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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$(Y)=1$?

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Yes. This is the usual way of "standardizing" a random variable. –  Robert Israel Apr 1 '12 at 19:26
    
Unless $\sigma = 0$... –  Qiaochu Yuan Apr 1 '12 at 20:38

2 Answers 2

Hint: $\mathrm E[Y]=\mathrm E[(X-\mu)/\sigma]=\mathrm E[X]/\sigma-\mathrm E[\mu/\sigma]$

$\mathrm {var}(X)=\mathrm E[X^2]-\mathrm E[X]^2$

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$$Var(Y)=\left(E\left[\frac{(X-\mu)^2}{\sigma^2}\right]-E\left[\frac{X-\mu}{\sigma}\right]^2\right)=\frac{1}{\sigma^2}\left(E[(X-\mu)^2]-E[X-\mu]^2\right)=\frac{1}{\sigma^2}\left(E[X^2]-2\mu EX+\mu^2-(E[X]^2-2\mu EX +\mu^2)\right)=\frac{1}{\sigma^2}\left(E[X^2]-E[X]^2\right)=\frac{Var(X)}{\sigma^2}=1$$ $$E(Y)=E\left(\frac{X-\mu}{\sigma}\right)=\frac{1}{\sigma}\left(E[X]-\mu\right)=0$$

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