proof that normal distribution is a distribution
ProofThatNormalDistributionIsADistribution
2003-02-24 01:25:26
2008-04-19 15:18:30
Proof
normal random variable
0
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\newcommand{\lint}{\int\limits}
\begin{eqnarray*}
\lint_{-\infty}^\infty \frac{e^{-\frac{(x-\mu)^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}} \, dx
&=& \sqrt{ \left( \lint_{-\infty}^\infty \frac{e^{-\frac{(x-\mu)^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}} \, dx \right)^2} \\
&=& \sqrt{ \lint_{-\infty}^\infty \frac{e^{-\frac{(x-\mu)^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}} \, dx
\lint_{-\infty}^\infty \frac{e^{-\frac{(y-\mu)^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}} \, dy} \\
&=& \sqrt{ \lint_{-\infty}^\infty \lint_{-\infty}^\infty \frac{e^{-\frac{(x-\mu)^2+(y-\mu)^2}{2\sigma^2}}}{\sigma^2 2\pi} \, dx \, dy}
\end{eqnarray*}
Substitute $x^\prime=x-\mu$ and $y^\prime=y-\mu$. Since the bounds are infinite, they do not change, and $dx^\prime=dx$ and $dy^\prime=dy$. Thus, we have
\begin{eqnarray*}
\sqrt{\lint_{-\infty}^\infty \lint_{-\infty}^\infty \frac{e^{-\frac{(x-\mu)^2+(y-\mu)^2}{2\sigma^2}}}{\sigma^2 2\pi} \, dx \, dy}
&=& \sqrt{\lint_{-\infty}^\infty \lint_{-\infty}^\infty \frac{e^{-\frac{(x^\prime)^2+(y^\prime)^2}{2\sigma^2}}}{\sigma^2 2\pi} \, dx^\prime \, dy^\prime}.
\end{eqnarray*}
Converting to polar coordinates, we obtain
\begin{eqnarray*}
\sqrt{\lint_{-\infty}^\infty \lint_{-\infty}^\infty \frac{e^{-\frac{(x^\prime)^2+(y^\prime)^2}{2\sigma^2}}}{\sigma^2 2\pi} \, dx^\prime \, dy^\prime}
&=& \sqrt{\lint_0^\infty \lint_0^{2\pi} \frac{re^{-\frac{r^2}{2\sigma^2}}}{\sigma^2 2\pi} \, dr \, d\theta} \\
&=& \sqrt{\lint_0^{2\pi} \frac{d\theta}{2\pi}} \sqrt{\lint_0^\infty \frac{re^{-\frac{r^2}{2\sigma^2}}}{\sigma^2} \, dr} \\
&=& \sqrt{\frac{\theta}{2\pi}\bigg|_0^{2\pi}} \sqrt{\frac{1}{\sigma^2}\lint_{0}^\infty r e^{-\frac{r^2}{2\sigma^2}} \, dr} \\
&=& \sqrt{\frac{2\pi}{2\pi}} \sqrt{\frac{\sigma^2}{\sigma^2}\left( -e^{-\frac{r^2}{2\sigma^2}} \right)\bigg|_0^\infty} \\
&=& \sqrt{1} \sqrt{1}\\
&=& 1.
\end{eqnarray*}