Mathematical Expectation

Chapter 4 Mathematical Expectation

1 R a R√ a 2 −y 2 1 a 2 −y 2 a 2

4.1 E(X) = 2

4.2 E(X) = x f (x) = (0)(27/64) + (1)(27/64) + (2)(9/64) + (3)(1/64) = 3/4.

x=0

4.3 µ = E(X) = (20)(1/5) + (25)(3/5) + (30)(1/5) = 25 cents.

4.4 Assigning wrights of 3w and w for a head and tail, respectively. We obtain P (H) = 3/4 and P (T ) = 1/4. The sample space for the experiment is S = {HH, HT, T H, T T }. Now if X represents the number of tails that occur in two tosses of the coin, we have

P (X = 0) = P (HH) = (3/4)(3/4) = 9/16, P (X = 1) = P (HT ) + P (T H) = (2)(3/4)(1/4) = 3/8, P (X = 2) = P (T T ) = (1/4)(1/4) = 1/16.

The probability distribution for X is then

f (x) 9/16 3/8 1/16

from which we get µ = E(X) = (0)(9/16) + (1)(3/8) + (2)(1/16) = 1/2.

4.5 µ = E(X) = (0)(0.41) + (1)(0.37) + (2)(0.16) + (3)(0.05) + (4)(0.01) = 0.88.

4.6 µ = E(X) = ($7)(1/12)+($9)(1/12)+($11)(1/4)+($13)(1/4)+($15)(1/6)+($17)(1/6) = $12.67.

4.7 Expected gain = E(X) = (4000)(0.3) + (−1000)(0.7) = $500.

4.8 Let X = profit. Then

µ = E(X) = (250)(0.22) + (150)(0.36) + (0)(0.28) + (−150)(0.14) = $88.

46 Chapter 4 Mathematical Expectation

4.9 Let c = amount to play the game and Y = amount won.

5−c3−c −c

f (y) 2/13 2/13 9/13

E(Y ) = (5 − c)(2/13) + (3 − c)(2/13) + (−c)(9/13) = 0. So, 13c = 16 which implies

c = $1.23. P

4.10 µ X = xg(x) = (1)(0.17) + (2)(0.5) + (3)(0.33) = 2.16, P µ Y =

yh(y) = (1)(0.23) + (2)(0.5) + (3)(0.27) = 2.04.

4.11 For the insurance of $200,000 pilot, the distribution of the claim the insurance company would have is as follows:

Claim Amount $200,000 $100,000 $50,000

0.01 0.1 0.888 So, the expected claim would be

f (x)

($200, 000)(0.002) + ($100, 000)(0.01) + ($50, 000)(0.1) + ($0)(0.888) = $6, 400. Hence the insurance company should charge a premium of $6, 400 + $500 = $6, 900.

4.12 E(X) = 0 2x(1 − x) dx = 1/3. So, (1/3)($5, 000) = $1, 667.67.

4.13 E(X) = ln 4

π 0 1+x 2 dx = π .

R 1 2x(x+2)

4.14 E(X) = 0 5 dx = 8 15 .

4.15 E(X) =

0 x dx + 1 x(2 − x) dx = 1. Therefore, the average number of hours per year is (1)(100) = 100 hours.

4.16 P (X 1 +X 2 = 1) = P (X 1 = 1, X 2 = 0) + P (X 1 = 0, X 2 = 1)

4.17 The probability density function is,

f (x) 1/6 1/2 1/3 g(x) 25 169 361

µ 2 g(X) = E[(2X + 1) ] = (25)(1/6) + (169)(1/2) + (361)(1/3) = 209.

4.18 E(X 2 ) = (0)(27/64) + (1)(27/64) + (4)(9/64) + (9)(1/64) = 9/8.

4.19 Let Y = 1200X − 50X 2 be the amount spent.

Solutions for Exercises in Chapter 4

y = g(x)

µ 2 Y = E(1200X − 50X ) = (0)(1/10) + (1150)(3/10) + (2200)(2/5) + (3150)(1/5) = $1, 855.

R ∞ 2x/3

4.20 E[g(X)] = E(e

e 0 −x/3 0 dx = 3.

4.21 E(X 2 )= 0 2x 2 (1 − x) dx = 1 6 . Therefore, the average profit per new automobile is (1/6)($5000.00) = $833.33.

4.22 E(Y ) = E(X + 4) = 1

0 32(x + 4) (x+4) 3 dx = 8 days.

