $u$, $$\int \sup_{u'\in V} \left|\frac{\partial \psi(u',x)}{\partial x}\right| \rmd x\leq \int [|x|+\sup_{u'\in V} |u'|] \rme^{-x^2/2+\sup_{u'\in V}|u'|^2/2} \rmd x<\infty$$ The proof is completed.

world/characteristic_function_gaussian.txt · Last modified: 2022/03/16 07:40 (external edit)