APPENDIX B
SOLUTIONS TO PROBLEMS
B.1 Before the student takes the SAT exam, we do not know – nor can we predict with certainty
– what the score will be. The actual score depends on numerous factors, many of which we
cannot even list, let alone know ahead of time. (The student’s innate ability, how the student
feels on exam day, and which particular questions were asked, are just a few.) The eventual SAT
score clearly satisfies the requirements of a random variable.
B.2 (i) P(X ≤ 6) = P[(X – 5)/2 ≤ (6 – 5)/2] = P(Z ≤ .5) ≈ .692, where Z denotes a Normal (0,1)
random variable. [We obtain P(Z ≤ .5) from Table G.1.]
(ii) P(X > 4) = P[(X – 5)/2 > (4 – 5)/2] = P(Z > −.5) = P(Z ≤ .5) ≈ .692.
(iii) P(|X – 5| > 1) = P(X – 5 > 1) + P(X – 5 < –1) = P(X > 6) + P(X < 4) ≈ (1 – .692) + (1 –
.692) = .616, where we have used answers from parts (i) and (ii).
B.3 (i) Let Yit be the binary variable equal to one if fund i outperforms the market in year t. By
assumption, P(Yit = 1) = .5 (a 50-50 chance of outperforming the market for each fund in each
year). Now, for any fund, we are also assuming that performance relative to the market is
independent across years. But then the probability that fund i outperforms the market in all 10
years, P(Yi1 = 1,Yi2 = 1, … , Yi,10 = 1), is just the product of the probabilities: P(Yi1 = 1) ⋅ P(Yi2 =
1) … P(Yi,10 = 1) = (.5)10 = 1/1024 (which is slightly less than .001). In fact, if we define a
binary random variable Yi such that Yi = 1 if and only if fund i outperformed the market in all 10
years, then P(Yi = 1) = 1/1024.
(ii) Let X denote the number of funds out of 4,170 that outperform the market in all 10 years.
Then X = Y1 + Y2 + … + Y4,170. If we assume that performance relative to the market is
independent across funds, then X has the Binomial (n,θ) distribution with n = 4,170 and θ =
1/1024. We want to compute P(X ≥ 1) = 1 – P(X = 0) = 1 – P(Y1 = 0, Y2 = 0, …, Y4,170 = 0) = 1 –
P(Y1 = 0)⋅ P(Y2 = 0)⋅⋅⋅P(Y4,170 = 0) = 1 – (1023/1024)4170 ≈ .983. This means, if performance
relative to the market is random and independent across funds, it is almost certain that at least
one fund will outperform the market in all 10 years.
(iii) Using the Stata command Binomial(4170,5,1/1024), the answer is about .385. So there
is a nontrivial chance that at least five funds will outperform the market in all 10 years.
B.4 We want P(X ≥.6). Because X is continuous, this is the same as P(X > .6) = 1 – P(X ≤ .6) =
F(.6) = 3(.6)2 – 2(.6)3 = .648. One way to interpret this is that almost 65% of all counties have
an elderly employment rate of .6 or higher.
B.5 (i) As stated in the hint, if X is the number of jurors convinced of Simpson’s innocence, then
X ~ Binomial(12,.20). We want P(X ≥ 1) = 1 – P(X = 0) = 1 – (.8)12 ≈ .931.
236
, (ii) Above, we computed P(X = 0) as about .069. We need P(X = 1), which we obtain from
(B.14) with n = 12, θ = .2, and x = 1: P(X = 1) = 12⋅ (.2)(.8)11 ≈ .206. Therefore, P(X ≥ 2) ≈ 1 –
(.069 + .206) = .725, so there is almost a three in four chance that the jury had at least two
members convinced of Simpson’s innocence prior to the trial.
3 3 3 3
3
B.6 E(X) = ∫ xf ( x)dx = ∫ x[(1/ 9) x ]dx = (1/9) ∫ x dx . But ∫ x dx = (1/4)x4 |
2 3 3
0
= 81/4.
0 0 0 0
Therefore, E(X) = (1/9)(81/4) = 9/4, or 2.25 years.
B.7 In eight attempts the expected number of free throws is 8(.74) = 5.92, or about six free
throws.
B.8 The weights for the two-, three-, and four-credit courses are 2/9, 3/9, and 4/9, respectively.
Let Yj be the grade in the jth course, j = 1, 2, and 3, and let X be the overall grade point average.