4.23 (a) E[g(X, Y )] = E(XY )=

xy f (x, y)

(b) µ X = E(X) = (2)(0.40) + (4)(0.60) = 3.20, µ Y = E(Y ) = (1)(0.25) + (3)(0.50) + (5)(0.25) = 3.00.

4.24 (a) E(X 2

Y − 2XY ) = (x y − 2xy)f(x, y) = (1 − 2)(18/70) + (4 − 4)(18/70) +

x=0 y=0

· · · + (8 − 8)(3/70) = −3/7. x

0 1 2 (b) g(x) 5/70 30/70 30/70 5/70

h(y) 15/70 40/70 15/70 µ X = E(X) = (0)(5/70) + (1)(30/70) + (2)(30/70) + (3)(5/70) = 3/2, µ Y = E(Y ) = (0)(15/70) + (1)(40/70) + (2)(15/70) = 1.

4.25 µ X+Y = E(X + Y ) = (x + y)f (x, y) = (0 + 0)(1/55) + (1 + 0)(6/55) + · · · + (0 +

0 0 4xy x 2 +y 2 dx dy = 3 0 [y(1 + y ) −y ] dy = 8(2 3/2 − 1)/15 = 0.9752.

4.26 E(Z) = E( 4 X +Y )=

4.27 E(X) = 1

2000 0 x exp(−x/2000) dx = 2000 0 y exp(−y) dy = 2000.

4.28 (a) The density function is shown next.

f(x)

48 Chapter 4 Mathematical Expectation

2 R 26.25 1 2 2 (b) E(X) = 5 23.75 x dx = 5 (26.25 − 23.75 ) = 25.

(c) The mean is exactly in the middle of the interval. This should not be surprised due to the symmetry of the density at 25.

4.29 (a) The density function is shown next

f(x) 1

(b) µ = E(X) =

3x −3

1 dx = 2 .

4.30 E(Y ) =

4.31 (a) µ = E(Y ) = 5 1 4 5

0 (1 − y) dy = 6 .

(b) P (Y > 1/6) =

1/6 5(1 − y) dy = − (1 − y) | 1/6 = (1 − 1/6) = 0.4019.

4.32 (a) A histogram is shown next.

f(x)f(x) 0.2

(b) µ = (0)(0.41) + (1)(0.37) + (2)(0.16) + (3)(0.05) + (4)(0.01) = 0.88. (c) E(X 2 ) = (0) 2 (0.41) + (1) 2 (0.37) + (2) 2 (0.16) + (3) 2 (0.05) + (4) 2 (0.01) = 1.62. (d) V ar(X) = 1.62 − 0.88 2 = 0.8456.

= E[(X − µ) ]=

4.33 µ = $500. So, σ 2 2

2 2 (x − µ) 2 f (x) = (−1500) (0.7) + (3500) (0.3) =

Solutions for Exercises in Chapter 4

4.34 µ = (−2)(0.3) + (3)(0.2) + (5)(0.5) = 2.5 and

E(X 2 2 ) = (−2) 2 (0.3) + (3) (0.2) + (5) 2 (0.5) = 15.5.

2 2 So, σ 2 = E(X )−µ = 9.25 and σ = 3.041.

2 2 2 2 2 E(X 2 ) = (2) (0.01) + (3) (0.25) + (4) (0.4) + (5) (0.3) + (6) (0.04) = 17.63. So, σ 2 = 17.63 − 4.11 2 = 0.74.

and E(X 2 ) = (0) 2 (0.4) + (1) 2 (0.3) + (2) 2 (0.2) + (3) 2 (0.1) = 2.0.

2 So, σ 2 = 2.0 − 1.0 = 1.0.

4.37 It is know µ = 1/3.

So, E(X )= 0 2x (1 − x) dx = 1/6 and σ = 1/6 − (1/3) = 1/18. So, in the actual profit, the variance is 1

4.38 It is known µ = 8/15.

Since E(X 1 2 R

0 5 x 2 (x + 2) dx = 11 30 , then σ 2 = 11/30 − (8/15) 2 = 37/450.

4.39 It is known µ = 1. Since E(X 2

)= 0 x 2 dx + 1 x 2 2 (2 − x) dx = 7/6, then σ 2 = 7/6 − (1) = 1/6.

4.40 µ = E[g(X)] = 0 (3x 2 + 4) g(X) 2x+4 5 dx = 1 5 0 (6x 3 + 12x 2 + 8x + 16) dx = 5.1.

