Use the definition in Expression (3.13) to prove that V(aX + b) = ?2. ?2x [ Hint: With h(X) = aX + b, E[h(X)] = aµ = b where µ = E(X).] Reference Expression (3.13 The variance of h(X) is the expected value of the squared difference between h(X) and its expected value: When h(X) = aX + b, a linear function, Substituting this into (3.13) gives a simple relationship between V[h(X)] and V(X):

Problem 41E Answer: Step1: We have the variance of h(X) is the expected value of the squared difference between h(X) and its expected value: V [h(x)]= 2 = {h(x) E[h(x)]} • P(x) ……….(1) h(x) x When h(X) = aX + b, a linear function, h(x) E[h(x)] = ax + b (a + b) = a(x ) Substituting this into (1) gives a simple relationship between V[h(X)] and V(X): We need to prove V (aX + b) = a • 2 x Step2: Let us assume that Y = aX + b W.k.t, V [h(x)]= 2 = {h(x) E[h(x)]} • P(x) h(x) x Now, V ar(Y ) = 2 = {y E[y]} • P(Y = y) h(x) y 2 = {y } y P(Y = y) y Substitute y = aX + b in above equation we get 2 V ar(Y ) = {(ax + b) ax+b} • P(X = x) x = {(ax + b) a + b}x• P(X = x) x = (ax a ) • P(X = x) x x 2 2 = {a (x ) }x• P(X = x) x 2 2 = a {(x ) } x P(X = x) x = a V ar(x) [from equ (1)] From the given information we have V [h(x)]= 2 h(x) 2 Hence, V ar(x) = x Therefore, V ar(Y ) = a V ar(x) = a • x V ar(Y ) = a • 2 x Hence the proof.