Functions of Random Variables

Chapter 7 Functions of Random Variables

7.1 From y = 2x − 1 we obtain x = (y + 1)/2, and given x = 1, 2, and 3, then

g(y) = f [(y + 1)/2] = 1/3,

for y = 1, 3, 5.

7.2 From y = x 2 , x = 0, 1, 2, 3, we obtain x = √y,

3− √ y

g(y) = f ( y) = √

fory = 0, 1, 4, 9.

7.3 The inverse functions of y 1 =x 1 +x 2 and y 2 =x 1 −x 2 are x 1 = (y 1 +y 2 )/2 and x 2 = (y 1 −y 2 )/2. Therefore,

2 , 2 ,2−y 1 4 3 12 where y 1 = 0, 1, 2, y 2 = −2, −1, 0, 1, 2, y 2 ≤y 1 and y 1 +y 2 = 0, 2, 4.

7.4 Let W = X 2 . The inverse functions of y = x 1 x 2 and w = x 2 are x 1 = y/w and x 2 = w, where y/w = 1, 2. Then

g(y, w) = (y/w)(w/18) = y/18, y = 1, 2, 3, 4, 6; w = 1, 2, 3, y/w = 1, 2. In tabular form the joint distribution g(y, w) and marginal h(y) are given by

g(y, w)

w 2 2/18

h(y)

86 Chapter 7 Functions of Random Variables The alternate solutions are:

P (Y = 1) = f (1, 1) = 1/18, P (Y = 2) = f (1, 2) + f (2, 1) = 2/18 + 2/18 = 2/9, P (Y = 3) = f (1, 3) = 3/18 = 1/6, P (Y = 4) = f (2, 2) = 4/18 = 2/9, P (Y = 6) = f (2, 3) = 6/18 = 1/3.

7.5 The inverse function of y = −2 ln x is given by x = e −y/2 from which we obtain

|J| = | − e −y/2 /2| = e /2. Now,

−y/2

g(y) = f (e y/2

)|J| = e −y/2 /2,

y > 0,

which is a chi-squared distribution with 2 degrees of freedom.

7.6 The inverse function of y = 8x 3 is x = y 1/3 /2, for 0 < y < 8 from which we obtain |J| = y −2/3 /6. Therefore,

g(y) = f (y

/2)|J| = 2(y −1/3 /2)(y /6) = y ,

7.7 To find k we solve the equation k v e −bv 2 dv = 1. Let x = bv 2 , then dx = 2bv dv

x −1/2

and dv = √ 2 b dx. Then the equation becomes

kΓ(3/2)

x 3/2−1 e −x dx = 1, or

4b Hence k = 3/2 Γ(1/2) .

Now the inverse function of w = mv /2 is v = 2w/m, for w > 0, from which we √ obtain |J| = 1/ 2mw. It follows that

4b 3/2

g(w) = f ( 2w/m)|J| =

(2w/m)e −2bw/m

w 3/2 3/2−1 e −(2b/m)w ,

(m/2b) Γ(3/2) for w > 0, which is a gamma distribution with α = 3/2 and β = m/2b.

7.8 (a) The inverse of y = x 2 √

is x = √y, for 0 < y < 1, from which we obtain |J| = 1/2 y. Therefore,

g(y) = f ( y)/2 y=y y)|J| = 2(1 − −1/2 − 1, 0 < y < 1.

(b) P (Y < 1) = 0 (y −1/2

− 1) dy = (2y 1/2 − y)

7.9 (a) The inverse of y = x + 4 is x = y − 4, for y > 4, from which we obtain |J| = 1. Therefore,

g(y) = f (y − 4)|J| = 32/y 3 , y > 4.

Solutions for Exercises in Chapter 7

7.10 (a) Let W = X. The inverse functions of z = x + y and w = x are x = w and y = z − w, 0 < w < z, 0 < z < 1, from which we obtain

Then g(w, z) = f (w, z − w)|J| = 24w(z − w), for 0 < w < z and 0 < z < 1. The marginal distribution of Z is

1 (z) =

24(z − w)w dw = 4z ,

0 < z < 1.

R 3/4 (b) P (1/2 < Z < 3/4) = 4 1/2 z 3 dz = 65/256.

7.11 The amount of kerosene left at the end of the day is Z = Y − X. Let W = Y . The inverse functions of z = y − x and w = y are x = w − z and y = w, for 0 < z < w and

0 < w < 1, from which we obtain

g(w, z) = g(w − z, w) = 2, 0 < z < w, 0 < w < 1,

and the marginal distribution of Z is

h(z) = 2

dw = 2(1 − z),

0 < z < 1.