2 2 R 1 2 2 So, σ 2x+4

= E[g(X) − µ] = 0 (3x + 4 − 5.1)

5 dx

= (9x 4 2 0 2x+4 − 6.6x + 1.21) 5 dx = 0.83.

4.41 It is known µ

g(X) = E[(2X + 1) ] = 209. Hence σ 2

P = [(2X + 1) 2 g(X) 2 − 209] g(x)

2 2 = (25 − 209) 2 √ (1/6) + (169 − 209) (1/2) + (361 − 209) (1/3) = 14, 144. So, σ g(X) =

4.42 It is known µ 2 g(X) = E(X ) = 1/6. Hence

2 R 1 2 1 2 σ g(X) = 0 2x − 6 (1 − x) dx = 7/180.

Y = E(3X − 2) = 4 0 (3x − 2)e

−x/4

dx = 10. So

Y = E{[(3X − 2) − 10] }= 4 0 (x − 4) e −x/4 dx = 144.

4.44 E(XY ) =

xyf (x, y) = (1)(1)(18/70) + (2)(1)(18/70)

+ (3)(1)(2/70) + (1)(2)(9/70) + (2)(2)(3/70) = 9/7; P P µ X =

xf (x, y) = (0)f (0, 1) + (0)f (0, 2) + (1)f (1, 0) + · · · + (3)f(3, 1) = 3/2,

xy

and µ Y = 1.

So, σ XY = E(XY ) − µ X µ Y = 9/7 − (3/2)(1) = −3/14.

50 Chapter 4 Mathematical Expectation P

4.45 µ X = xg(x) = 2.45, µ Y =

yh(y) = 2.10, and

P y P E(XY ) =

xyf (x, y) = (1)(0.05) + (2)(0.05) + (3)(0.10) + (2)(0.05)

+ (4)(0.10) + (6)(0.35) + (3)(0) + (6)(0.20) + (9)(0.10) = 5.15. So, σ XY = 5.15 − (2.45)(2.10) = 0.005.

3 4.46 From previous exercise, k = 2

3 R , with

, and g(x) = k 20x 98000 +

50 R 50

µ X = E(X) =

30 xg(x) dx = k 30 20x + 3 x dx = 40.8163.

Similarly, µ Y = 40.8163. On the other hand, R

50 R 50

E(XY ) = k 30 30 xy(x 2 +y 2 ) dy dx = 1665.3061.

Hence, σ 2 XY = E(XY ) − µ X µ Y = 1665.3061 − (40.8163) = −0.6642.

4.47 g(x) = 3 0 (x + 2y) dy = 3 (x + 1, for 0 < x < 1, so µ X = 3 0 x(x + 1) dx = 9 ;

2 R 1 2 1 2 R 1 1 11 h(y) = 3 0 (x + 2y) dx = 3 2 + 2y , so µ Y = 3 0 y 2 + 2y dy = 18 ; and

2 R 1 R 1 1 E(XY ) = 3 0 0 xy(x + 2y) dy dx = .

So, σ XY = E(XY ) − µ X µ Y = 3 − 9 18 = −0.0062.

4.48 Since σ XY = Cov(a + bX, X) = bσ 2

and σ 2 Y =b 2 σ 2 X , then

ρ= σ XY

= |b| = sign of b.

σ bσ = X b

Hence ρ = 1 if b > 0 and ρ = −1 if b < 0.

4.49 E(X) = (0)(0.41) + (1)(0.37) + (2)(0.16) + (3)(0.05) + (4)(0.01) = 0.88 and E(X 2 ) = (0) 2 (0.41) + (1) 2 (0.37) + (2) 2 (0.16) + (3) 2 (0.05) + (4) 2 (0.01) = 1.62.

So, V ar(X) = 1.62 − 0.88 = 0.8456 and σ = 0.8456 = 0.9196.

4.50 E(X) = 2 0 x(1 − x) dx = 2 2 − 3 = and

R 1 x 2 x 3 )=2 4 x

E(X 2

= 0 1 (1 − x) dx = 2 3 − 4 . Hence,

V ar(X) = 6 − 3 = 18 , and σ = 1/18 = 0.2357.

4.51 Previously we found µ = 4.11 and σ 2 = 0.74, Therefore, µ g(X) = E(3X − 2) = 3µ − 2 = (3)(4.11) − 2 = 10.33 and σ g(X) = 9σ 2 = 6.66.

