let x be a continuous type random variable with p.d.f. f(x) where f(x) > 0 for a < x < b,...
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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) .
Section 5.1
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) =
P(X ey) =
ey
–
f(x) dx =
ey
1
1— dx = x2
1– — = x
x = 1
ey
1 – e–y
The distribution function of Y is G(y) =if y < 0
if 0 y
0
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)
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:
Section 5.1
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)] =
a
f(x) dx =
u–1(y)
F(u–1(y)) – F(a) = G(y) .
We now find the p.d.f. of Y to be g(y) = G /(y) =
F(u–1(y)) – F(a) = d—dy
d[u–1(y)]f(u–1(y)) ————
dy
Look at Class Exercise #1(a)(ii) while doing this derivation.
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)] =
b
f(x) dx =
u–1(y)
F(b) – F(u–1(y)) = G(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)) ————
dyF(b) – F(u–1(y)) =
This must be negative since u and u–1 are both decreasing functions.
Look at Class Exercise #1(b)(ii) while doing this derivation.
In either case, we find that the p.d.f of Y is
g(y) =
d[u–1(y)]f(u–1(y)) ————
dy
u(a) < y < u(b) (in Case 1)if u(b) < y < u(a) (in Case 2)
Return to Class Exercise #1(a)
(ii) the change-of-variable method.
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
The p.d.f. of Y is g(y) = d
f(u–1(y)) — u–1(y) = dy
1—— | ey | = (ey)2
e–y if 0 < y
Return to Class Exercise #1(a)
(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) =
P(1 / X y) =
P(X 1 / y) =
1 / y
f(x) dx =
1 / y
1— dx = x2
1– — = x
x = 1 / y
y
The distribution function of Y is G(y) =
if y < 0
if 0 y < 1
if 1 y
0
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
The p.d.f. of Y is g(y) = d
f(u–1(y)) — u–1(y) = dy
1 —— | – 1 / y2 | =(1 / y)2
1 if 0 < y < 1
(c)
(i)
Find the p.d.f. of Y = X2 by using
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) =
P(X y) =
y
–
f(x) dx =
y
1
1— dx = x2
1– — = x
x = 1
y
11 – — y
The distribution function of Y is G(y) =if y < 1
if 1 y
0 11 – — y
The p.d.f. of Y is g(y) = G (y) = 1— if 1 < y2y3/2
We find that Y has a distribution not in one of the families we studied.
Do part (c) for homework!
(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——2y
The p.d.f. of Y is g(y) = d
f(u–1(y)) — u–1(y) = dy
1 1—— —— =(y)2 2y
1— if 1 < y2y3/2
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 ;
For 1 < x , we have F(x) = P(X x)
=
x
–
f(t) dt =
x
1
1— dt = t2
1– — = t
t = 1
x
11 – —
x
The distribution function for X is F(x) =if x < 1
if 1 x
0 11 – — 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 calculatex1 = 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 ;
For 1 < y , we have P(Y y) =
y
–
g(t) dt =
y
1
2— dt = t3
1– — = t2
t = 1
y
11 – —
y2
The distribution function for Y is G(y) =if y < 1
if 1 y
0 11 – — 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 calculatey1 = (1 – u1)–1/2 , y2 = (1 – u2)–1/2 , y3 = (1 – u3)–1/2 , y4 = (1 – u4)–1/2 .
Skip to part (c), and do part (b) for homework!
(c) we can simulate 4 independent observations of a random variable W having a Weibull distribution with = 5/3 and = 1/27.
The distribution function for W is G(w) =if w 0
if 0 < w
0
1 – e–243w5/3
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 calculatew1 = [–ln(1 – u1)]3/5 / 27 , … , w4 = [–ln(1 – u4)]3/5 / 27.
The random variable W has p.d.f. g(w) = H /(w)e–H(w) if w > 0 , where H(w) = 243w5/3.
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) = (4v + 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 / 2w 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.
X =
2 if
3 if
5 if
7 if
0.0000 U 0.1374
0.1375 U 0.3249
0.3250 U 0.6124
0.6125 U 0.9999
Y =
1 if
2 if
3 if
4 if
.
.
.
0.0000 U 0.4999
0.5000 U 0.74990.7500 U 0.87490.8750 U 0.9374
.
.
.
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
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
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 ;
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
W =
2 if
3 if
6 if
0.0000 U 0.4999
0.5000 U 0.8332
0.8333 U 0.9999
Go to Class Exercise #2(d).
Do #2(d) for homework!
1 if
0 if
0.0000 U 0.6999
0.7000 U 0.9999
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
xkk = 1
(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
(f) we can simulate one observation of a random variable with a b(5,0.7) distribution.
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.
What type of distribution does X have?
Explain why the change-of-variable method cannot be used to find the p.d.f. of Y.
Use the distribution function method to find the p.d.f. of Y.
X has a distribution.U(–5 , 5)
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.
The space of Y = X2 is {y : 0 < y < 25}.
If 0 < y < 25, the distribution function for Y is
G(y) = P(Y y) =
P(X2 y) =
P(– y X y) =
y
–y
f(x) dx =
Skip to #5, and do #4 for homework!
y
– y
f(x) dx =
y
–y
1— dx =10
x— =10
x = –y
y
y— 5
The distribution function of Y is G(y) =
if y < 0
if 0 y < 25
if 25 y
0
y— 5
1
The p.d.f. of Y is g(y) = G (y) = 1—— if 0 < y < 2510y1/2
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.
Use the distribution function method 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.
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) =
1+y
1–y
f(x) dx =
1+y
1–y
1—— dx =4x
x— = 2
x = 1–y
1+y
(1+y)1/2 – (1 –y)1/2
———————— 2
If 1 < y < 9, P(–1 X – 1 y) = P(0 X 1+y) =
1+y
0
f(x) dx =
1+y
0
1—— dx =4x
x— = 2
x = 0
1+y
(1+y)1/2
————2
if 0 < y 1
g(y) =
if 1 < y < 9
The distribution function of Y is if y < 0
if 0 y < 1
if 1 y < 9
if 9 y
0
(1+y)1/2 – (1 –y)1/2
———————— 2
1
G(y) =
(1+y)1/2
———— 2
1 1—————— + ——————8 (1+y)1/2 y 8 (1–y)1/2 y
1——————8 (1+y)1/2 y
The p.d.f. of Y is G (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 .
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) =
1 / y
f(x) dx =
4
1 / y
1—— dx =4x
x— = 2x = 1 / y
4
11 – ——
2y
The distribution function of Y is G(y) =if y < 1/4
if 1/4 y
0 11 – —— 2y
The p.d.f. of Y is g(y) = G (y) = 1— if 1/4 < y4y3/2
Do #6 and #7 for homework!
(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
The p.d.f. of Y is g(y) = d
f(u–1(y)) — u–1(y) = dy
1 —— | – 1 / y2 | =41/y
1— if 1/4 < y4y3/2
(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) =
P(X y) =
P(X y2) =
y2
–
f(x) dx =
y2
0
1—— dx =4x
x— = 2x = 0
y2
y— 2
The distribution function of Y is
if y < 0
if 0 y < 2
if 2 y
0
y— 2
1
G(y) =
The p.d.f. of Y is g(y) = G (y) =
1— if 0 < y < 2 2
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
The p.d.f. of Y is g(y) = d
f(u–1(y)) — u–1(y) = dy
1 —— | 2y | =4y2
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 .
The p.d.f. of Y is g(y) = 1———— if – 2 < y < 24 y + 2