Page 214 - Elements of Distribution Theory
P. 214

P1: JZX
            052184472Xc07  CUNY148/Severini  May 24, 2005  3:59





                            200             Distribution Theory for Functions of Random Variables

                            yielding the distribution of Y.For instance, if X has a discrete distribution with frequency
                            function p X , then the distribution of Y is discrete with frequency function

                                                      p Y (y) =      p X (x).
                                                             x∈X: g(x)=y
                            However, in more general cases, difficulties may arise when attempting to implement this
                            approach. For instance, the set {x ∈ X:g(x) ∈ A} is often complicated, making computation
                            of the integral in (7.1) difficult. The analysis is simplified if g is a one-to-one function.

                            Theorem 7.1. Let X denote a real-valued random variable with range X and distribution
                            function F X . Suppose that Y = g(X) where g is a one-to-one function on X. Let Y = g(X)
                            and let h denote the inverse of g.
                               (i) Let F Y denote the distribution function of Y. If g is an increasing function, then

                                                        F Y (y) = F X (h(y)), y ∈ Y
                                  If g is a decreasing function, then
                                                     F Y (y) = 1 − F X (h(y)−), y ∈ Y.

                               (ii) If X has a discrete distribution with frequency function p X , then Y has a discrete
                                  distribution with frequency function p Y , where

                                                        p Y (y) = p X (h(y)), y ∈ Y.

                              (iii) Suppose that X has an absolutely continuous distribution with density function p X
                                  and let g denote a continuously differentiable function. Assume that there exists

                                  an open subset X 0 ⊂ X with Pr(X ∈ X 0 ) = 1 such that |g (x)| > 0 for all x ∈ X 0
                                  and let Y 0 = g(X 0 ). Then Y has an absolutely continuous distribution with density
                                  function p Y , where

                                                     p Y (y) = p X (h(y))|h (y)|, y ∈ Y 0 .
                            Proof. If g is increasing on X, then, for y ∈ Y,

                                            F Y (y) = Pr(Y ≤ y) = Pr(X ≤ h(y)) = F X (h(y)).
                            Similarly, if g is decreasing on X, then

                                     Pr(Y ≤ y) = Pr(X ≥ h(y)) = 1 − Pr(X < h(y)) = 1 − F(h(y)−).
                              Part (ii) follows from the fact that, for a one-to-one function g,

                                                    Pr(Y = y) = Pr(X = g(y)).
                              Consider a bounded function f defined on Y 0 = g(X 0 ). Since Pr(X ∈ X 0 ) = 1,

                                                  E[ f (Y)] =  f (g(x))p X (x) dx.
                                                             X 0
                            By the change-of-variable formula for integration,


                                                E[ f (Y)] =  f (y)p X (h(y))|h (y)| dy.
                                                          Y 0
   209   210   211   212   213   214   215   216   217   218   219