APPENDIX B
Exercise Solutions
419
EXERCISE B.1
(a)
()2
()00.2510.5020.251i i i E X x P X x ====×+×+×=∑
(b) ()()2
2
22
220
0.2510.5020.25 1.5i
i
i E X
x P X x ====×+×+×=∑
(c) ()()()()()2
2
2
2
2
0var()()010.25110.50210.250.5i i i X x E X P X x ==?==?×+?×+?×=∑
(d) ()(32)3()25E g X E X E X ??=+=+=??
()2var var(32)3var() 4.5g X X X ??=+==??
EXERCISE B.2
(a)
Probabilities of a single roll of a fair die
X ()f x
1 16
2 16
3 16
4 16
5 1
6 6
16
(b)
()416P X ==
()()()4 or 545161613P X X P X P X ====+==+=
(c)
6
1().()116216316416516616 3.5x E X x f x ===×+×+×+×+×+×=∑
() 3.5E X = is the average face value, if the dice is rolled a large number of times.
(d) 6
2
22222221
91().()112163164165166166x E X x f x ===×+×+×+×+×+×=∑
(e) The variance is given by
()()
()()()()()()6
2
12222
2
2
var()()1 3.5162 3.5163 3.514 3.5165 3.516 3.512.917
i i i X x E X P X x ==??==?×+?×+?×+?×+?×+?×=∑
(f) Average values after 5 rolls, 10 rolls and 20 rolls.
EXERCISE B.3
(a) A sketch of the density function
()2010otherwise x x f x ≤≤?=??
is as follows
(b) When 1/2,()1x f x ==. The probability (00.5)P X ≤≤ is the area of a triangle with base = 0.5
and height = 1. Since area of a triangle = 0.5base×height ×, the probability is given by (00.5)0.50.510.25P X ≤≤=××=
(c) When 1/4,()2/4x f x ==. When 3/4,()6/4x f x ==. The probability (0.250.75)P X ≤≤ is
given by the area of a rectangle with base = 0.5 and height = 0.5 plus the area of a triangle with base = 0.5 and height = (1.50.5)?. That is, (0.250.75)(0.50.5)(0.50.51)0.5P X ≤≤=×+××=
f (x ) x 1 2
EXERCISE B.4
(a) Using the pdf ()()11x
x f x p p ?=? for x = 0,1:
The mean of the binary random variable, X , is
()()()()()
11
100111x
E X x f x f f p p p ?==×+×=?=∑
The variance of X is
()()()()()()()
()()()()()()()()()
2
2
2
122
1011
012
2var 111111111x
x x
x
X E X E X x p f x x p p p p p p p p p p p p p p p p p p p ???=?=?=??=??+??=?+?=?+?=?∑∑
(b)
1212()()()()()n n E B E X X X E X E X E X =+++=+++""
p p p np =+++="
()()12var var ...n B X X X =+++ ()()()12=var var var n X X X +++", since 1,...,n X X are independent
()()()()1111p p p p p p np p =?+?++?=?"
(c) 1
()()B np E Y E E B p n n
n ??====????
()()()()22
111
var var var np p p p B Y B n n n n ????====????
EXERCISE B.5
(a) The marginal probability density function of Y is h (y ) where 1111(1)8482h =++=
1111(3)244243
h =
++=
111(9)012126
h =
++= (b)
The conditional probability density function for y given 2X = is (|2)(2,)(2)f y X f y g ==,
where ()g x is the marginal probability density function for x .
Now,
1111
(2)824124
g =++=
and so (|2)f y X = is given by
(2,1)1/81
(1|2)(2)1/42f f X g ==== (2,3)1/241
(3|2)(2)1/46
f f X
g ==
==
(2,9)1/121
(9|2)(2)1/43
f f X
g ==
==
(c) ()cov ,[()][()]X Y E X E X Y E Y =??
()1114242464E X =×+×+×=
()1112361393E Y =×+×+×=
()()()()
()()()()()()()()11118242412cov ,43,24132433643364930
x
y
X Y x y f x y =??=??+??++??+??=∑∑"
(d)
This is an example where X and Y are not independent, despite the fact that their covariance is
zero. If X and Y are independent, then (,)()()f x y g x h y =. To show that independence does not hold, consider the example when x = 2 and y = 3,
()()()11112,323244312
f g h =
≠=×=
EXERCISE B.6
E Y=×+×=
(a) ()00.610.40.4
22
Y=?×+?×=
var()(00.4)0.6(10.4)0.40.24
(b) ()10.120.230.340.43
E X=×+×+×+×=
2222
X=?×+?×+?×+?×= var()(13)0.1(23)0.2(33)0.3(43)0.41
E XY=××+××++××+××=
()000.1020.1130.1140.11
"
()()()()()()
=??=?=?×=?cov,[][]130.40.2 X Y E X E X Y E Y E XY E X E Y
(c) ρ0.408
===?
EXERCISE B.7
(a) ()()()121211
...()()()n n E X E X X X E X E X E X n n
??=+++=+++????"
