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Business Statistics: Midterm Solutions N Vera Chau The University of Chicago: Booth November 9, 2017 1 True False (2 points each) 1. F 2. T 3. F 4. F 5. T 6. F 7. T 8. F 9. T 10. F 2 Multiple Choice (2 points each) 1. c 2. d 3. d 4. 3,4,1,2 3 Long Answer Questions 3.1 a. (2 pts) P (r i+1 = -0.1) = 0.35 P (r i+1 = 0) = 0.3 P (r i+1 =0.1) = 0.35 1

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N Vera ChauThe University of Chicago: Booth

November 9, 2017

1 True False(2 points each)

1. F

2. T

3. F

4. F

5. T

6. F

7. T

8. F

9. T

10. F

2 Multiple Choice(2 points each)

1. c

2. d

3. d

4. 3,4,1,2

a. (2 pts)

P (ri+1 = −0.1) = 0.35

P (ri+1 = 0) = 0.3

P (ri+1 = 0.1) = 0.35

1

Business Stats Midterm Solutions N Vera Chau

b. (4 pts)

E [ri=1] = −0.1× 0.35 + 0.1× 0.35 = 0

V ar(ri+1) = 0.35× (−0.1− 0)2 + 0.35× (0.1− 0)2 = 0.007

c. (3 pts)

yes. You can verify that P (ri = x) = P (ri+1 = x) for every x.

d. (3 pts)

No.

P (r + i = 0, ri+1 = 0) = 0.15

P (ri = 0)P (ri+1 = 0) = 0.35× 0.35

e. (2 pts)

P (ri+1 = −0.1|ri = 0) =0.95

0.3= 0.1667

P (ri+1 = 0|ri = 0) =0.2

0.3= 0.667

P (ri+1 = 0.1|ri = 0) =0.05

0.3= 0.1667

f. (2 pts)

P (ri+1 = −0.1|ri = 0.1) =0.15

0.35= 0.4286

P (ri+1 = 0|ri = 0.1) =0.05

0.35= 0.1429

P (ri+1 = 0.1|ri = 0.1) =0.15

0.35= 0.4286

g. (2 pts)

P (ri+1 = −0.1|ri = −0.1) =0.15

0.35= 0.4286

P (ri+1 = 0|ri = −0.1) =0.05

0.35= 0.1429

P (ri+1 = 0.1|ri = −0.1) =0.15

0.35= 0.4286

h. (3 pts)

E [ri+1|ri = −0.1] = −0.1× 0.4286 + 0.1× 0.4286 = 0

E [ri+1|ri = 0] = −0.1× 0.1667 + 0.1× 0.1667 = 0

E [ri=1|ri = 0.1] = −0.1× 0.4286 + 0.1× 0.4286 = 0

i. (3 pts)

V ar(ri+1|ri = 0) = 0.1667× (−0.1)2 + 0.1667× (−0.1)2 = 0.0033

V ar(ri+1|ri = 0.1) = 0.4286× (−0.1)2 + 0.4286× (0.1)2 = 0.00857

November 9, 2017 Prof. Russell Fall, 2017

Business Stats Midterm Solutions N Vera Chau

j. (4 pts)

Variance is higher if the return moves in the previous period. In particular, when ri = 0, the variancein the next period is lower than when ri either decreases or increases.

k. (3 pts)

P (0.01)P (−.01|.01)P (.01| − .01)P (.01|.01)= 0.35× 0.42863 = 0.0276

l. (3 pts)

P (0)P (0|0)3 = 0.3× 0.6673 = 0.089

3.2a. (6 pts)

E [0.75 ∗A+ 0.25H] = 0.75(0.00016) + 0.25(0.00009) = 0.0001425

V ar(0.75A+ 0.25H) = 0.752(0.0018) + 0.252(0.0027) + 2(0.25 ∗ 0.75).0018 = 0.001856

b. (6 pts)

E [1.5A− 0.5H] = 1.5(0.00016)− 0.5(0.00009) = 0.000195

V ar(1.5A− 0.5H) = 1.52(0.0018) + (−0.5)2(0.0027)− 2(1.5 ∗ 0.5).0018 = 0.002025

c. (6 pts)

U1 = 0.0001425− 7(0.001856) = −0.01284

U2 = 0.000195− 7(0.002025) = −0.01398

Portfolio 1 is preferred.

3.3(3 points each)

a. 0.2875

b. 1− 0.8057 = 0.1943

c. 0.8057− 0.2827 = 0.523

d. Notice that 0.0049 ≈ 0.005 so ≈ −0.035

November 9, 2017 Prof. Russell Fall, 2017