Then X = (2/9)Y1 + (3/9)Y2 + (4/9)Y3 and the expected value is E(X) = (2/9)E(Y1) + (3/9)E(Y2) +
(4/9)E(Y3) = (2/9)(3.5) + (3/9)(3.0) + (4/9)(3.0) = (7 + 9 + 12)/9 ≈ 3.11.
B.9 If Y is salary in dollars then Y = 1000 ⋅ X, and so the expected value of Y is 1,000 times the
expected value of X, and the standard deviation of Y is 1,000 times the standard deviation of X.
Therefore, the expected value and standard deviation of salary, measured in dollars, are $52,300
and $14,600, respectively.
B.10 (i) E(GPA|SAT = 800) = .70 + .002(800) = 2.3. Similarly, E(GPA|SAT = 1,400) = .70 +
.002(1400) = 3.5. The difference in expected GPAs is substantial, but the difference in SAT
scores is also rather large.
(ii) Following the hint, we use the law of iterated expectations. Since E(GPA|SAT) = .70 +
.002 SAT, the (unconditional) expected value of GPA is .70 + .002 E(SAT) = .70 + .002(1100) =
2.9.
237
, APPENDIX C
SOLUTIONS TO PROBLEMS
C.1 (i) This is just a special case of what we covered in the text, with n = 4: E(Y ) = µ and
Var(Y ) = σ2/4.
(ii) E(W) = E(Y1)/8 + E(Y2)/8 + E(Y3)/4 + E(Y4)/2 = µ[(1/8) + (1/8) + (1/4) + (1/2)] = µ(1 +
1 + 2 + 4)/8 = µ, which shows that W is unbiased. Because the Yi are independent,
Var(W) = Var(Y1)/64 + Var(Y2)/64 + Var(Y3)/16 + Var(Y4)/4
= σ2[(1/64) + (1/64) + (4/64) + (16/64)] = σ2(22/64) = σ2(11/32).
(iii) Because 11/32 > 8/32 = 1/4, Var(W) > Var(Y ) for any σ2 > 0, so Y is preferred to W
because each is unbiased.
C.2 (i) E(Wa) = a1E(Y1) + a2E(Y2) + … + anE(Yn) = (a1 + a2 + … + an)µ. Therefore, we must
have a1 + a2 + … + an = 1 for unbiasedness.
(ii) Var(Wa) = a12 Var(Y1) + a22 Var(Y2) + … + an2 Var(Yn) = ( a12 + a22 + … + an2 )σ2.
(iii) From the hint, when a1 + a2 + … + an = 1 – the condition needed for unbiasedness of Wa
– we have 1/n ≤ a12 + a22 + … + an2 . But then Var( Y ) = σ2/n ≤ σ2( a12 + a22 + … + an2 ) =
Var(Wa).
C.3 (i) E(W1) = [(n – 1)/n]E(Y ) = [(n – 1)/n]µ, and so Bias(W1) = [(n – 1)/n]µ – µ = –µ/n.
Similarly, E(W2) = E(Y )/2 = µ/2, and so Bias(W2) = µ/2 – µ = –µ/2. The bias in W1 tends to
zero as n → ∞, while the bias in W2 is –µ/2 for all n. This is an important difference.
(ii) plim(W1) = plim[(n – 1)/n] ⋅ plim(Y ) = 1 ⋅ µ = µ. plim(W2) = plim(Y )/2 = µ/2. Because
plim(W1) = µ and plim(W2) = µ/2, W1 is consistent whereas W2 is inconsistent.
(iii) Var(W1) = [(n – 1)/n]2Var(Y ) = [(n – 1)2/n3]σ2 and Var(W2) = Var(Y )/4 = σ2/(4n).
(iv) Because Y is unbiased, its mean squared error is simply its variance. On the other hand,
MSE(W1) = Var(W1) + [Bias(W1)]2 = [(n – 1)2/n3]σ2 + µ2/n2. When µ = 0, MSE(W1) = Var(W1) =
[(n – 1)2/n3]σ2 < σ2/n = Var(Y ) because (n – 1)/n < 1. Therefore, MSE(W1) is smaller than
Var(Y ) for µ close to zero. For large n, the difference between the two estimators is trivial.
C.4 (i) Using the hint, E(Z|X) = E(Y/X|X) = E(Y|X)/X = θX/X = θ. It follows by Property CE.4,
the law of iterated expectations, that E(Z) = E(θ) = θ.
238
, (ii) This follows from part (i) and the fact that the sample average is unbiased for the
population average: write
n n
W1 = n −1 ∑ (Yi / X i ) = n −1 ∑ Z i ,
i =1 i =1
where Zi = Yi/Xi. From part (i), E(Zi) = θ for all i.
