# 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.

-

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 $\sigma_z^2 = E[Z^2] - E[Z]^2 = 1 - 0^2 = 1$.

-
Thanks @martini!! – user1527227 Jun 5 '13 at 19:35

When dealing with variances, as we know that for a random variable X

$$\operatorname{Var} (X + c) = \operatorname{Var}(X)$$

for any constant $c \in \Bbb R$, we see that without loss of generality we may assume that $\mu_x = 0$ and hence

$$\operatorname{Var}(Z) = \operatorname{Var} \left( \frac{X}{\sigma_x}\right)= \frac{1}{\sigma_x^2} \operatorname{Var}(X) = \frac{\sigma_x^2}{\sigma_x^2} = 1$$

While I'm aware this doesn't address your method directly, I would think that this is a much easier way of going about it - you don't have to find the variance directly from the distribution of $Z$, and instead you can just use the properties of the variance.

-
"The properties of the variance" were what user1527227 was trying to prove, surely? – Billy Jun 5 '13 at 20:07
@Billy I would assume that if the OP was proving the properties of the variance, then the idea would be to try and show that $\operatorname{Var}(aX+b) = a^2 \operatorname{Var}(X)$ rather than individually specifying that $a$ and $b$ are constants specifically related to $X$. – Andrew D Jun 5 '13 at 20:13