# What is the variance of the standard normal distribution?

Apr 11, 2018

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: $m a t h \boldsymbol{P} \left(z\right) = \frac{1}{\sqrt{2 \pi}} {e}^{- {z}^{2} / 2}$

It has mean value:

$\mu = {\int}_{- \infty}^{\infty} \mathrm{dz} \setminus z \setminus m a t h \boldsymbol{P} \left(z\right) = \frac{1}{\sqrt{2 \pi}} {\int}_{- \infty}^{\infty} \setminus \mathrm{dz} \setminus z {e}^{- {z}^{2} / 2}$

$= \frac{1}{\sqrt{2 \pi}} {\int}_{- \infty}^{\infty} \setminus d \left(- {e}^{- {z}^{2} / 2}\right)$

$= \frac{1}{\sqrt{2 \pi}} {\left[{e}^{- {z}^{2} / 2}\right]}_{\infty}^{- \infty} = 0$

It follows that:

$V a r \left(z\right) = {\int}_{- \infty}^{\infty} \mathrm{dz} \setminus {\left(z - \mu\right)}^{2} m a t h \boldsymbol{P} \left(z\right)$

$= \frac{1}{\sqrt{2 \pi}} {\int}_{- \infty}^{\infty} \setminus \mathrm{dz} \setminus {z}^{2} {e}^{- {z}^{2} / 2}$

This time, use IBP:

$V a r \left(z\right) = - \frac{1}{\sqrt{2 \pi}} {\int}_{- \infty}^{\infty} \setminus d \left({e}^{- {z}^{2} / 2}\right) \setminus z$

$= - \frac{1}{\sqrt{2 \pi}} \left(\left[z {e}^{- {z}^{2} / 2}\right] {\setminus}_{- \infty}^{\infty} - {\int}_{- \infty}^{\infty} \mathrm{dz} \setminus {e}^{- {z}^{2} / 2}\right)$

$= - \frac{1}{\sqrt{2 \pi}} \left(\left[z {e}^{- {z}^{2} / 2}\right] {\setminus}_{- \infty}^{\infty} - {\int}_{- \infty}^{\infty} \mathrm{dz} \setminus {e}^{- {z}^{2} / 2}\right)$

Because $\left[z {e}^{- {z}^{2} / 2}\right] {\setminus}_{- \infty}^{\infty} = 0$

$= \frac{1}{\sqrt{2 \pi}} {\int}_{- \infty}^{\infty} \mathrm{dz} \setminus {e}^{- {z}^{2} / 2}$

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

$V a r \left(z\right) = \frac{1}{\sqrt{2 \pi}} \sqrt{2 \pi} = 1$