(iii) In general, the average of the ratios, Yi/Xi, is not the ratio of averages, W2 = Y / X . (This
non-equivalence is discussed a bit on page 676.) Nevertheless, W2 is also unbiased, as a simple
application of the law of iterated expectations shows. First, E(Yi|X1,…,Xn) = E(Yi|Xi) under
random sampling because the observations are independent. Therefore, E(Yi|X1,…,Xn) = θ X i
and so
n n
E(Y | X 1 ,..., X n ) = n −1 ∑ E(Yi | X 1 ,..., X n ) = n −1 ∑θ X i
i =1 i =1
n
= θ n −1 ∑ X i = θ X .
i =1
Therefore, E(W2 | X 1 ,..., X n ) = E(Y / X | X 1 ,..., X n ) = θ X / X = θ , which means that W2 is actually
unbiased conditional on ( X 1 ,..., X n ) , and therefore also unconditionally unbiased.
(iv) For the n = 17 observations given in the table – which are, incidentally, the first 17
observations in the file CORN.RAW – the point estimates are w1 = .418 and w2 = 120.43/297.41
= .405. These are pretty similar estimates. If we use w1, we estimate E(Y|X = x) for any x > 0 as
E(Y | X = x) = .418 x. For example, if x = 300 then the predicted yield is .418(300) = 125.4.
C.5 (i) While the expected value of the numerator of G is E(Y ) = θ, and the expected value of
the denominator is E(1 – Y ) = 1 – θ, the expected value of the ratio is not the ratio of the
expected value.
(ii) By Property PLIM.2(iii), the plim of the ratio is the ratio of the plims (provided the plim
of the denominator is not zero): plim(G) = plim[Y /(1 – Y )] = plim(Y )/[1 – plim(Y )] = θ/(1 –
θ) = γ.
C.6 (i) H0: µ = 0.
(ii) H1: µ < 0.
(iii) The standard error of y is s / n = 466.4/30 ≈ 15.55. Therefore, the t statistic for
testing H0: µ = 0 is t = y /se( y ) = –32.8/15.55 ≈ –2.11. We obtain the p-value as P(Z ≤ –2.11),
where Z ~ Normal(0,1). These probabilities are in Table G.1: p-value = .0174. Because the p-
239
SOLUTIONS TO PROBLEMS
B.1 Before the student takes the SAT exam, we do not know – nor can we predict with certainty
– what the score will be. The actual score depends on numerous factors, many of which we
cannot even list, let alone know ahead of time. (The student’s innate ability, how the student
feels on exam day, and which particular questions were asked, are just a few.) The eventual SAT
score clearly satisfies the requirements of a random variable.
B.2 (i) P(X ≤ 6) = P[(X – 5)/2 ≤ (6 – 5)/2] = P(Z ≤ .5) ≈ .692, where Z denotes a Normal (0,1)
random variable. [We obtain P(Z ≤ .5) from Table G.1.]
(ii) P(X > 4) = P[(X – 5)/2 > (4 – 5)/2] = P(Z > −.5) = P(Z ≤ .5) ≈ .692.
(iii) P(|X – 5| > 1) = P(X – 5 > 1) + P(X – 5 < –1) = P(X > 6) + P(X < 4) ≈ (1 – .692) + (1 –
.692) = .616, where we have used answers from parts (i) and (ii).
B.3 (i) Let Yit be the binary variable equal to one if fund i outperforms the market in year t. By
assumption, P(Yit = 1) = .5 (a 50-50 chance of outperforming the market for each fund in each
year). Now, for any fund, we are also assuming that performance relative to the market is
independent across years. But then the probability that fund i outperforms the market in all 10
years, P(Yi1 = 1,Yi2 = 1, … , Yi,10 = 1), is just the product of the probabilities: P(Yi1 = 1) ⋅ P(Yi2 =
1) … P(Yi,10 = 1) = (.5)10 = 1/1024 (which is slightly less than .001). In fact, if we define a
binary random variable Yi such that Yi = 1 if and only if fund i outperformed the market in all 10
years, then P(Yi = 1) = 1/1024.
(ii) Let X denote the number of funds out of 4,170 that outperform the market in all 10 years.
Then X = Y1 + Y2 + … + Y4,170. If we assume that performance relative to the market is
independent across funds, then X has the Binomial (n,θ) distribution with n = 4,170 and θ =
1/1024. We want to compute P(X ≥ 1) = 1 – P(X = 0) = 1 – P(Y1 = 0, Y2 = 0, …, Y4,170 = 0) = 1 –
P(Y1 = 0)⋅ P(Y2 = 0)⋅⋅⋅P(Y4,170 = 0) = 1 – (1023/1024)4170 ≈ .983. This means, if performance
relative to the market is random and independent across funds, it is almost certain that at least
one fund will outperform the market in all 10 years.
