Let \(X\) be a continuous random variable with mean \(\mu (X)\) and variance \(\sigma^2 (X)\), and let \(X^*=\frac{(X-\mu)}{\sigma}\) be its standardized version. Verify directly that \(\mu (X^*)=0\) and \(\sigma^2 (X^*)=1\).
Given, \[X^*=\frac{(X-\mu)}{\sigma}\]
\[E(X^*)=\frac{1}{\sigma}(E(X)-\mu)\] \[E(X^*)=\frac{1}{\sigma}(\mu - \mu) \] \[E(X^*)=\mu (X^*)=0 \]
\[\sigma^2(X^*)=E(\frac{X-\mu}{\sigma})^2\] \[\sigma^2(X^*)=\frac{1}{\sigma^2} \sigma^2 \] \[\sigma^2(X^*)=1 \]