4.52 Previously we found µ = 1 and σ 2 = 1. Therefore,

µ g(X) = E(5X + 3) = 5µ + 3 = (5)(1) + 3 = 8 and σ g(X) = 25σ 2 = 25.

4.53 Let X = number of cartons sold and Y = profit. We can write Y = 1.65X + (0.90)(5 − X) − 6 = 0.75X − 1.50. Now E(X) = (0)(1/15) + (1)(2/15) + (2)(2/15) + (3)(3/15) + (4)(4/15) + (5)(3/15) = 46/15, and E(Y ) = (0.75)E(X) − 1.50 = (0.75)(46/15) − 1.50 = $0.80.

4.54 µ X = E(X) = 4 0 xe −x/4 dx = 4.

Therefore, µ Y = E(3X − 2) = 3E(X) − 2 = (3)(4) − 2 = 10.

2 1 R Since E(X ∞ )= 2 −x/4

4 0 x e dx = 32, therefore, σ X = E(X )−µ X = 32 − 16 = 16.

Hence σ 2 Y = 9σ 2 X = (9)(16) = 144.

Solutions for Exercises in Chapter 4

4.55 E(X) = (−3)(1/6) + (6)(1/2) + (9)(1/3) = 11/2,

E(X 2 2 (1/6) + (6) ) = (−3) 2 (1/2) + (9) 2 (1/3) = 93/2. So,

2 E[(2X + 1) 2 ] = 4E(X ) + 4E(X) + 1 = (4)(93/2) + (4)(11/2) + 1 = 209.

4.56 Since E(X) = 0 x 2 dx + 1 x(2 − x) dx = 1, and

2 R 1 3 R E(X 2 )= 2

0 x 2 dx + 1 x (2 − x) dx = 7/6,then E(Y ) = 60E(X 2 ) + 39E(X) = (60)(7/6) + (39)(1) = 109 kilowatt hours.

2 4.57 The equations E[(X − 1) 2 ] = 10 and E[(X − 2) ] = 6 may be written in the form:

2 E(X 2 ) − 2E(X) = 9, E(X ) − 4E(X) = 2.

Solving these two equations simultaneously we obtain

E(X) = 7/2, 2 and E(X ) = 16. Hence µ = 7/2 and σ 2 = 16 − (7/2) 2 = 15/4.

4.58 E(X) = (2)(0.40) + (4)(0.60) = 3.20, and E(Y ) = (1)(0.25) + (3)(0.50) + (5)(0.25) = 3. So,

(a) E(2X − 3Y ) = 2E(X) − 3E(Y ) = (2)(3.20) − (3)(3.00) = −2.60. (b) E(XY ) = E(X)E(Y ) = (3.20)(3.00) = 9.60.

2 2 2 4.59 E(2XY 2 −X Y ) = 2E(XY ) − E(X Y ). Now,

E(XY )=

xy f (x, y) = (1)(1) (3/14) = 3.14, and

x=0 y=0

E(X 2 Y)=

x 2 yf (x, y) = (1) 2 (1)(3/14) = 3.14.

x=0 y=0

Therefore, E(2XY 2 −X 2 Y ) = (2)(3/14) − (3/14) = 3/14.

4.60 Using µ = 60 and σ = 6 and Chebyshev’s theorem

P (µ − kσ < X < µ + kσ) ≥ 1 −

since from µ + kσ = 84 we obtain k = 4. So, P (X < 84) ≥ P (36 < X < 84) ≥ 1 − 1

4 2 = 0.9375. Therefore,

P (X ≥ 84) ≤ 1 − 0.9375 = 0.0625.

Since 1000(0.0625) = 62.5, we claim that at most 63 applicants would have a score as

84 or higher. Since there will be 70 positions, the applicant will have the job.

4.61 µ = 900 hours and σ = 50 hours. Solving µ − kσ = 700 we obtain k = 4. So, using Chebyshev’s theorem with P (µ − 4σ < X < µ + 4σ) ≥ 1 − 1/4 2 = 0.9375,

we obtain P (700 < X < 1100) ≥ 0.9375. Therefore, P (X ≤ 700) ≤ 0.03125.

52 Chapter 4 Mathematical Expectation

4.62 µ = 52 and σ = 6.5. Solving µ + kσ = 71.5 we obtain k = 3. So,

1 P (µ − 3σ < X < µ + 3σ) ≥ 1 − 2 = 0.8889,

which is

P (32.5 < X < 71.5) ≥ 0.8889. we obtain P (X > 71.5) < 1−0.8889 2 = 0.0556 using the symmetry.

