5-2 (p. 159)
a.
x P(X|Y=30) 0 1/29 1 3/29 2 18/29 3 7/29E(X|Y=30) = (0)(1/29) + (1)(3/29) + (2)(18/29) + (3)(7/29)
b.
x P(X|Y=45) 0 2/37 1 6/37 2 21/37 3 8/37E(X|Y=45) = (0)(2/37) + (1)(6/37) + (2)(21/37) + (3)(8/37)
x P(X|Y=70) 0 5/34 1 10/34 2 15/34 3 4/34E(X|Y=70) = (0)(5/34) + (1)(10/34) + (2)(15/34) + (3)(4/34)
The average level of education increased as the age of the group decreased.
c. Education and age aren't independent, because P(X) is different for different values of Y.
5-8 (p. 162)
y
0 2 4
0 0.1 0.1 0
x 2 0.1 0.4 0.1
4 0 0.1 0.1
R = X^2 + Y^2
r p(r) 0 0.1 4 0.2 8 0.4 16 0 20 0.2 32 0.1a.E(R) = (0)(0.1) + (4)(0.2) + (8)(0.4) + (16)(0) + (20)(0.2) + (32)(0.1) = 11.2
b.E(R) = E(X^2 + Y^2)
E(R) = (0)(0.1) + (4)(0.1) + (16)(0) + (4)(0.1) + (8)(0.4) + (20)(0.1) + (16)(0) + (20)(0.1) + (32)(0.1)
E(R) = 0.4 + 0.4 + 3.2 + 2.0 + 2.0 + 3.2 = 11.2
c.Finding E(X) and E(Y) helps a lot with finding the expected value for the functions later.
E(X) = (2)(0.1) + (2)(0.4) + (2)(0.1) + (4)(0.1) + (4)(0.1)
E(X) = 0.2 + 0.8 + 0.2 + 0.4 = 2.0
E(Y) = (2)(0.1) + (2)(0.4) + (2)(0.1) + (4)(0.1) + (4)(0.1)
E(Y) = 0.2 + 0.8 + 0.2 + 0.4 + 0.4 = 2.0
(i) E[(X-2)(Y-2)] = E(XY) - 2E(Y) - 2E(X) + 4
Need to find E(XY). Also note that E[(X-2)(Y-2)] is Cov(X,Y), since E(X)=2 and E(Y)=2. Also, we can tell that covariance should be positive, because the non-zero probabilities tend to fall along the diagonal from upper left to lower right. In this case, combinations of X above its mean and Y below its mean (and vice-versa) actually have zero probability. (That is, P(X=0 and Y=4)=0 and P(X=4 and Y=0)=0.)
E(XY) = (4)(0.4) + (8)(0.1) + (8)(0.1) + (16)(0.1)
E(XY) = 1.6 + 0.8 + 0.8 + 1.6 = 4.8
E[(X-2)(Y-2)] = E(XY) - 2E(Y) - 2E(X) + 4 = 4.8-4-4+4 = 0.8
(ii) E[(X-2)(X-2)] = E(X^2) - 4E(X) + 4
Need to find E(X^2). Also note that E[(X-2)(X-2)] is Var(X), since E(X)=2.
E(X^2) = (4)(0.1) + (4)(0.4) + (4)(0.1) + (16)(0.1) + (16)(0.1)
E(X^2) = 0.4 + 1.6 + 0.4 + 1.6 + 1.6 = 5.6
E[(X-2)(X-2)] = E(X^2) - 4E(X) + 4 = 5.6-8+4 = 1.6
(iii) E(4X + 2Y) = 4E(X) + 2E(Y) = 8 + 4 = 12
5-14 (p. 168)
a.
E(H|X=2.5) = (0)(0.03) + (1)(0.12) + (2)(0.07) = 0.26
E(H|X=7.5) = (0)(0.02) + (1)(0.13) + (2)(0.11) = 0.35
E(H|X=12.5) = (0)(0.01) + (1)(0.13) + (2)(0.14) = 0.41
E(H|X=17.5) = (0)(0.01) + (1)(0.09) + (2)(0.14) = 0.37
b.
