A125lineartransformations


 * Linear Transformations on a Continuous Random Variable **

We can apply the same linear transformations on E(X) and Var(X) that we used with Discrete Random Variables


 * E(aX + b) = aE(X) + b
 * E(X + Y) = E(X) + E(Y)
 * Var(aX + b) = a 2 Var(X)
 * Var(X +Y) = Var (X) + Var (Y)


 * Example 1 **

The continuous random variable Z has a mean of 5 and a variance of 2

Find ... ... ... (a) .. E(3Z – 1) ... ... ... (b) .. Var(3Z – 1) ... ... ... (c) .. E(Z 2 ) ... ... ... (d) .. E(3Z 2 – 1)

__**Solution:**__

math . \qquad \textbf{(a) } \; E(3Z - 1) = 3E(Z) - 1 \qquad. \\ . \\ . \qquad \qquad \qquad \qquad \quad = 3 \times 5 - 1 \\. \\ . \qquad \qquad \qquad \qquad \quad = 14 math

math . \qquad \textbf{(b) } \; Var(3Z - 1) = 3^2Var(Z) \qquad. \\ . \\ . \qquad \qquad \qquad \qquad \qquad = 9 \times 2 \\. \\ . \qquad \qquad \qquad \qquad \qquad = 18 math

math . \qquad \textbf{(c) } \; \text{Use } \text{Var} \big( X \big) = E \big( X^2 \big) - \Big[ E \big( X \big) \Big]^2 \qquad. math

math . \qquad \qquad \text{Var} \big( Z \big) = E \big( Z^2 \big) - \Big[ E \big( Z \big) \Big]^2 \qquad. \\ . \\ . \qquad \qquad 2 = E \big( Z^2 \big) - \big( 5 \big)^2 \\. \\ . \qquad \qquad 2 = E \big( Z^2 \big) - 25 \\. \\ . \qquad \qquad E \big( Z^2 \big) = 27 math

math . \qquad \textbf{(d) } \; E(3Z^2 - 1) = 3E(Z^2) - 1 \qquad. \\ . \\ . \qquad \qquad \qquad \qquad \qquad = 3 \times 27 - 1 \\. \\ . \qquad \qquad \qquad \qquad \qquad = 80 math

.