4.63 n = 500, µ = 4.5 and σ = 2.8733. Solving µ + k(σ/ 500) = 5 we obtain

4.64 σ 2 Z =σ 2 −2X+4Y −3 = 4σ 2 X + 16σ 2 Y = (4)(5) + (16)(3) = 68.

2 2 2 4.65 σ 2 Z =σ −2X+4Y −3 = 4σ X + 16σ Y − 16σ XY = (4)(5) + (16)(3) − (16)(1) = 52.

4.66 (a) P (6 < X < 18) = P [12 − (2)(3) < X < 12 + (2)(3)] ≥ 1 − 1

1 (b) P (3 < X < 21) = P [12 − (3)(3) < X < 12 + (3)(3)] ≥ 1 − 8

4.67 (a) P (|X − 10| ≥ 3) = 1 − P (|X − 10| < 3)

h i = 1 − P [10 − (3/2)(2) < X < 10 + (3/2)(2)] ≤ 1 − 1 1−

4 (b) P (|X − 10| < 3) = 1 − P (|X − 10| ≥ 3) ≥ 1 − 5

(c) P (5 < X < 15) = P [10 − (5/2)(2) < X < 10 + (5/2)(2)] ≥ 1 − 1 (5/2) 2 = 21 25 .

(d) P (|X − 10| ≥ c) ≤ 0.04 implies that P (|X − 10| < c) ≥ 1 − 0.04 = 0.96.

Solving 0.96 = 1 − 1 k 2 we obtain k = 5. So, c = kσ = (5)(2) = 10.

4.68 µ = E(X) = 6 x 2 2

)=6 x 0 3 (1 − x) dx = 0.5, E(X 0 (1 − x) dx = 0.3, which imply

2 σ 2 = 0.3 − (0.5) = 0.05 and σ = 0.2236. Hence,

P (µ − 2σ < X < µ + 2σ) = P (0.5 − 0.4472 < X < 0.5 + 0.4472) Z 0.9472

= P (0.0528 < X < 0.9472) = 6

x(1 − x) dx = 0.9839,

compared to a probability of at least 0.75 given by Chebyshev’s theorem.

4.69 It is easy to see that the expectations of X and Y are both 3.5. So, (a) E(X + Y ) = E(X) + E(Y ) = 3.5 + 3.5 = 7.0.

(b) E(X − Y ) = E(X) − E(Y ) = 0.

Solutions for Exercises in Chapter 4

53 (c) E(XY ) = E(X)E(Y ) = (3.5)(3.5) = 12.25.

4.70 E(Z) = E(XY ) = E(X)E(Y ) =

0 2 16xy(y/x ) dx dy = 8/3.

3 2 3 4.71 E[g(X, Y )] = E(X/Y 2 +X Y ) = E(X/Y ) + E(X Y ).

3 R 2 R 1 2x(x+2y)

E(X/Y )= 1 0 7y 3 dx dy = 2 1 1 7 15 1 3y 3 + y 2 dy = 84 ;

2 R 2 R 1 2x 2 y(x+2y)

Hence, E[g(X, Y )] = 84 + 252 = 63 .

4.72 µ =µ = 3.5. σ 2 =σ 2 = [(1) 2 + (2) 2 2 2 X 35 Y X Y + · · · + (6) ](1/6) − (3.5) = 12 .

2 2 (a) σ 175

2X−Y = 4σ X +σ Y = 12 ;

2 2 (b) σ 175

X+3Y −5 =σ X + 9σ Y = 6 .

4.73 (a) µ = 1 2 2 2 1 2 2 5 2 0 x dx = 2.5, σ = E(X )−µ = x x dx − 2.5 = 2.08. √

So, σ = σ 2 = 1.44.

(b) By Chebyshev’s theorem, P [2.5 − (2)(1.44) < X < 2.5 + (2)(1.44)] = P (−0.38 < X < 5.38) ≥ 0.75.

Using integration, P (−0.38 < X < 5.38) = 1 ≥ 0.75; P [2.5 − (3)(1.44) < X < 2.5 + (3)(1.44)] = P (−1.82 < X < 6.82) ≥ 0.89. Using integration, P (−1.82 < X < 6.82) = 1 ≥ 0.89.

