# proof that normal distribution is a distribution

 $\displaystyle\int\limits_{-\infty}^{\infty}\frac{e^{-\frac{(x-\mu)^{2}}{2% \sigma^{2}}}}{\sigma\sqrt{2\pi}}\,dx$ $\displaystyle=$ $\displaystyle\sqrt{\left(\int\limits_{-\infty}^{\infty}\frac{e^{-\frac{(x-\mu)% ^{2}}{2\sigma^{2}}}}{\sigma\sqrt{2\pi}}\,dx\right)^{2}}$ $\displaystyle=$ $\displaystyle\sqrt{\int\limits_{-\infty}^{\infty}\frac{e^{-\frac{(x-\mu)^{2}}{% 2\sigma^{2}}}}{\sigma\sqrt{2\pi}}\,dx\int\limits_{-\infty}^{\infty}\frac{e^{-% \frac{(y-\mu)^{2}}{2\sigma^{2}}}}{\sigma\sqrt{2\pi}}\,dy}$ $\displaystyle=$ $\displaystyle\sqrt{\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty% }\frac{e^{-\frac{(x-\mu)^{2}+(y-\mu)^{2}}{2\sigma^{2}}}}{\sigma^{2}2\pi}\,dx\,dy}$

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

 $\displaystyle\sqrt{\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty% }\frac{e^{-\frac{(x-\mu)^{2}+(y-\mu)^{2}}{2\sigma^{2}}}}{\sigma^{2}2\pi}\,dx\,dy}$ $\displaystyle=$ $\displaystyle\sqrt{\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty% }\frac{e^{-\frac{(x^{\prime})^{2}+(y^{\prime})^{2}}{2\sigma^{2}}}}{\sigma^{2}2% \pi}\,dx^{\prime}\,dy^{\prime}}.$

Converting to polar coordinates, we obtain

 $\displaystyle\sqrt{\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty% }\frac{e^{-\frac{(x^{\prime})^{2}+(y^{\prime})^{2}}{2\sigma^{2}}}}{\sigma^{2}2% \pi}\,dx^{\prime}\,dy^{\prime}}$ $\displaystyle=$ $\displaystyle\sqrt{\int\limits_{0}^{\infty}\int\limits_{0}^{2\pi}\frac{re^{-% \frac{r^{2}}{2\sigma^{2}}}}{\sigma^{2}2\pi}\,dr\,d\theta}$ $\displaystyle=$ $\displaystyle\sqrt{\int\limits_{0}^{2\pi}\frac{d\theta}{2\pi}}\sqrt{\int% \limits_{0}^{\infty}\frac{re^{-\frac{r^{2}}{2\sigma^{2}}}}{\sigma^{2}}\,dr}$ $\displaystyle=$ $\displaystyle\sqrt{\frac{\theta}{2\pi}\bigg{|}_{0}^{2\pi}}\sqrt{\frac{1}{% \sigma^{2}}\int\limits_{0}^{\infty}re^{-\frac{r^{2}}{2\sigma^{2}}}\,dr}$ $\displaystyle=$ $\displaystyle\sqrt{\frac{2\pi}{2\pi}}\sqrt{\frac{\sigma^{2}}{\sigma^{2}}\left(% -e^{-\frac{r^{2}}{2\sigma^{2}}}\right)\bigg{|}_{0}^{\infty}}$ $\displaystyle=$ $\displaystyle\sqrt{1}\sqrt{1}$ $\displaystyle=$ $\displaystyle 1.$
Title proof that normal distribution  is a distribution  ProofThatNormalDistributionIsADistribution 2013-03-22 13:29:12 2013-03-22 13:29:12 Wkbj79 (1863) Wkbj79 (1863) 7 Wkbj79 (1863) Proof msc 62E15 AreaUnderGaussianCurve