Section 5.1 Let X be a continuous type random variable with p.d.f. f(x) where f(x) > 0 for a < x < b, where a = – and/or b = + is possible; we also let F(x) = P(X  x) be the distribution function for X (i.e., F /(x) = f(x)). Suppose Y = u(X) where u is a continuous function. One method for finding the p.d.f. of the random variable Y is the distribution function method. To use this method, one first attempts to find a formula for the distribution function of Y, that is, a formula for G(y) = Then, the p.d.f. of Y is obtained by taking the derivative of the distribution function, that is, the p.d.f. of Y is g(y) = P(Y  y) . G /(y) .

1. (a) (i) Suppose X is a random variable with p.d.f. f(x) = 1/x2 if 1 < x . Find the p.d.f. of Y = ln(X) by using the distribution function method, X has p.d.f. f(x) = 1 / x2 if 1 < x . The space of Y = ln(X) is {y : 0 < y}. If 0 < y, the distribution function for Y is G(y) = P(Y  y) = P(ln(X)  y) = ey ey ey 1 — dx = x2 1 – — = x P(X  ey) = f(x) dx = 1 – e–y x = 1 –  1 if y < 0 if 0  y The distribution function of Y is G(y) = 1 – e–y The p.d.f. of Y is g(y) = G (y) = e–y if 0 < y We recognize that Y has an distribution. exponential(1)

Section 5.1 Let X be a continuous type random variable with p.d.f. f(x) where f(x) > 0 for a < x < b, where a = – and/or b = + is possible; we also let F(x) = P(X  x) be the distribution function for X (i.e., F /(x) = f(x)). Suppose Y = u(X) where u is a continuous function. One method for finding the p.d.f. of the random variable Y is the distribution function method. To use this method, one first attempts to find a formula for the distribution function of Y, that is, a formula for G(y) = Then, the p.d.f. of Y is obtained by taking the derivative of the distribution function, that is, the p.d.f. of Y is g(y) = P(Y  y) . G /(y) . Another method for finding the p.d.f. of the random variable Y is the change-of-variable method. This method can be used when u is either a strictly increasing function from a to b (i.e., u /(x) > 0 for a < x < b) or a strictly decreasing function from a to b (i.e., u /(x) < 0 for a < x < b) The two cases can be described as follows:

Case 1: Suppose u(x) is strictly increasing (u /(x) > 0) for a < x < b. The space of Y = u(X) is {y | u(a) < y < u(b)} . P(Y  y) = P(u(X)  y) = P[u–1(u(X))  u–1(y)] = P[X  u–1(y)] = Look at Class Exercise #1(a)(ii) while doing this derivation. u–1(y) f(x) dx = F(u–1(y)) – F(a) = G(y) . a We now find the p.d.f. of Y to be g(y) = G /(y) = d — dy d[u–1(y)] f(u–1(y)) ———— dy F(u–1(y)) – F(a) =

Case 2: Suppose u(x) is strictly decreasing (u /(x) < 0) for a < x < b. The space of Y = u(X) is {y | u(b) < y < u(a)} . P(Y  y) = P(u(X)  y) = P[u–1(u(X))  u–1(y)] = P[X  u–1(y)] = Look at Class Exercise #1(b)(ii) while doing this derivation. b f(x) dx = F(b) – F(u–1(y)) = G(y) . u–1(y) We now find the p.d.f. of Y to be g(y) = G /(y) = d — dy d[u–1(y)] – f(u–1(y)) ———— dy F(b) – F(u–1(y)) = This must be negative since u and u–1 are both decreasing functions.

In either case, we find that the p.d.f of Y is g(y) = u(a) < y < u(b) (in Case 1) if u(b) < y < u(a) (in Case 2) d[u–1(y)] f(u–1(y)) ———— dy Return to Class Exercise #1(a)

(ii) the change-of-variable method. Return to Class Exercise #1(a) X has p.d.f. f(x) = 1 / x2 if 1 < x . Y = ln(X) u(x) = ln(x) is for 1 < x . increasing The space of Y is {y : u(1) < y} = {y : 0 < y}. y = u(x) = ln(x) x = u–1(y) = ey d — u–1(y) = dy ey d f(u–1(y)) — u–1(y) = dy 1 —— | ey | = (ey)2 The p.d.f. of Y is g(y) = e–y if 0 < y