4.74 P = I 2 R with R = 50, µ I = E(I) = 15 and σ 2 I = V ar(I) = 0.03.

2 2 2 E(P ) = E(I 2 R) = 50E(I ) = 50[V ar(I) + µ

I ] = 50(0.03 + 15 ) = 11251.5. If we use the approximation formula, with g(I) = I 2 ,g ′ (I) = 2I and g ′′ (I) = 2, we obtain,

E(P ) ≈ 50 g(µ I )+2

h ∂g(i) i 2

Since V ar[g(I)] ≈ 2

I , we obtain

∂i

i=µ

2 2 2 2 2 2 V ar(P ) = 50 2 V ar(I ) = 50 (2µ

I ) σ I = 50 (30) (0.03) = 67500.

P ∞ x−1

4.75 For 0 < a < 1, since g(a) = 1 a =

1−a ,g (a) =

xa

2 and

(1−a)

x=0

x=1

g ′′ (a) =

x(x − 1)a 2 =

x−2

(1−a) 3 .

x=2

54 Chapter 4 Mathematical Expectation

(a) E(X) = (3/4)

x(1/4) 2 = (3/4)(1/4) x(1/4) = (3/16)[1/(1 − 1/4) ]

= 1/3, and E(Y ) = E(X) = 1/3.

E(X 2 ) − E(X) = E[X(X − 1)] = (3/4) x x(x − 1)(1/4)

x=2

x(x − 1)(1/4) x−2 = (3/4 3 )[2/(1 − 1/4) 3 ] = 2/9.

x=2

So, V ar(X) = E(X 2 ) − [E(X)] 2 = [E(X 2 ) − E(X)] + E(X) − [E(X)] 2 2/9 + 1/3 − (1/3) 2 = 4/9, and V ar(Y ) = 4/9.

(b) E(Z) = E(X) + E(Y ) = (1/3) + (1/3) = 2/3, and

V ar(Z) = V ar(X + Y ) = V ar(X) + V ar(Y ) = (4/9) + (4/9) = 8/9, since X and Y are independent (from Exercise 3.79).

4.76 (a) g(x) =

(x +y ) dy = (3x + 1) for 0 < x < 1 and

h(y) = 2 (3y 2 + 1) for 0 < y < 1.

Since f (x, y) 6= g(x)h(y), X and Y are not independent.

(b) E(X + Y ) = E(X) + E(Y ) = 2E(X) = 0 x(3x + 1) dx = 3/4 + 1/2 = 5/4.

3 R 1 R 1 2 2 3 R 1 1 y 2 E(XY ) = 2 0 0 xy(x +y ) dx dy = 2 0 y 4 + 2 dy

(c) V ar(X) = E(X 2 2

) − [E(X)] = 1 2 2 5 2 7 25 2 73 0 x (3x + 1) dx − 8 = 15 − 64 = 960 , and

3 5 2 V ar(Y ) = 1 960 . Also, Cov(X, Y ) = E(XY ) − E(X)E(Y ) = 8 − 8 =− 64 .

73 1 (d) V ar(X + Y ) = V ar(X) + V ar(Y ) + 2Cov(X, Y ) = 2 29 960 −2 64 = 240 . R ∞

4.77 (a) E(Y ) = 0 ye −y/4 dy = 4.

(b) E(Y )=

y e 0 −y/4 dy = 32 and V ar(Y ) = 32 − 4 = 16.

4.78 (a) The density function is shown next.

f(x)

R 8 1 2 2 15 (b) E(Y ) = 7 y dy = 2 [8 −7 ]= 2 = 7.5,

7 y dy = 3 [8 −7 ]= 3 , and V ar(Y ) = 3 − 2 = 12 .

4.79 Using the exact formula, we have

y E(e 8 )= e dy = e |

Solutions for Exercises in Chapter 4

Using the approximation, since g(y) = e y , so g ′′ (y) = e . Hence, using the approxima- tion formula,

The approximation is very close to the true value.

4.80 Using the exact formula, E(Z 2

)= 7 e 2y

dy = 8

2 e 2y | 7 = 3841753.12. Hence,

2 V ar(Z) = E(Z 2 ) − [E(Z)] = 291091.3.

Using the approximation formula, we have

V ar(e ) = (e )

V ar(Y ) =

The approximation is not so close to each other. One reason is that the first order approximation may not always be good enough.