Cov(X, Y) = E(XY) - E(X)E(Y)
E(XY) = (2.5)(0.12) + (5)(0.07) + (7.5)(0.13) + (15)(0.11) + (12.5)(0.13) + (25)(0.14) + (17.5)(0.09) + (35)(0.14)
E(XY) = 14.875
y
0 1 2
2.5 0.03 0.12 0.07 | 0.22
x 7.5 0.02 0.13 0.11 | 0.26
12.5 0.01 0.13 0.14 | 0.28
17.5 0.01 0.09 0.14 | 0.24
------------------------
0.07 0.47 0.46
x p(x) y p(y)
2.5 0.22 0 0.07
7.5 0.26 1 0.47
12.5 0.28 2 0.46
17.5 0.24
E(X) = (2.5)(0.22) + (7.5)(0.26) + (12.5)(0.28) + (17.5)(0.24) = 10.2
E(Y) = (0)(0.07) + (1)(0.47) + (2)(0.46) = 1.39
E(X^2) = (2.5^2)(0.22) + (7.5^2)(0.26) + (12.5^2)(0.28) + (17.5^2)(0.24) = 133.25
E(Y^2) = (0^2)(0.07) + (1^2)(0.47) + (2^2)(0.46) = 2.31
Cov(X,Y) = E(XY) - E(X)E(Y) = 14.875 - (10.2)(1.39) = 0.697
Var(X) = E(X^2) - [E(X)]^2 = 133.25 - (10.2^2) = 29.21
Var(Y) = E(Y^2) - [E(Y)]^2 = 2.31 - (1.39^2) = 0.3779
Corr(X,Y) = rho = Cov(X,Y)/Sqrt[Var(X)Var(Y)] = 0.697/Sqrt[(29.21)(0.3779)] = 0.2098 (approx)
(i) True. Generally, H increases as X increase (although this isn't exactly true as X goes from 12.5 to 17.5).
(ii) True.
(iii) False. Corelation does not imply causation. It may be true that the rich tend to be happier than the poor, but we don't know that money actually causes people to be happier.
5-20 (p. 177)
a. P(X=3 and Z=1) = p(3,1) = 7/100 = 0.07
z
0 1
0 0.04 0.05
x 1 0.10 0.12
2 0.23 0.29
3 0.10 0.07
b.
P(X=3) = 0.17
x p(x) 0 0.09 1 0.22 2 0.52 3 0.17
c. P(X>=2) = 0.52 + 0.17 = 0.69
d. E(X) = (0)(0.09) + (1)(0.22) + (2)(0.52) + (3)(0.17) = 1.77
e.
z p(z) 0 0.47 1 0.53E(Z) = (0)(0.47) + (1)(0.53) = 0.53
5-21
a.
X P(X|Z=0) 0 4/47 1 10/47 2 23/47 3 10/47b. Education X is dependent on sex Z, because P(X) is different from P(X|Z=0).
c. E(X|Z=0) = (0)(4/47) + (1)(10/47) + (2)(23/47) + (3)(10/47) = 86/47 = 1.83 (approx)
d.
X P(X|Z=1) 0 5/53 1 12/53 2 29/53 3 7/53E(X|Z=1) = (0)(5/53) + (1)(12/53) + (2)(29/53) + (3)(7/53) = 91/53 = 1.72 (approx)
e. E(X) = 1.77 falls between E(X|Z=0)=86/47 and E(X|Z=1)=91/53.
E(X) = P(Z=0)E(X|Z=0) + P(Z=1)E(X|Z=1)
1.77 = (0.47)(86/47) + (0.53)(91/53)
1.77 = 1.77
5-22
a.
z
0 1
0 10 14
x 1 12 16
2 14 18
3 16 20
s p(s)
10 0.04
12 0.10
14 0.28
16 0.22
18 0.29
20 0.07
b. E(S) = (10)(0.04) + (12)(0.10) + (14)(0.28) + (16)(0.22) + (18)(0.29) + (20)(0.07) = 15.66c. E(S) = 10 + 2E(X) + 4E(Z) = 10 + (2)(1.77) + (4)(0.53) = 15.66
d. Answers in parts b and c must agree with each other because we learned that:
E(aX + b) = aE(X) + b and E(aX + bY) = aE(X) + bE(Y)
5-23
a. E(10 + 10XZ) = 10 + 10E(XZ)
E(XZ) = (1)(0.12) + (2)(0.29) + (3)(0.07) = 0.91
E(10 + 10XZ) = 10 + (10)(0.91) = 19.1
b. E(10 + 3X - 2Z) = 10 + 3E(X) - 2E(Z) = 10 + (3)(1.77) - (2)(0.53) = 14.25
c. E[(2X+40)/5] = 0.4E(X) + 8 = (0.4)(1.77) + 8 = 8.708
d. E(10 + X^2) = 10 + E(X^2)
E(X^2) = (1)(0.22) + (4)(0.52) + (9)(0.17) = 0.22 + 2.08 + 1.53 = 3.83
E(10 + X^2) = 10 + 3.83 = 13.83