Chapter 9 Excercise 1
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 the standardized version of a random variable \(X^* = \frac{{X - \mu}}{{\sigma}}\):
\[ \begin{align*} \mu(X^*) &= \mathbb{E}(X^*) \\ &= \mathbb{E}\left(\frac{{X - \mu}}{{\sigma}}\right) \\ &= \frac{1}{\sigma} \cdot \mathbb{E}(X - \mu) \quad \text{(by linearity of expectation)} \\ &= \frac{1}{\sigma} \cdot (\mathbb{E}(X) - \mu) \quad \text{(since } \mu \text{ is constant)} \\ &= \frac{1}{\sigma} \cdot (\mu - \mu) \quad \text{(since } \mathbb{E}(X) = \mu) \\ &= \frac{1}{\sigma} \cdot 0 \\ &= 0. \end{align*} \]
So, \(\mu(X^*) = 0\).
\[ \begin{align*} \sigma^2(X^*) &= \text{Var}(X^*) \\ &= \text{Var}\left(\frac{{X - \mu}}{{\sigma}}\right) \\ &= \left(\frac{1}{\sigma}\right)^2 \cdot \text{Var}(X - \mu) \quad \text{(since } \text{Var}(aX) = a^2\text{Var}(X) \text{ for constant } a) \\ &= \left(\frac{1}{\sigma}\right)^2 \cdot \text{Var}(X) \quad \text{(since } \mu \text{ is constant)} \\ &= \left(\frac{1}{\sigma}\right)^2 \cdot \sigma^2 \quad \text{(since } \text{Var}(X) = \sigma^2) \\ &= \frac{1}{\sigma^2} \cdot \sigma^2 \\ &= 1. \end{align*} \]
Thus, \(\sigma^2(X^*) = 1\).
Therefore, we have verified that \(\mu(X^*) = 0\) and \(\sigma^2(X^*) = 1\), as required.