Question

How do you derive the variance of a gaussian distribution?

Answer 1

Hint: The probability density function of a gaussian distribution is $\dfrac{1}{\sigma \sqrt{2\pi }}{{e}^{-\dfrac{1}{2}{{\left( \dfrac{x-\alpha }{\sigma } \right)}^{2}}}}$ 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 $\int\limits_{-\infty }^{\infty }{xf\left( x \right)dx}$ and $\int\limits_{-\infty }^{\infty }{{{x}^{2}}f\left( x \right)dx}-{{\left( \int\limits_{-\infty }^{\infty }{xf\left( x \right)dx} \right)}^{2}}$ so variance is equal to $\int\limits_{-\infty }^{\infty }{{{x}^{2}}f\left( x \right)dx}-{{\alpha }^{2}}$

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

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

Note: The definition of gamma function is $\Gamma \left( x \right)=\int\limits_{0}^{\infty }{{{t}^{x-1}}{{e}^{-t}}dt}$ gamma of any positive integer is equal to factorial of the preceding number of that number. We can define gamma as $\Gamma \left( x \right)=\left( x-1 \right)\Gamma \left( x-1 \right)$ . 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 $\dfrac{1}{\sigma \sqrt{2\pi }}{{e}^{-\dfrac{1}{2}{{\left( \dfrac{x-\alpha }{\sigma } \right)}^{2}}}}$ , the curve of this function is symmetric with respect to straight line x = $\alpha $.