Question

How do you derive the variance of a gaussian distribution?

Answer 1

Hint: The probability density function of a gaussian distribution is ${\displaystyle \frac{1}{\sigma \sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{x-\alpha }{\sigma }}\right)}^2}$ where $\alpha$ is the means of distribution and $\sigma$ is the variance of standard deviation of distribution. So ${\sigma }^2$ is the variance of distribution. If f(x) is the probability density function of any distribution, the mean of the distribution is equal to $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}xf\left(x\right)dx$ and $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^{2}f\left(x\right)dx-{\left(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}xf\left(x\right)dx\right)}^2$ so variance is equal to $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^{2}f\left(x\right)dx-{\alpha }^2$

Complete step by step solution:
We have to derive variance of normal distribution ${\displaystyle \frac{1}{\sigma \sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{x-\alpha }{\sigma }}\right)}^2}$
Variance of the distribution = $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{\displaystyle \frac{1}{\sigma \sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{x-\alpha }{\sigma }}\right)}^2}dx-{\alpha }^2$
If we put ${\displaystyle \frac{x-\alpha }{\sigma }}=t$ we get x equal to $\alpha +t\sigma$ and the limit of t is $-\mathrm{\infty }$ to $\mathrm{\infty }$ and $dx=\sigma dt$
$\Rightarrow \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{\displaystyle \frac{1}{\sigma \sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{x-\alpha }{\sigma }}\right)}^2}dx-{\alpha }^2=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{(\alpha +t\sigma )}^2{\displaystyle \frac{1}{\sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{t}^2}dt-{\alpha }^2$
$\Rightarrow \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{\displaystyle \frac{1}{\sigma \sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{x-\alpha }{\sigma }}\right)}^2}dx-{\alpha }^2=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}({\alpha }^2+t^2{\sigma }^2+2t\alpha \sigma ){\displaystyle \frac{1}{\sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{t}^2}dt-{\alpha }^2$
$\Rightarrow \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{\displaystyle \frac{1}{\sigma \sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{x-\alpha }{\sigma }}\right)}^2}dx-{\alpha }^2=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\alpha }^2{\displaystyle \frac{1}{\sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{t}^2}dt+\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{2t\alpha \sigma }{\sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{t}^2}dt+\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{t^2{\sigma }^2}{\sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{t}^2}dt-{\alpha }^2$
${\displaystyle \frac{2t\alpha \sigma }{\sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{t}^2}$is an odd function, so $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{2t\alpha \sigma }{\sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{t}^2}dt$ is equal to 0.
We know that $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-{\displaystyle \frac{1}{2}}{t}^2}$ is equal to $2\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-{\displaystyle \frac{1}{2}}{t}^2}$ because it is an even function$\Rightarrow \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{\displaystyle \frac{1}{\sigma \sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{x-\alpha }{\sigma }}\right)}^2}dx-{\alpha }^2=2\underset{0}{\overset{\mathrm{\infty }}{\int }}{\alpha }^2{\displaystyle \frac{1}{\sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{t}^2}dt+2\underset{0}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{t^2{\sigma }^2}{\sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{t}^2}dt-{\alpha }^2$
$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-{\displaystyle \frac{1}{2}}{t}^2}$ is equal to ${\displaystyle \frac{1}{\sqrt{2}}}\mathrm{\Gamma }{\displaystyle \frac{1}{2}}$ = $\sqrt{{\displaystyle \frac{\pi }{2}}}$
We can calculate $\underset{0}{\overset{\mathrm{\infty }}{\int }}{t}^2{e}^{-{\displaystyle \frac{1}{2}}{t}^2}$ is equal to $\sqrt{2}\mathrm{\Gamma }{\displaystyle \frac{3}{2}}$ = $\sqrt{{\displaystyle \frac{\pi }{2}}}$
Replacing these values, we get
$\Rightarrow \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{\displaystyle \frac{1}{\sigma \sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{x-\alpha }{\sigma }}\right)}^2}dx-{\alpha }^2={\displaystyle \frac{2{\alpha }^2\sqrt{\pi }}{\sqrt{2\pi }\sqrt{2}}}+{\displaystyle \frac{2{\sigma }^2\sqrt{\pi }}{\sqrt{2\pi }\sqrt{2}}}-{\alpha }^2$
$\Rightarrow \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{\displaystyle \frac{1}{\sigma \sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{x-\alpha }{\sigma }}\right)}^2}dx-{\alpha }^2={\alpha }^2+{\sigma }^2-{\alpha }^2$
$\Rightarrow \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{\displaystyle \frac{1}{\sigma \sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{x-\alpha }{\sigma }}\right)}^2}dx-{\alpha }^2={\sigma }^2$

So the variance is equal to ${\sigma }^2$

Note: The definition of gamma function is $\mathrm{\Gamma }\left(x\right)=\underset{0}{\overset{\mathrm{\infty }}{\int }}{t}^{x-1}{e}^{-t}dt$ gamma of any positive integer is equal to factorial of the preceding number of that number. We can define gamma as $\mathrm{\Gamma }\left(x\right)=(x-1)\mathrm{\Gamma }(x-1)$ . Definition of even function is f( x ) = f( -x). The graph of an even function is always symmetric with respect to y axis and the odd function is f( x) = -f (x ). The probability density function of gaussian or normal distribution which ${\displaystyle \frac{1}{\sigma \sqrt{2\pi }}}{e}^{-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{x-\alpha }{\sigma }}\right)}^2}$ , the curve of this function is symmetric with respect to straight line x = $\alpha$.