(iii) Using the Stata command Binomial(4170,5,1/1024), the answer is about .385. So there
is a nontrivial chance that at least five funds will outperform the market in all 10 years.
B.4 We want P(X ≥.6). Because X is continuous, this is the same as P(X > .6) = 1 – P(X ≤ .6) =
F(.6) = 3(.6)2 – 2(.6)3 = .648. One way to interpret this is that almost 65% of all counties have
an elderly employment rate of .6 or higher.
B.5 (i) As stated in the hint, if X is the number of jurors convinced of Simpson’s innocence, then
X ~ Binomial(12,.20). We want P(X ≥ 1) = 1 – P(X = 0) = 1 – (.8)12 ≈ .931.
236
, (ii) Above, we computed P(X = 0) as about .069. We need P(X = 1), which we obtain from
(B.14) with n = 12, θ = .2, and x = 1: P(X = 1) = 12⋅ (.2)(.8)11 ≈ .206. Therefore, P(X ≥ 2) ≈ 1 –
(.069 + .206) = .725, so there is almost a three in four chance that the jury had at least two
members convinced of Simpson’s innocence prior to the trial.
3 3 3 3
3
B.6 E(X) = ∫ xf ( x)dx = ∫ x[(1/ 9) x ]dx = (1/9) ∫ x dx . But ∫ x dx = (1/4)x4 |
2 3 3
0
= 81/4.
0 0 0 0
Therefore, E(X) = (1/9)(81/4) = 9/4, or 2.25 years.
B.7 In eight attempts the expected number of free throws is 8(.74) = 5.92, or about six free
throws.
B.8 The weights for the two-, three-, and four-credit courses are 2/9, 3/9, and 4/9, respectively.
Let Yj be the grade in the jth course, j = 1, 2, and 3, and let X be the overall grade point average.
Then X = (2/9)Y1 + (3/9)Y2 + (4/9)Y3 and the expected value is E(X) = (2/9)E(Y1) + (3/9)E(Y2) +
(4/9)E(Y3) = (2/9)(3.5) + (3/9)(3.0) + (4/9)(3.0) = (7 + 9 + 12)/9 ≈ 3.11.
B.9 If Y is salary in dollars then Y = 1000 ⋅ X, and so the expected value of Y is 1,000 times the
expected value of X, and the standard deviation of Y is 1,000 times the standard deviation of X.
Therefore, the expected value and standard deviation of salary, measured in dollars, are $52,300
and $14,600, respectively.
B.10 (i) E(GPA|SAT = 800) = .70 + .002(800) = 2.3. Similarly, E(GPA|SAT = 1,400) = .70 +
.002(1400) = 3.5. The difference in expected GPAs is substantial, but the difference in SAT
scores is also rather large.
(ii) Following the hint, we use the law of iterated expectations. Since E(GPA|SAT) = .70 +
.002 SAT, the (unconditional) expected value of GPA is .70 + .002 E(SAT) = .70 + .002(1100) =
2.9.
237
, APPENDIX C
SOLUTIONS TO PROBLEMS
C.1 (i) This is just a special case of what we covered in the text, with n = 4: E(Y ) = µ and
Var(Y ) = σ2/4.
(ii) E(W) = E(Y1)/8 + E(Y2)/8 + E(Y3)/4 + E(Y4)/2 = µ[(1/8) + (1/8) + (1/4) + (1/2)] = µ(1 +
1 + 2 + 4)/8 = µ, which shows that W is unbiased. Because the Yi are independent,
Var(W) = Var(Y1)/64 + Var(Y2)/64 + Var(Y3)/16 + Var(Y4)/4
= σ2[(1/64) + (1/64) + (4/64) + (16/64)] = σ2(22/64) = σ2(11/32).
(iii) Because 11/32 > 8/32 = 1/4, Var(W) > Var(Y ) for any σ2 > 0, so Y is preferred to W
because each is unbiased.
C.2 (i) E(Wa) = a1E(Y1) + a2E(Y2) + … + anE(Yn) = (a1 + a2 + … + an)µ. Therefore, we must
have a1 + a2 + … + an = 1 for unbiasedness.
(ii) Var(Wa) = a12 Var(Y1) + a22 Var(Y2) + … + an2 Var(Yn) = ( a12 + a22 + … + an2 )σ2.