4.81 Define I 1 = {x i | |x i − µ| < kσ} and I 2 = {x i | |x i − µ| ≥ kσ}. Then

2 2 X 2 X 2 X 2 σ = E[(X − µ) ]=

f (x i )=k σ P (|X − µ| ≥ kσ),

x i ∈I 2 x i ∈I 2

which implies

1 P (|X − µ| ≥ kσ) ≤ .

Hence, P (|X − µ| < kσ) ≥ 1 − 1

4.82 E(XY ) = 0 0 xy(x+y) dx dy = 3 , E(X) = 0 0 x(x+y) dx dy = 12 and E(Y ) = 12 . Therefore, σ 2

XY = E(XY ) − µ X µ Y = 1 7 3 1 − 12 =− 144 .

4.83 E(Y − X) = 0 0 2(y − x) dx dy = 0 y dy = 3 . Therefore, the average amount of kerosene left in the tank at the end of each day is (1/3)(1000) = 333 liters.

4.84 (a) E(X) =

0 5 e dx = 50, so V ar(X) = 50 − 5 = 25, and σ = 5. (c) E[(X + 5) 2 2 2 ] = E{[(X − 5) + 10] 2 } = E[(X − 5) ] + 10 + 20E(X − 5) = V ar(X) + 100 = 125.

−x/5

R 1 R 1−y 2 2 R 1 2 3 2

4.85 E(XY ) = 24 0 0 x y dx dy = 8 0 y (1 − y) dy = 15 ,

R 1 R 1−y 2 2 R 1 R 1−y 2 2 µ X = 24 0 0 x y dx dy = 5 and µ Y = 24 0 0 xy dx dy = 5 . Therefore,

XY = E(XY ) − µ X µ Y = 15 − 5 =− 75 .

56 Chapter 4 Mathematical Expectation

R 1 R 1−y

4.86 E(X + Y ) = 4

0 0 24(x + y)xy dx dy = 5 .

4.87 (a) E(X) = 0 900 e −x/900 dx = 900 hours.

(b) E(X 2

0 900 e −x/900 dx = 1620000 hours . (c) V ar(X) = E(X 2 2 ) − [E(X)] 2 = 810000 hours and σ = 900 hours.

2 4.88 It is known g(x) = 1

3 (x + 1), for 0 < x < 1, and h(y) = 3 (1 + 4y), for 0 < y < 1.

R 1 2 5 R 1 1 11 (a) µ X = 0 3 x(x + 1) dx = 9 and µ Y = 0 3 y(1 + 4y) dy = 18 .

1 (b) E[(X + Y )/2] = 7

2 [E(X) + E(Y )] = 12 .

4.89 Cov(aX, bY ) = E[(aX − aµ X )(bY − bµ Y )] = abE[(X − µ X )(Y − µ Y )] = abCov(X, Y ).

4.90 It is known µ = 900 and σ = 900. For k = 2,

P (µ − 2σ < X < µ + 2σ) = P (−900 < X < 2700) ≥ 0.75

using Chebyshev’s theorem. On the other hand,

P (µ − 2σ < X < µ + 2σ) = P (−900 < X < 2700) = 1 − e −3 = 0.9502. For k = 3, Chebyshev’s theorem yields

P (µ − 3σ < X < µ + 3σ) = P (−1800 < X < 3600) ≥ 0.8889,

while P (−1800 < X < 3600) = 1 − e −4 = 0.9817. R 1 16y

4.91 g(x) = 0 x 3 dy = x 3 , for x > 2, with µ X = 2 x 2 dx = − x 2 = 4, R ∞ 16y

R 1 2 h(y) = 2 x 3

dx = − 2

8y ∞

x 2 2 = 2y, for 0 < y < 1, with µ Y = 0 2y = 3 , and

R ∞ R 1 16y 2 16 R ∞ 1 8

E(XY ) = 2 0 x 2 dy dx = 3 2 x 2 dx =

. Hence,

σ XY = E(XY ) − µ X µ Y = 3 − (4) 3 = 0.

4.92 Since σ 1 = 1, σ

X = 5 and σ Y = 3, we have ρ = XY XY σ X σ Y = √ (5)(3) = 0.2582.

4.93 (a) From Exercise 4.37, we have σ 2 = 1/18, so σ = 0.2357. (b) Also, µ X = 1/3 from Exercise 4.12. So,

P (µ − 2σ < X < µ + 2σ) = P [1/3 − (2)(0.2357) < X < 1/3 + (2)(0.2357)]

Z 0.8047

= P (0 < X < 0.8047) =

2(1 − x) dx = 0.9619.