(b) (i) Find the p.d.f. of Y = 1 / X by using the distribution function method, X has p.d.f. f(x) = 1 / x2 if 1 < x. The space of Y = 1 / X is {y : 0 < y < 1}. If 0 < y < 1, the distribution function for Y is G(y) = P(Y  y) =    1 — dx = x2 1 – — = x P(1 / X  y) = P(X  1 / y) = f(x) dx = y 1 / y 1 / y x = 1 / y if y < 0 if 0  y < 1 if 1  y y The distribution function of Y is G(y) = 1 The p.d.f. of Y is g(y) = G (y) = 1 if 0 < y < 1 We recognize that Y has a distribution. U(0,1)

(ii) the change-of-variable method. X has p.d.f. f(x) = 1 / x2 if 1 < x . Y = 1 / X u(x) = 1 / x is for 1 < x . decreasing The space of Y is {y : 0 < y < u(1)} = {y : 0 < y < 1}. y = u(x) = 1 / x x = u–1(y) = 1 / y d — u–1(y) = dy – 1 / y2 d f(u–1(y)) — u–1(y) = dy 1 —— | – 1 / y2 | = (1 / y)2 The p.d.f. of Y is g(y) = 1 if 0 < y < 1

(c) (i) Find the p.d.f. of Y = X2 by using Do part (c) for homework! the distribution function method, X has p.d.f. f(x) = 1 / x2 if 1 < x . The space of Y = X2 is {y : 1 < y}. If 1 < y, the distribution function for Y is G(y) = P(Y  y) = P(X2  y) = y y y 1 — dx = x2 1 – — = x 1 1 – — y P(X  y) = f(x) dx = –  1 x = 1 if y < 1 if 1  y The distribution function of Y is G(y) = 1 1 – — y 1 — if 1 < y 2y3/2 The p.d.f. of Y is g(y) = G (y) = We find that Y has a distribution not in one of the families we studied.

(ii) the change-of-variable method. X has p.d.f. f(x) = 1 / x2 if 1 < x . Y = X2 u(x) = x2 is for 1 < x . increasing The space of Y is {y : u(1) < y} = {y : 1 < y}. y = u(x) = x2 x = u–1(y) = y d — u–1(y) = dy 1 —— 2y d f(u–1(y)) — u–1(y) = dy 1 1 —— —— = (y)2 2y The p.d.f. of Y is g(y) = 1 — if 1 < y 2y3/2

Important Theorems in the Text: Suppose F(x) is the distribution function of a continuous type random variable whose space is a < x < b, where a = – and/or b = + is possible (i.e., F(x) is strictly increasing on a < x < b). Then if Y is U(0, 1), the random variable X = F–1(Y) is a continuous type random variable with distribution function F(x). Theorem 5.1-1 Since Y is U(0, 1), then if 0 < y < 1, the distribution function for Y is P(Y  y) = y . We need to show that P(X  x) = F(x) . The distribution function for X is P(X  x) = P[F–1(Y)  x] = P[F(F–1(Y))  F(x)] = P[Y  F(x)] = F(x) . Go to Class Exercise #2(a).

2. (a) Suppose values of a U(0,1) random variable U are read with four decimal place accuracy from a random number table. Indicate how we can simulate 4 independent observations of a random variable X with p.d.f. f(x) = 1 / x2 if 1 < x ; x x x For 1 < x , we have F(x) = P(X  x) = 1 — dt = t2 1 – — = t 1 1 – — x f(t) dt = –  1 t = 1 if x < 1 if 1  x The distribution function for X is F(x) = 1 1 – — x Let u = F(x) = 1 – 1/x . Then, x = F –1(u) = 1 / (1 – u) To simulate 4 values of X, we first obtain 4 values of U, say u1 , u2 , u3 , u4 , from the random number table, and then we calculate x1 = 1 / (1 – u1) , x2 = 1 / (1 – u2) , x3 = 1 / (1 – u3) , x4 = 1 / (1 – u4) . Compare this exercise with the proof of Theorem 5.1-1

Suppose X is a continuous type random variable whose space is a < x < b, where a = – and/or b = + is possible , and whose distribution function is F(x) (i.e., F(x) is strictly increasing on a < x < b). Then the random variable Y = F(X) is U(0, 1). Theorem 5.1-2 To show that Y is U(0, 1), we must show that if 0 < y < 1, the distribution function for Y is P(Y  y) = y . P(Y  y) = P(F(X)  y) = P[F–1(F(X))  F–1(y)] = P[X  F–1(y)] = F (F–1(y)) = y . Compare Class Exercise #1(b) with the proof of this theorem.

