Q:

How does standard random variable have variance of 1? 

Let X be a discrete random variable and define $Z = \cfrac{X - \mu_x}{\sigma_x}=\cfrac{1}{\sigma_x} \cdot X - \cfrac{\mu_x}{\sigma_x}$ which is a linear transformation of $X$.  
How do you get a variance of 1 assuming this?   I tried working it out but couldn't get it.   I am stuck on the last step:
$\mu_z=E[Z] = \cfrac{\mu_x}{\sigma_x} - \cfrac{\mu_x}{\sigma_x} =0$
$Z^2 = \left(\cfrac{X-\mu_x}{\sigma_x}\right)^2= \cfrac{X^2}{\sigma_x^2} - \cfrac{-2 X \mu_x}{\sigma_x^2} + \cfrac{\mu_x^2}{\sigma_x^2} $
$E[Z^2] = \cfrac{1}{\sigma_x^2}(E[X^2] - 2 E[X] \mu_x + \mu_x^2)$
$E[Z^2] = \cfrac{1}{\sigma_x^2}(E[X^2] -E[X]^2)$
By defn:  
$\sigma_z^2 = E[Z^2] - E[Z]^2=E[Z^2]=\cfrac{1}{\sigma_x^2}(E[X^2] -E[X]^2)$
I see $E[X]^2 = \sigma_x^2$ but how do you simplify $E[X^2]$?   Thank you in advance. 

A:

As $\sigma_x^2 = E[X^2] - E[X]^2$ by definition of $\sigma_x$, you have in your computation of $E[Z^2]$, that
$$ E[Z^2] = \frac 1{\sigma_x^2} \bigl(E[X^2] - E[X]^2\bigr) = 1 $$
So $\sigma_z^2 = E[Z^2] - E[Z]^2 = 1 - 0^2 = 1$.

