Consider the continuous random variable \(X\) and let&nbsp;\(a, b\) be non-zero constants.
\[\text{Then}\hspace{0.5em} \mathrm{E}(aX + b) = a\mathrm{E}(X) + b \text{ and}\hspace{0.5em} \mathrm{Var}(aX + b) = a^2 \mathrm{Var}(X).\]
Proof:
\[\begin{aligned} \mathrm{E}(aX + b) &= \int_{\text{all}\hspace{0.5em} x} (ax + b) \mathrm{f}(x) \, \mathrm{d}x \\ &= \int_{\text{all}\hspace{0.5em} x} ax\mathrm{f}(x) + b\mathrm{f}(x) \, \mathrm{d}x \\ &= a \mathrm{E}(X) + b \end{aligned}\]
\[\begin{aligned} \mathrm{Var}(aX + b) &= \mathrm{E}\left((aX + b)^2\right) - [\mathrm{E}(aX + b)]^2 \\ &= \mathrm{E}(a^2X^2 + 2abX + b^2) - (a\mathrm{E}(X) + b)^2 \\ &= a^2\mathrm{E}(X^2) + 2ab\mathrm{E}(X) + b^2 - (a^2[\mathrm{E}(X)]^2 + 2ab\mathrm{E}(X) + b^2) \\ &= a^2\mathrm{E}(X^2) - a^2[\mathrm{E}(X)]^2 \\ &= a^2 \mathrm{Var}(X) \end{aligned}\]
This result applies to both discrete and continuous random variables.

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

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

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

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

Expected value and variance of a linearly transformed CRV

Expected value and variance of a linearly transformed CRV