(b) we can simulate 4 independent observations of a random variable Y with p.d.f g(y) = 2 / y3 if 1 < y ; Skip to part (c), and do part (b) for homework! y y y 2 — dt = t3 1 – — = t2 1 1 – — y2 For 1 < y , we have P(Y  y) = g(t) dt = t = 1 –  1 if y < 1 if 1  y The distribution function for Y is G(y) = 1 1 – — y2 Let u = G(y) = 1 – 1/y2 . Then, y = G –1(u) = (1 – u)–1/2 To simulate 4 values of Y, we first obtain 4 values of U, say u1 , u2 , u3 , u4 from the random number table, and then we calculate y1 = (1 – u1)–1/2 , y2 = (1 – u2)–1/2 , y3 = (1 – u3)–1/2 , y4 = (1 – u4)–1/2 .

(c) we can simulate 4 independent observations of a random variable W having a Weibull distribution with  = 5/3 and  = 1/27. The random variable W has p.d.f. g(w) = H /(w)e–H(w) if w > 0 , where H(w) = 243w5/3. if w  0 if 0 < w The distribution function for W is G(w) = –243w5/3 1 – e Let u = G(w) = 1 – e . Then, w = G –1(u) = –243w5/3 [–ln(1 – u)]3/5 / 27 To simulate 4 values of W, we first obtain 4 values of U, say u1 , u2 , u3 , u4 , from the random number table, and then we calculate w1 = [–ln(1 – u1)]3/5 / 27 , … , w4 = [–ln(1 – u4)]3/5 / 27.

3. (a) (b) Suppose X is a random variable with p.m.f. f(x) = (4x + 3) / 80 if x = 2, 3, 5, 7 , and Y is a random variable with p.m.f. g(y) = 1 / 2y if y = any positive integer . Find the p.m.f. for each of V = X2 and W = Y2. The space of V is {4, 9, 25, 49}. The p.m.f. of V is h1(v) = (4v + 3) / 80 if v = 4, 9, 25, 49 . The space of W is {1, 4, 9, … }. The p.m.f. of W is h2(w) = 1 / 2w if w = 1, 4, 9, … . Suppose values of a U(0,1) random variable U are read with four decimal place accuracy from a random number table. Indicate how we can

simulate 3 independent observations of X, simulate 3 independent observations of Y. To simulate 3 values of X, we first obtain 3 values of U, say u1 , u2 , u3 , from the random number table, and then we obtain the corresponding values of X from letting 2 if 3 if 5 if 7 if 0.0000  U  0.1374 0.1375  U  0.3249 X = 0.3250  U  0.6124 0.6125  U  0.9999 To simulate 3 values of Y, we first obtain 3 values of U, say u1 , u2 , u3 , from the random number table, and then we obtain the corresponding values of Y from letting 1 if 2 if 3 if 4 if . 0.0000  U  0.4999 0.5000  U  0.7499 0.7500  U  0.8749 Y = 0.8750  U  0.9374 . Go to Class Exercise #2(d).

(d) we can simulate 3 independent observations of a random variable W with p.m.f h(w) = 1 / w if w = 2, 3, 6 ; Go to Class Exercise #2(d). To simulate 3 values of W, we first obtain 3 values of U, say u1 , u2 , u3 , from the random number table, and then we obtain the corresponding values of W from letting Do #2(d) for homework! 2 if 3 if 6 if 0.0000  U  0.4999 W = 0.5000  U  0.8332 0.8333  U  0.9999

(e) we can simulate 5 independent Bernoulli trials with the probability of a one (1) being equal to 0.7 ; First obtain 5 values of U, say u1 , u2 , u3 , u4 , u5 , from the random number table, and then we determine the corresponding values of the 5 independent Bernoulli trials to be 1 if 0 if 0.0000  U  0.6999 0.7000  U  0.9999 (f) we can simulate one observation of a random variable with a b(5,0.7) distribution. An easy way to do this is to first simulate 5 independent Bernoulli trials with the probability of a one (1) being equal to 0.7, say x1 , x2 , x3 , x4 , x5 , and then calculate y = 5  xk k = 1

4. (a) (b) (c) Suppose X is a random variable with p.d.f. f(x) = 1/10 if –5 < x < 5, and let Y = X2. Skip to #5, and do #4 for homework! What type of distribution does X have? X has a distribution. U(–5 , 5) Explain why the change-of-variable method cannot be used to find the p.d.f. of Y. The function u(x) = x2 needs to be either always increasing or always decreasing for –5 < x < 5 in order to use the change-of-variables method, and neither of these is true. Use the distribution function method to find the p.d.f. of Y. The space of Y = X2 is {y : 0 < y < 25}. y If 0 < y < 25, the distribution function for Y is G(y) = P(Y  y) = P(X2  y) = P(– y  X  y) = f(x) dx = –y

y y y 1 — dx = 10 x — = 10 y — 5 f(x) dx = – y –y x = –y if y < 0 if 0  y < 25 if 25  y y — 5 The distribution function of Y is G(y) = 1 1 —— if 0 < y < 25 10y1/2 The p.d.f. of Y is g(y) = G (y) = We find that Y has a distribution not in one of the families we studied.

