g(T) = pel is a monotone increasing function. fr(t) is a continuous on the support of T and g-1(y) = = In - has a continuous derivative on (0, too) (the support of Y) avg (y) = qy q 19. fr(g-'(y)) = 1 1 Bq Ba p. Ba 1 Ba - none of the above 20. Therefore, fy(y) =? 1 1 Bq Bay ,y20 1 1 Ba Bay ,yzp 1 1 Bq Bq y 20 1 Bq Bq y zp - none of the aboveTheorem 2.1.5 Let X have pdf fx (x) and let Y = g(X), where g is a monotone function. Let A and y be defined by (2.1.7). Suppose that fx (x) is continuous on X and that g-(y) has a continuous derivative on V. Then the pdf of Y is given by d (2.1.10) fy(y) = fx (9-1(y)) 29 1(y) yEy otherwise. Proof: From Theorem 2.1.3 we have, by the chain rule, d fx (g-1 (y) ) d fy (y) = dy Fy (y) = dy g-(y) if g is increasing -fx(g (y) ) d dy g (y) if g is decreasing, which can be expressed concisely as (2.1.10)