()1n n n
μ
=
μ+μ++μ==μ" (b) ()()121var var n X X X X n ??
=+++????"
()()()()122
1
var var var n X X X n =
+++"
2221n n n
σ=σ=
Since
12,,...,n X X X are independent random variables, their covariances are zero.
EXERCISE B.8
(a) ()()()33
11
1113333i i i i E Y E Y E Y ==??===μ=μ????∑∑
(b)
()()3123111
var var var 39i i Y Y Y Y Y =??==++????∑
()()()()()()()
123121323222221
var var +var 2cov(,)2cov(,)2cov(,)9
1
3320.59
113323
Y Y Y Y Y Y Y Y Y =++++=
σ+×σ=σ+σσ=
EXERCISE B.9
(a) (
)()()1 1.3571 1.3570.087P X P Z P Z P Z ?
<=<=
(b) ()()()4 2.940.78610.7860.2161.4P X P Z P Z P Z ??
?≥=>=>=?<=????
(c) ()()()2 2.920.6430.6430.7401.4P X P Z P Z P Z ??
?≥=>=>?=<=????
(d)
()()2.5 2.9
4 2.92.540.2860.7861.4 1.4P X P Z P Z ????<<=<<=?<
()()
()()()0.7860.2860.78610.2860.396
P Z P Z P Z P Z =
(e) We want X 0 such that P (X < X 0) = 0.05. Now, ( 1.645)0.95P Z <= and hence
( 1.645)0.05P Z
Thus, it follows that a suitable X 0 is such that
0 2.9
1.645 1.4
X ??=
Solving for X 0 yields
()()0 2.9 1.645 1.40.597X =?= (which is approximately 7 months)
EXERCISE B.10
(a)
The probability function of X is shown below.
1
2
3
45
6
7
(b) The probability that, on a given Monday, either 2, or 3, or 4 students will be absent is
4
2
()(2)(3)(4)0.3100.3400.2200.87x f x f f f ==
++=++=∑
(c) The probability that, on a given Monday, more than 3 students are absent is
7
4()(4)(5)(6)(7)0.2200.0800.0190.0010.32x f x f f f f ==
+++=+++=∑
(d) 7
().()00.00510.02520.31030.340x
E X x f x ===×+×+×+×∑
40.22050.08060.01970.001
3.066
+×+×+×+×=
Based on information over many Mondays, the average number of students absent on Mondays is 3.066. (e) 22var()()[()]X E X E X =?
7
2
222220
2222()()00.00510.02520.31030.340
40.22050.08060.01970.001 = 10.578
x E X x f x ===×+×+×+×+×+×+×+×∑
22var()10.578(3.066) 1.17764X =σ=?=
1.085
2.σ==
(f) ()(73)7()37 3.066324.462E Y E X E X =+=+=×+=
2var()var(73)7var()49 1.1776457.704Y X X =+==×=
EXERCISE B.11
(a) The probability density function is shown below.
(b) Total area of the triangle is half the base multiplied by the height; i.e., the area is 0.5211××=
(c) When x = 1, ()(1)0.5.f x f ==
Then, (1)P X ≥ is given by the area to the right of 1 which is (10.50.251)0.5P X ××=≥=
(d) When x = 0.5, ()0.5f = 0.75.
1
(0.5)1(0.5)1 1.50.750.43752
P X P X ≤=?>=?××=
(e) For a continuous random variable the probability of observing a single point is zero. Thus, ( 1.5)0P X ==.
EXERCISE B.12
(a) 2~(0.10,0.04)X N
00.1(0)( 2.5)10.99380.00620.04P X P Z P Z ??
?<=<=
?
(b)
0.150.1(0.15)( 1.25)10.89440.10560.04P X P Z P Z ??
?>=>=>=?=????
(c) Now
2~(0.12,0.05)X N
00.12(0)( 2.4)10.99180.00820.05P X P Z P Z ??
?<=<=
?
0.150.12(0.15)(0.6)10.72570.27430.05P X P Z P Z ??
?>=>=>=?=???
?
The calculations show that the probability of a negative return has increased from 0.62% to
0.82%, while the probability of a return greater than 15% has increased from 10.56% to 27.43%. Whether fund managers should or should not change their portfolios depends on their risk preferences.
EXERCISE B.13
(a)
()()()()0.250.750.250.75A B A B E P E R R E R E R =+=+
0.250.7513.25815=×+×=
(b) ()()2var var 0.250.75P A B P R R =σ=+
()()()220.25var 0.75var 20.250.75cov ,A B A B R R R R =++×××
Now,
1ρ==
Hence,
cov(,)1222264A B A B R R =σσ=×=
2222var()0.25120.752220.250.75264380.25P =×+×+×××=
19.5P σ==
(c) When
0.5ρ==
cov(,)0.50.51222132A B A B R R =×σσ=××=
2222var()0.25120.752220.250.75132330.75P =×+×+×××=
18.1865P σ==
(d) When ρ0=, cov(,)0A B R R =, and the variance and standard deviation of the portfolio are 2222var()0.25120.7522281.25P =×+×=
16.7705P σ==