(iii) From the hint, when a1 + a2 + … + an = 1 – the condition needed for unbiasedness of Wa
– we have 1/n ≤ a12 + a22 + … + an2 . But then Var( Y ) = σ2/n ≤ σ2( a12 + a22 + … + an2 ) =
Var(Wa).
C.3 (i) E(W1) = [(n – 1)/n]E(Y ) = [(n – 1)/n]µ, and so Bias(W1) = [(n – 1)/n]µ – µ = –µ/n.
Similarly, E(W2) = E(Y )/2 = µ/2, and so Bias(W2) = µ/2 – µ = –µ/2. The bias in W1 tends to
zero as n → ∞, while the bias in W2 is –µ/2 for all n. This is an important difference.
(ii) plim(W1) = plim[(n – 1)/n] ⋅ plim(Y ) = 1 ⋅ µ = µ. plim(W2) = plim(Y )/2 = µ/2. Because
plim(W1) = µ and plim(W2) = µ/2, W1 is consistent whereas W2 is inconsistent.
(iii) Var(W1) = [(n – 1)/n]2Var(Y ) = [(n – 1)2/n3]σ2 and Var(W2) = Var(Y )/4 = σ2/(4n).
(iv) Because Y is unbiased, its mean squared error is simply its variance. On the other hand,
MSE(W1) = Var(W1) + [Bias(W1)]2 = [(n – 1)2/n3]σ2 + µ2/n2. When µ = 0, MSE(W1) = Var(W1) =
[(n – 1)2/n3]σ2 < σ2/n = Var(Y ) because (n – 1)/n < 1. Therefore, MSE(W1) is smaller than
Var(Y ) for µ close to zero. For large n, the difference between the two estimators is trivial.
C.4 (i) Using the hint, E(Z|X) = E(Y/X|X) = E(Y|X)/X = θX/X = θ. It follows by Property CE.4,
the law of iterated expectations, that E(Z) = E(θ) = θ.
238
, (ii) This follows from part (i) and the fact that the sample average is unbiased for the
population average: write
n n
W1 = n −1 ∑ (Yi / X i ) = n −1 ∑ Z i ,
i =1 i =1
where Zi = Yi/Xi. From part (i), E(Zi) = θ for all i.
(iii) In general, the average of the ratios, Yi/Xi, is not the ratio of averages, W2 = Y / X . (This
non-equivalence is discussed a bit on page 676.) Nevertheless, W2 is also unbiased, as a simple
application of the law of iterated expectations shows. First, E(Yi|X1,…,Xn) = E(Yi|Xi) under
random sampling because the observations are independent. Therefore, E(Yi|X1,…,Xn) = θ X i
and so
n n
E(Y | X 1 ,..., X n ) = n −1 ∑ E(Yi | X 1 ,..., X n ) = n −1 ∑θ X i
i =1 i =1
n
= θ n −1 ∑ X i = θ X .
i =1
Therefore, E(W2 | X 1 ,..., X n ) = E(Y / X | X 1 ,..., X n ) = θ X / X = θ , which means that W2 is actually
unbiased conditional on ( X 1 ,..., X n ) , and therefore also unconditionally unbiased.
(iv) For the n = 17 observations given in the table – which are, incidentally, the first 17
observations in the file CORN.RAW – the point estimates are w1 = .418 and w2 = 120.43/297.41
= .405. These are pretty similar estimates. If we use w1, we estimate E(Y|X = x) for any x > 0 as
E(Y | X = x) = .418 x. For example, if x = 300 then the predicted yield is .418(300) = 125.4.
C.5 (i) While the expected value of the numerator of G is E(Y ) = θ, and the expected value of
the denominator is E(1 – Y ) = 1 – θ, the expected value of the ratio is not the ratio of the
expected value.
(ii) By Property PLIM.2(iii), the plim of the ratio is the ratio of the plims (provided the plim
of the denominator is not zero): plim(G) = plim[Y /(1 – Y )] = plim(Y )/[1 – plim(Y )] = θ/(1 –
θ) = γ.
C.6 (i) H0: µ = 0.
(ii) H1: µ < 0.
(iii) The standard error of y is s / n = 466.4/30 ≈ 15.55. Therefore, the t statistic for
testing H0: µ = 0 is t = y /se( y ) = –32.8/15.55 ≈ –2.11. We obtain the p-value as P(Z ≤ –2.11),
where Z ~ Normal(0,1). These probabilities are in Table G.1: p-value = .0174. Because the p-
239
