Stat 5101 Lecture Notes - School of Statistics

1.2. CHANGE OF VARIABLES 13where f is the density for the model (Lindgren would say p. d. f.)So far (one sentence) this section looks much like the section on discreterandom variables. The only difference is that (1.18) has an integral where(1.13) has a sum. But the next equation (1.14) in the section on discrete randomvariables has no useful analog for continuous random variables. In factP ({x}) =0,for all x(p. 32 in Lindgren). Because of this there is no simple analog of Theorem 1.2for continuous random variables.There is, however, an analog of Theorem 1.3.Theorem 1.4. If X is a continuous random variable with density f X and samplespace S, ifg:S→Tis an invertible transformation with differentiableinverse h = g −1 , and Y = g(X), then Y is a continuous random variable withdensity f Y defined byf Y (y) =f X(h(y))·|h ′ (y)|, y ∈ T. (1.19)The first term on the right hand side in (1.19) is the same as the right handside in (1.15), the only difference is that we have written h for g −1 . The secondterm has no analog in the discrete case. Here summation and integration, andhence discrete and continuous random variables, are not analogous.We won’t bother to prove this particular version of the theorem, since it isa special case of a more general theorem we will prove later (the multivariablecontinuous change of variable theorem).Example 1.2.5.SupposeX ∼ Exp(λ).What is the distribution of Y = X 2 ?This is just like Example 1.2.4 except now we use the continuous change ofvariable theorem.The transformation in question is g :(0,∞)→(0, ∞) defined byg(x) =x 2 , x > 0.The inverse transformation is, of course,h(y) =g −1 (y)=y 1/2 , y > 0,and it also maps from (0, ∞) to(0,∞). Its derivative ish ′ (y) = 1 2 y−1/2 , y > 0.The density of X isf X (x) =λe −λx , x > 0.