5. (a) (b) Suppose X is a random variable with p.d.f. f(x) = x–1/2/4 if 0 < x < 4, and let Y = (X – 1)2. Explain why the change-of-variable method cannot be used to find the p.d.f. of Y. The function u(x) = (x – 1)2 needs to be either always increasing or always decreasing for 0 < x < 4 in order to use the change-of-variables method, and neither of these is true. Use the distribution function method to find the p.d.f. of Y. The space of Y = (X – 1)2 is {y : 0 < y < 9}. If 0 < y < 9, the distribution function for Y is G(y) = P(Y  y) = P((X – 1)2  y) = P(–y  X – 1  y) if 0 < y < 1 P(–1  X – 1  y) if 1 < y < 9

If 0 < y < 1, P(–y  X – 1  y) = P(1–y  X  1+y) = —— dx = 4x x — = 2 (1+y)1/2 – (1 –y)1/2 ———————— 2 f(x) dx = 1–y 1–y x = 1–y If 1 < y < 9, P(–1  X – 1  y) = P(0  X  1+y) = 1+y 1+y 1+y 1 —— dx = 4x x — = 2 (1+y)1/2 ———— 2 f(x) dx = x = 0

The distribution function of Y is if y < 0 if 0  y < 1 if 1  y < 9 if 9  y (1+y)1/2 – (1 –y)1/2 ———————— 2 G(y) = (1+y)1/2 ———— 2 1 The p.d.f. of Y is G (y) = 1 1 —————— + —————— 8 (1+y)1/2 y 8 (1–y)1/2 y if 0 < y  1 g(y) = if 1 < y < 9 1 —————— 8 (1+y)1/2 y

6. (a) (i) Suppose X is a random variable with p.d.f. f(x) = x–1/2/4 if 0 < x < 4, and let Y = 1 / X . Do #6 and #7 for homework! Find the p.d.f. of Y = 1 / X by using the distribution function method, X has p.d.f. f(x) = x–1/2/4 if 0 < x < 4 . The space of Y = 1 / X is {y : 1/4 < y }. If 1/4 < y , the distribution function for Y is G(y) = P(Y  y) = P(1 / X  y) = P(X  1 / y) =  4 4 1 —— dx = 4x x — = 2 1 1 – —— 2y f(x) dx = x = 1 / y 1 / y 1 / y if y < 1/4 if 1/4  y The distribution function of Y is G(y) = 1 1 – —— 2y 1 — if 1/4 < y 4y3/2 The p.d.f. of Y is g(y) = G (y) =

(ii) the change-of-variable method. X has p.d.f. f(x) = x–1/2/4 if 0 < x < 4 . Y = 1 / X u(x) = 1 / x is for 0 < x < 4 . decreasing The space of Y is {y : u(4) < y < u(0)} = {y : 1/4 < y}. y = u(x) = 1 / x x = u–1(y) = 1 / y d — u–1(y) = dy – 1 / y2 d f(u–1(y)) — u–1(y) = dy 1 —— | – 1 / y2 | = 41/y The p.d.f. of Y is g(y) = 1 — if 1/4 < y 4y3/2

(b) (i) Find the p.d.f. of Y = X by using the distribution function method, X has p.d.f. f(x) = x–1/2/4 if 0 < x < 4 . The space of Y = X is {y : 0 < y < 2}. If 0 < y < 2, the distribution function for Y is G(y) = P(Y  y) = y2 y2 y2 1 —— dx = 4x x — = 2 y — 2 P(X  y) = P(X  y2) = f(x) dx = –  x = 0 The distribution function of Y is The p.d.f. of Y is g(y) = G (y) = if y < 0 if 0  y < 2 if 2  y 1 — if 0 < y < 2 2 y — 2 G(y) = 1 We recognize that Y has a distribution. U(0,2)

(ii) the change-of-variable method. X has p.d.f. f(x) = x–1/2/4 if 0 < x < 4 . The space of Y is {y : u(0) < y < u(4)} = {y : 0 < y < 2}. y = u(x) = x x = u–1(y) = y2 d — u–1(y) = dy 2y d f(u–1(y)) — u–1(y) = dy 1 —— | 2y | = 4y2 The p.d.f. of Y is g(y) = 1 — if 0 < y < 2 2

7. Suppose X is a random variable with p.d.f. f(x) = x–1/2/4 if 0 < x < 4. Find the p.d.f. of Y = X – 2 . 1 ———— if – 2 < y < 2 4 y + 2 The p.d.f. of Y is g(y) =