Using Chebyshev’s theorem, the probability of this event should be larger than

0.75, which is true.

R 1 (c) P (profit > $500) = P (X > 0.1) = 0.1 2(1 − x) = 0.81.

Solutions for Exercises in Chapter 4

4.94 Since g(0)h(0) = (0.17)(0.23) 6= 0.10 = f(0, 0), X and Y are not independent.

4.95 E(X) = (−5000)(0.2) + (10000)(0.5) + (30000)(0.3) = $13, 000.

4.96 (a) f (x) = 3

x (0.15) x (0.85) 3−x , for x = 0, 1, 2, 3.

f (x) 0.614125 0.325125 0.057375 0.003375 (b) E(X) = 0.45.

(c) E(X 2 ) = 0.585, so V ar(X) = 0.585 − 0.45 2 = 0.3825. (d) P (X ≤ 2) = 1 − P (X = 3) = 1 − 0.003375 = 0.996625.

(e) 0.003375. (f) Yes.

4.97 (a) E(X) = (−$15k)(0.05)+($15k)(0.15)+($25k)(0.30)+($40k)(0.15)+($50k)(0.10)+ ($100k)(0.05) + ($150k)(0.03) + ($200k)(0.02) = $33.5k.

(b) E(X 2 2 p

) = 2, 697, 500, 000 dollars . So, σ = E(X 2 ) − [E(X)] 2 = $39.689k.

3 R 50

4.98 (a) E(X) =

3 x(50 2 4×50 2 −50 −x ) dx = 0.

R 50

(b) E(X 2 )= 3

4×50 3 −50 x 2 (50 2 −x 2 ) dx = 500.

(c) σ = E(X 2 ) − [E(X)] 2 = 500 − 0 = 22.36.

4.99 (a) The marginal density of X is

f X 1 (x 1 )

(b) The marginal density of Y is

f X 2 (x 2 )

(c) Given X 2 = 3, the conditional density function of X 1 is f (x 1 , 3)/0.15. So

X 2 (x 2 ) 15 5 15 5 15

(d) E(X 1 ) = (0)(0.13) + (1)(0.21) + (2)(0.31) + (3)(0.23) + (4)(0.12) = 2. (e) E(X 2 ) = (0)(0.10) + (1)(0.30) + (2)(0.39) + (3)(0.15) + (4)(0.06) = 1.77.

7 1 1 1 1 18 (f) E(X 6

1 |X 2 = 3) = (0) 15 + (1) 5 + (2) 15 + (3) 5 + (4) 15 = 15 = 5 = 1.2. (g) E(X 2 1 ) = (0) 2 (0.13) + (1) 2 (0.21) + (2) 2 (0.31) + (3) 2 (0.23) + (4) 2 (0.12) = 5.44.

So, σ X 1 = E(X 1 ) − [E(X 1 )] 2 =

4.100 (a) The marginal densities of X and Y are, respectively,

58 Chapter 4 Mathematical Expectation

h(y) 0.26 0.35 0.39 The conditional density of X given Y = 2 is

g(x) 0.2

4 5 f 30

X|Y =2 (x|2) 39 39 39

(b) E(X) = (0)(0.2) + (1)(0.32) + (2)(0.48) = 1.28,

2 2 2 E(X 2 ) = (0) (0.2) + (1) (0.32) + (2) (0.48) = 2.24, and

V ar(X) = 2.24 − 1.28 2 = 0.6016.

(c) E(X|Y = 2) = (1) 125

39 + (2) 39 = 39 and E(X |Y = 2) = (1) 39 + (2) 39 = 39 . So,

V ar(X) = 50

4.101 The profit is 8X + 3Y − 10 for each trip. So, we need to calculate the average of this quantity. The marginal densities of X and Y are, respectively,

0 1 2 3 4 5 g(x) 0.34 0.32 0.34

h(y) 0.05 0.18 0.15 0.27 0.19 0.16 So, E(8X +3Y −10) = (8)[(1)(0.32)+(2)(0.34)]+(3)[(1)(0.18)+(2)(0.15)+(3)(0.27)+

(4)(0.19) + (5)(0.16)] − 10 = $6.55. P k h ∂h(x ,x ,...,x ) i 2 1 2 k

4.102 Using the approximation formula, V ar(Y ) ≈ 2