Let $(\Omega,\mathcal F,\mathbf P)$ be a probability space. Suppose a random variable $X$ has expectation $\mu=\mathbf E(X)$ and variance $\sigma^2=\mathrm{Var}(X)$. Does the random variable $Y$ given by $Y=\frac{X-\mu}\sigma$ satisfy $\mathbf E(Y)=0$ and Var$(Y)=1$?
2 Answers
Hint: $\mathrm E[Y]=\mathrm E[(X-\mu)/\sigma]=\mathrm E[X]/\sigma-\mathrm E[\mu/\sigma]$
$\mathrm {var}(X)=\mathrm E[X^2]-\mathrm E[X]^2$
Yes, it does, since \begin{align*} \operatorname{\mathbb{V}ar}[Y] &= \mathbb{E}\left[\frac{(X-\mu)^2}{\sigma^2}\right]-\mathbb{E}\left[\frac{X-\mu}{\sigma}\right]^2 =\frac{\operatorname{\mathbb{V}ar}[X]}{\sigma^2} =1, \\ \mathbb{E}[Y]&=\mathbb{E}\left[\frac{X-\mu}{\sigma}\right]=\frac{1}{\sigma}\left(\mathbb{E}[X]-\mu\right)=0. \end{align*}