proof that normal distribution is a distribution

${\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{\displaystyle \frac{e^{-\frac{{(x-\mu )}^2}{2{\sigma }^2}}}{\sigma \sqrt{2\pi }}}𝑑x$ $=$ $\sqrt{{\left({\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{\displaystyle \frac{e^{-\frac{{(x-\mu )}^2}{2{\sigma }^2}}}{\sigma \sqrt{2\pi }}}𝑑x\right)}^2}$

$=$ $\sqrt{{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{\displaystyle \frac{e^{-\frac{{(x-\mu )}^2}{2{\sigma }^2}}}{\sigma \sqrt{2\pi }}}𝑑x{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{\displaystyle \frac{e^{-\frac{{(y-\mu )}^2}{2{\sigma }^2}}}{\sigma \sqrt{2\pi }}}𝑑y}$

$=$ $\sqrt{{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{\displaystyle \frac{e^{-\frac{{(x-\mu )}^2+{(y-\mu )}^2}{2{\sigma }^2}}}{{\sigma }^{2}2\pi }}𝑑x𝑑y}$

Substitute $x^{\prime }=x-\mu$ and $y^{\prime }=y-\mu$. Since the bounds are infinite, they do not change, and $d{x}^{\prime }=dx$ and $d{y}^{\prime }=dy$. Thus, we have

$\sqrt{{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{\displaystyle \frac{e^{-\frac{{(x-\mu )}^2+{(y-\mu )}^2}{2{\sigma }^2}}}{{\sigma }^{2}2\pi }}𝑑x𝑑y}$ $=$ $\sqrt{{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{\displaystyle \frac{e^{-\frac{{(x^{\prime })}^2+{(y^{\prime })}^2}{2{\sigma }^2}}}{{\sigma }^{2}2\pi }}𝑑x^{\prime }𝑑y^{\prime }}.$

Converting to polar coordinates, we obtain

$\sqrt{{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{\displaystyle \frac{e^{-\frac{{(x^{\prime })}^2+{(y^{\prime })}^2}{2{\sigma }^2}}}{{\sigma }^{2}2\pi }}𝑑x^{\prime }𝑑y^{\prime }}$ $=$ $\sqrt{{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}{\displaystyle \underset{0}{\overset{2\pi }{\int }}}{\displaystyle \frac{r{e}^{-\frac{r^2}{2{\sigma }^2}}}{{\sigma }^{2}2\pi }}𝑑r𝑑\theta }$

$=$ $\sqrt{{\displaystyle \underset{0}{\overset{2\pi }{\int }}}{\displaystyle \frac{d\theta }{2\pi }}}\sqrt{{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}{\displaystyle \frac{r{e}^{-\frac{r^2}{2{\sigma }^2}}}{{\sigma }^2}}𝑑r}$

$=$ $\sqrt{{{\displaystyle \frac{\theta }{2\pi }}|}_0^{2\pi }}\sqrt{{\displaystyle \frac{1}{{\sigma }^2}}{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}r{e}^{-\frac{r^2}{2{\sigma }^2}}𝑑r}$

$=$ $\sqrt{{\displaystyle \frac{2\pi }{2\pi }}}\sqrt{{{\displaystyle \frac{{\sigma }^2}{{\sigma }^2}}\left(-e^{-\frac{r^2}{2{\sigma }^2}}\right)|}_0^{\mathrm{\infty }}}$

$=$ $\sqrt{1}\sqrt{1}$

$=$ $1.$

Title	proof that normal distribution is a distribution
Canonical name	ProofThatNormalDistributionIsADistribution
Date of creation	2013-03-22 13:29:12
Last modified on	2013-03-22 13:29:12
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	7
Author	Wkbj79 (1863)
Entry type	Proof
Classification	msc 62E15
Related topic	AreaUnderGaussianCurve