An Elementary Introduction to Mathematical Finance (3rd Edition, 2011) – Solutions Manual – Ross

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SOLUTIONS MANUAL

, 1

1.1 (a) 1 − p0 − p1 − p2 − p3 = 0.05 (b) p0 + p1 + p2 = 0.80
1.2 P {C ∪ R} = P {C} + P {R} − P {C ∩ R} = 0.4 + 0.3 − 0.2 = 0.5
8 7 56 6 5 30 6 8 8 6 96
1.3 (a) 14 13
= 182
(b) 14 13
= 182
(c) 14 13
+ 14 13
= 182

1.4 (a) 27/58 (b) 27/35
1.5

1. The probability that their child will develop cystic fibrosis is the probability that
the child receives a CF gene from each of his parents, which is 1/4.

2. Given that his sibling died of the disease, each of the parents much have exactly
one CF gene. Let A denote the event that he possesses one CF gene and B that he
does not have the disease (since he is 30 years old). Then

P (A ∩ B) P (A) 2/4 2
P (A|B) = = = =
P (B) P (B) 3/4 3

1.6 Let A be the event that they are both aces and B the event they are of different
suits. Then
4 3
P (A ∩ B) P (A) 1
P (A|B) = = = 523951 =
P (B) P (B) 51
169

1.7

(a) P (AB c ) = P (A) − P (AB)
= P (A) − P (A)P (B)
= P (A)(1 − P (B)
= P (A)P (B c )

Part (b) follows from part (a) since from (a) A and B c are independent, implying from
(a) that so are Ac and B c .
1.8 If the gambler loses both the bets, then X = −3. If he wins the first bet, or loses
the first bet and wins the second bet, X = 1. Therefore,
20 2 100
P {X = −3} = ( ) =
38 361
18 20 18 261
P {X = 1} = + =
38 38 38 361
261
1. P {X > 0} = P {X = 1} = 361

2. E[X] = 1 261
361
− 3 100
361
= −39
361


1.9

,2

1. E[X] is larger since a bus with more students is more likely to be chosen than a bus
with less students.

2.
1 5882
E[X] = (392 + 332 + 462 + 342 ) = ≈ 38.697
152 152
1
E[Y ] = (39 + 33 + 46 + 34) = 38
4

1.10 Let N denote the number of sets played. Then it is clear that P {N = 2} =
P {N = 3} = 1/2.

1. E[N ] = 2.5

2. Var(N ) = 21 (2 − 2.5)2 + 21 (3 − 2.5)2 = 1
4


1.11 Let µ = E[X].

Var(X) = E[(X − µ)2 ]
= E[X 2 − 2µX + µ2 ]
= E[X 2 ] − 2µE[X] + µ2
= E[X 2 ] − µ2


1.12 Let F be her fee if she takes the fixed amount and X when she takes the contin-
gency amount.
E[F ] = 5, 000, SD(F ) = 0
E[X] = 25, 000(.3) + 0(.7) = 7, 500
E[X 2 ] = (25, 000)2 (.3) + 0(.7) = 1.875 × 108
Therefore,
q q √
SD(X) = Var(X) = 1.875 × 108 − (7, 500)2 = 1.3125 × 104


1.13
n
1X
(a) E[X̄] = E[Xi ]
n i=1
1
= nµ = µ
n

, 3

1 X n
(b) Var(X̄) = ( )2 Var(Xi )
n i=1
1
= ( )2 nσ 2 = σ 2 /n
n
X
n X
n
2
(c) (Xi − X̄) = (Xi2 − 2Xi X̄ + X̄)2 )
i=1 i=1
Xn X
n
= Xi2 − 2X̄ Xi + nX̄ 2
i=1 i=1
Xn
= Xi2 − 2X̄nX̄ + nX̄ 2
i=1
Xn
= Xi2 − nX̄ 2
i=1

X
n
(d) E[(n − 1)S 2 ] = E[ Xi2 ] − E[nX̄ 2 ]
i=1
= nE[X12 ] − nE[X̄ 2 ]
= n(Var(X1 ) + E[X1 ]2 ) − n(Var(X̄) + E[X̄]2 )
= nσ 2 + nµ2 − n(σ 2 /n) − nµ2
= (n − 1)σ 2

1.14
Cov(X, Y ) = E [(X − E[X])(Y − E[Y ])]
= E [XY − XE[Y ] − E[X]Y + E[X]E[Y ])]
= E[XY ] − E[Y ]E[X] − E[X]E[Y ] + E[X]E[Y ]
= E[XY ] − E[Y ]E[X]

1.15
(a) Cov(X, Y ) = E [(X − E[X])(Y − E[Y ])]
= E [(Y − E[Y ])(X − E[X])]

(b) Cov(X, X) = E[(X − E[X])2 ] = Var(X)

(c) Cov(cX, Y ) = E [(cX − E[cX])(Y − E[Y ])]
= cE [(X − E[X])(Y − E[Y ])]
= cCov(X, Y )

(d) Cov(c, Y ) = E [(c − E[c])(Y − E[Y ])] = 0