1. Let X be a random variable of mean $\mu$ and variance $\sigma^2$. Use the properties of expectation to show that
    $$Z=\frac{X-\mu}{\sigma}$$ has mean 0 and variance 1.

  2. Let Z be a random variable of mean 0 and variance 1. Show that $$X=\sigma Z+\mu$$ has mean $\mu$ and variance $\sigma^2$.

I think I have to use $f_Z(z)=\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}$ to help me prove this but I have no idea where to start? I know what expectation is and how to calculate it but which properties specifically is the question talking about?

  • 3
    $\begingroup$ I think you should use that $E(aX+b) = aE(X)+b$ and similar properties for the variance (which can be derived from this linearity property of $E$.) That means you e.g. have to show that $E(Z) = 0$ an similarly that the variance of $Z$ is $1$ $\endgroup$
    – flawr
    Nov 28, 2015 at 18:33
  • $\begingroup$ There is no mention in the question of the hypothesis that the random variables involved are "normally distributed" and the solution does not depend on this hypothesis. A typo? $\endgroup$
    – Did
    Nov 28, 2015 at 20:41

2 Answers 2


There is no need to use PDFs.

  1. Since you're only allowed to use properties of expectation, then $$E[Z] = E\left[\frac{X-\mu}{\sigma}\right] = \frac{1}{\sigma}\left[E[X] - \mu\right] = \frac{1}{\sigma}[\mu-\mu] = 0.$$ Then for the variance \begin{align*} \text{Var}[Z^2] &= E[Z^2]-E^2[Z] \\ &= E\left[\left(\frac{X-\mu}{\sigma}\right)^2\right] - 0^2\\ &= \frac{1}{\sigma^2}\left[E[X^2]-2\mu E[X] +\mu^2\right] \\ &= \frac{1}{\sigma^2}[\{E[X^2]-E^2[X]\} + E^2[X]-2\mu E[X]+\mu^2]\\ &= \frac{1}{\sigma^2}[\{\text{Var}[X]\}+\mu^2-2\mu^2+\mu^2]\\ &= \frac{1}{\sigma^2}[\sigma^2 +0] \\ &= 1. \end{align*}

  2. Similar as 1.

Addendum: If $X,Y$ (not necessarily independent) are random variables and $a,b, c$ are some constants, then the properties it is referring to are

  • Scaling: $$E[cX] =cE[X].$$

  • Addition: $$E[X+Y] = E[X]+E[Y].$$ Then $$E[aX+bY+c] = aE[X]+bE[Y]+ c.$$ The instructions imply that you should be familiar with this by now.


For question 1, you cans use $E[\frac{X-\mu}{\sigma}]=E[\frac{X}{\sigma}]-\frac{\mu}{\sigma}=\frac{\mu}{\sigma}-\frac{\mu}{\sigma}=0$. For the variance, use the variance properties: $Var(\frac{X-\mu}{\sigma})=Var(\frac{X}{\sigma}-\frac{\mu}{\sigma})=Var(\frac{X}{\sigma})=\frac{1}{\sigma^2}Var(X)=\frac{\sigma^2}{\sigma^2}=1$.

Question 2 can be answered in a similar manner.

  • 1
    $\begingroup$ From the wording of the question (use properties of expectation) maybe for variance one is expected to use the definition of variance as $E((X-\mu)^2)$. $\endgroup$ Nov 28, 2015 at 18:41

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