What is the variance of the standard normal distribution?

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Answer 1 · 2018-04-11T10:16:28

See below. The standard normal is the normal set up such that $mu, sigma = 0,1$ so we know the results beforehand.

The PDF for the standard normal is: $mathbb P(z) = 1/sqrt(2 pi) e^(- z^2/2)$

It has mean value:

$mu = int_(-oo)^(oo) dz \ z \ mathbb P(z) = 1/sqrt(2 pi) int_(-oo)^(oo) \ dz \ z e^(- z^2/2)$

$=1/sqrt(2 pi) int_(-oo)^(oo) \ d( - e^(- z^2/2))$

$=1/sqrt(2 pi) [ e^(- z^2/2) ]_(oo)^(-oo) = 0$

It follows that:

$Var(z) = int_(-oo)^(oo) dz \ (z - mu)^2 mathbb P(z)$

$= 1/sqrt(2 pi) int_(-oo)^(oo) \ dz \ z^2 e^(- z^2/2)$

This time, use IBP:

$Var(z) = - 1/sqrt(2 pi) int_(-oo)^(oo) \ d( e^(- z^2/2)) \ z$

$= - 1/sqrt(2 pi) ( [ z e^(- z^2/2) ]_(-oo)^(oo) - int_(-oo)^(oo) dz \ e^(- z^2/2) )$

Because $[ z e^(- z^2/2) ]_(-oo)^(oo) = 0$

$= 1/sqrt(2 pi) int_(-oo)^(oo) dz \ e^(- z^2/2)$

This integral is well known. It can be done using a polar sub, but here the result is stated.

$Var(z) = 1/sqrt(2 pi) sqrt(2 pi) = 1$