Probability mass function

To verify that f(x) is a valid pmf we must verify that (1) it is non-negative everywhere, and (2) that the total integral is equal to 1. The first is obvious, so we move on to verify the second.

${\displaystyle \int _{-\infty }^{\infty }{\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-(x-\mu )^{2}/2\sigma ^{2}}dx={\frac {1}{\sigma {\sqrt {2\pi }}}}\int _{-\infty }^{\infty }e^{-(x-\mu )^{2}/2\sigma ^{2}}dx}$

Now let ${\displaystyle w={x-\mu \over \sigma {\sqrt {2}}}}$. We see that ${\displaystyle dw={dx \over \sigma {\sqrt {2}}}}$.

${\displaystyle {\frac {1}{\sqrt {\pi }}}\int _{-\infty }^{\infty }e^{-w^{2}}dw}$

Now we use the Gaussian integral that ${\displaystyle \int _{-\infty }^{\infty }e^{-w^{2}}\,dw={\sqrt {\pi }}}$

${\displaystyle {\frac {1}{\sqrt {\pi }}}{\sqrt {\pi }}=1}$

Mean

We derive the mean as follows

${\displaystyle \operatorname {E} [X]=\int _{-\infty }^{\infty }x\cdot f(x)dx}$

${\displaystyle =\int _{-\infty }^{\infty }x{\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-(x-\mu )^{2}/2\sigma ^{2}}dx}$

${\displaystyle =\int _{-\infty }^{\infty }[(x-\mu )+\mu ]{\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-(x-\mu )^{2}/2\sigma ^{2}}dx}$

${\displaystyle =\int _{-\infty }^{\infty }(x-\mu ){\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-(x-\mu )^{2}/2\sigma ^{2}}dx+\int _{-\infty }^{\infty }\mu {\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-(x-\mu )^{2}/2\sigma ^{2}}dx}$

${\displaystyle ={\frac {1}{\sigma {\sqrt {2\pi }}}}(-\sigma ^{2})\int _{-\infty }^{\infty }{-x+\mu \over \sigma ^{2}}e^{-(x-\mu )^{2}/2\sigma ^{2}}dx+\mu \int _{-\infty }^{\infty }{\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-(x-\mu )^{2}/2\sigma ^{2}}dx}$

We now see that the right integral is the complete integral over a normal pmf. This is therefore 1.

${\displaystyle \operatorname {E} [X]={\frac {1}{\sigma {\sqrt {2\pi }}}}(-\sigma ^{2})\left[e^{-(x-\mu )^{2}/2\sigma ^{2}}\right]_{-\infty }^{\infty }+\mu }$

${\displaystyle ={\frac {1}{\sigma {\sqrt {2\pi }}}}(-\sigma ^{2})[0-0]+\mu }$

${\displaystyle =\mu }$

Variance

${\displaystyle \operatorname {Var} (X)=\operatorname {E} [(X-{E}[X])^{2}]=\int _{-\infty }^{\infty }(x-\mu )^{2}\cdot f(x)dx=\int _{-\infty }^{\infty }(x-\mu )^{2}{\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\cdot \left({\frac {x-\mu }{\sigma }}\right)^{2}}dx}$

We let ${\displaystyle w={x-\mu \over \sigma {\sqrt {2}}}}$

${\displaystyle \operatorname {Var} (X)=\int _{-\infty }^{\infty }\sigma ^{2}2w^{2}{\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-w^{2}}\sigma {\sqrt {2}}dw={\frac {2\sigma ^{2}}{\sqrt {\pi }}}\int _{-\infty }^{\infty }w\cdot we^{-w^{2}}dw}$

We now use integration by parts with u=w and v=(-1/2)e^(-w^2)

${\displaystyle \operatorname {Var} (X)={\frac {2\sigma ^{2}}{\sqrt {\pi }}}\left(\left[w{-1 \over 2}e^{-w^{2}}\right]_{-\infty }^{\infty }-\int _{-\infty }^{\infty }{-1 \over 2}e^{-w^{2}}dw\right)}$

We see that the bracketed term is zero by L'Hôpital's rule.

${\displaystyle \operatorname {Var} (X)={\frac {2\sigma ^{2}}{\sqrt {\pi }}}\left({1 \over 2}\int _{-\infty }^{\infty }e^{-w^{2}}dw\right)}$

Now we use the Gaussian integral again

${\displaystyle \operatorname {Var} (X)={\frac {2\sigma ^{2}}{\sqrt {\pi }}}{1 \over 2}{\sqrt {\pi }}=\sigma ^{2}}$

External links

• Interactive Normal (Gaussian) Distribution Web Applet (Java)