Consider the continuous random variable X and let a, b be non-zero constants. Then E(aX + b) = aE(X) + b and Var(aX + b) = a2 Var(X). **Proof:** E(aX + b) = _all x (ax + b) f(x), dx = _all x axf(x) + bf(x), dx = a E(X) + b Var(aX + b) = E((aX + b)2) - [E(aX + b)]2 = E(a2X2 + 2abX + b2) - (aE(X) + b)2 = a2E(X2) + 2abE(X) + b2 - (a2[E(X)]2 + 2abE(X) + b2) = a2E(X2) - a2[E(X)]2 = a2 Var(X) This result applies to both discrete and continuous random variables.

Consider the continuous random variable \(X\) and let \(a, b\) be non-zero constants.

\[\text{Then } \mathrm{E}(aX + b) = a\mathrm{E}(X) + b \text{ and } \mathrm{Var}(aX + b) = a^2 \mathrm{Var}(X).\]

\(\begin{aligned} \mathrm{E}(aX + b) &= \int_{\text{all } x} (ax + b) \mathrm{f}(x) \, \mathrm{d}x \\ &= \int_{\text{all } x} ax\mathrm{f}(x) + b\mathrm{f}(x) \, \mathrm{d}x \\ &= a \mathrm{E}(X) + b \end{aligned}\)

\(\begin{aligned} \mathrm{Var}(aX + b) &= \mathrm{E}\left((aX + b)^2\right) - [\mathrm{E}(aX + b)]^2 \\ &= \mathrm{E}(a^2X^2 + 2abX + b^2) - (a\mathrm{E}(X) + b)^2 \\ &= a^2\mathrm{E}(X^2) + 2ab\mathrm{E}(X) + b^2 - (a^2[\mathrm{E}(X)]^2 + 2ab\mathrm{E}(X) + b^2) \\ &= a^2\mathrm{E}(X^2) - a^2[\mathrm{E}(X)]^2 \\ &= a^2 \mathrm{Var}(X) \end{aligned}\)

This result applies to both discrete and continuous random variables.

Continuous Random Variables

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Expected value and variance of a linearly transformed CRV

Expected value and variance of a linearly transformed CRV