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) .

2. 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)

3. 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:

4. 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) =

5. 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.

6. 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)

7. (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

8. (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)

9. (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

10. (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.

11. (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

12. 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).

13. 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

14. 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.

15. (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 .

16. (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.

17. 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

18. 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
 U 
 U 
X =
 U 
 U 
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 .
 U 
 U 
 U 
Y =
 U  .
Go to Class Exercise #2(d).

19. (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
 U 
W =
 U 
 U 

20. (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
 U 
 U 
(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

21. 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

22. 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.

23. 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

24. 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

25. 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) =
—————— + ——————
8 (1+y)1/2 y (1–y)1/2 y
if 0 < y  1
g(y) =
if 1 < y < 9 1
——————
8 (1+y)1/2 y

26. 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) =

27. (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

28. (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)

29. (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

30. 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) =