7.12 Since X 1 and X 2 are independent, the joint probability distribution is

1 +x 2 f (x )

1 ,x 2 ) = f (x 1 )f (x 2 )=e −(x

, x 1 > 0, x 2 > 0. The inverse functions of y 1 =x 1 +x 2 and y 2 =x 1 /(x 1 +x 2 ) are x 1 =y 1 y 2 and

x 2 =y 1 (1 − y 2 ), for y 1 > 0 and 0 < y 2 < 1, so that ∂x 1 /∂y 1 ∂x 1 /∂y 2 y 2 y 1

J=

∂x 2 /∂y 1 ∂x 2 /∂y 2 1−y 2 −y 1

Then, g(y 1 ,y 2 ) = f (y 1 y 2 ,y 1 (1 − y 2 ))|J| = y 1 e −y 1 , for y 1 > 0 and 0 < y 2 < 1. Therefore,

y 1 e dy 1 = Γ(2) = 1, 0<y 2 < 1.

Since g(y 1 ,y 2 ) = g(y 1 )g(y 2 ), the random variables Y 1 and Y 2 are independent.

88 Chapter 7 Functions of Random Variables

7.13 Since I and R are independent, the joint probability distribution is

f (i, r) = 12ri(1 − i), 0 < i < 1, 0 < r < 1.

2 p Let V = R. The inverse functions of w = i r and v = r are i = w/v and r = v, for

w < v < 1 and 0 < w < 1, from which we obtain

∂i/∂w ∂i/∂v

J=

∂r/∂w ∂r/∂v

2 vw

Then,

1 p g(w, v) = f ( w/v, v)|J| = 12v w/v(1 − w/v) √

= 6(1 − w/v),

2 vw

for w < v < 1 and 0 < w < 1, and the marginal distribution of W is Z 1

h(w) = 6

(1 − w/v) dv = 6 (v − 2 wv) v=w = 6 + 6w − 12 w,

v=1

0 < w < 1.

7.14 The inverse functions of y = x 2 are given by x = √y and x

1 2 =− y from which we

obtain J 1 = 1/2√y and J 2 = 1/2√y. Therefore, √

1 + √y

1 1− y

g(y) = f ( y)|J 1 | + f(− y)|J 2 |=

7.15 The inverse functions of y = x are x 1 = √y, x 2 =− y for 0 < y < 1 and x 1 = √y

√ for 0 < y < 4. Now |J 1 | = |J 2 | = |J 3 | = 1/2 y, from which we get

2(√y + 1)

1 2(− y + 1)

g(y) = f ( y)|J 1 | + f(− y)|J 2 |=

g(y) = f ( y)|J 3 |=

7.16 Using the formula we obtain

β Γ(α + r) ∞ x α+r−1 e −x/β µ r = E(X )=

x e −x/β

Γ(α) since the second integrand is a gamma density with parametersα + r and β.

Solutions for Exercises in Chapter 7

7.17 The moment-generating function of X is

by summing the geometric series of k terms.

7.18 The moment-generating function of X is

1 − qe by summing an infinite geometric series. To find out the moments, we use

(1 − qe t ) 2 pe t + 2pqe 2t

7.19 The moment-generating function of a Poisson random variable is

µ(e µ=M t −1)+t

2 =M X (0) = µe −1)+t (µe + 1)

7.20 From M t (t) = e 4(e

X −1) we obtain µ = 6, σ 2 = 4, and σ = 2. Therefore,

P (µ − 2σ < X < µ + 2σ) = P (0 < X < 8) = p(x; 4) = 0.9489 − 0.0183 = 0.9306.

x=1

90 Chapter 7 Functions of Random Variables

7.21 Using the moment-generating function of the chi-squared distribution, we obtain

x f (x) dx + · · · = 1 + µt + µ 1 +···+µ r +···.

7.23 The joint distribution of X and Y is f X,Y (x, y) = e −x−y for x > 0 and y > 0. The inverse functions of u = x + y and v = x/(x + y) are x = uv and y = u(1 − v) with

vu J=

1 − v −u

f U,V (u, v) = ue −uv −u(1−v) ·e = ue −u ,

for u > 0 and 0 < v < 1.

(a) f U (u) =

dv = ue for u > 0, which is a gamma distribution with parameters 2 and 1.

R ∞ (b) f V (v) = 0 ue −u du = 1 for 0 < v < 1. This is a uniform (0,1) distribution.