Question

What is the variance of the standard normal distribution?

Answer 1

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

Explanation:

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) )`

`= - 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`