Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 1 down vote accepted

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

share|improve this answer
    
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.

share|improve this answer
    
"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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.