978-0123865496 Chapter 4 Solution Manual

subject Type Homework Help
subject Pages 7
subject Words 346
subject Authors Jean-Pierre Danthine, John B. Donaldson

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page-pf1
CHAPTER 5.
5.1. For full investment in the risky asset the first order condition has to satisfy the following:
0rr
~
r
~
1Y'UE f0
~
page-pf2
1 2 3
13
31
13
1
1 1 1
1 2 3
U Y 1 exp bY
W a 1 exp b 2 a 1 exp b 2 1 exp b 2 a
W' a bexp b 2 a bexp b 2 a 0
ln b b 2 a ln b b 2 a
ln ln
a2b
1
U Y Y
1
1 1 1
W a 2 a 2 2 a
1 1 1
W'



13
1/ 1/
31
1/ 1/
13
a 2 a 2 a 0
a2



a
page-pf3
*
da
page-pf4
f0A
ff0A
ff0
ff0
r1YR
rr
~
ar1YR
rr
~
ar1YU
rr
~
ar1YU
Hence
ff0f0Aff0 rr
~
ar1YUr1YRrr
~
ar1YU
Since
0rr
~f
for this case,
(i)
fff0f0Afff0 rr
~
rr
~
ar1YUr1YRrr
~
rr
~
ar1YU
Case 2:
f
rr
~
.r1YR
rr
~
ar1YR
rr
~
ar1YU
rr
~
ar1YU
f0A
ff0A
ff0
ff0
Thus,
~
~
page-pf5
a
π
1 π
b
5.4. Let
BA R
~
and R
~
represent the gross returns on the two assets ;
BABABA R
~
UEa1R
~
UaER
~
Ua1R
~
aUER
~
a1R
~
aUE
BA R
~
UER
~
UE
since the utility function is concave and gross asset returns are identically distributed.
This result means that the investor is going to invest in both securities it is never
optimal in this situation to invest only in one of the two assets. He thus chooses to
diversify. Moreover if the investor cares only about the first two moments he will invest
equal amounts in the assets to minimize variance. To show this note that the expected
5.5
1Y0
payouts =
riskless asset: 1
risky asset:
page-pf6
0
da
dx
)b1)(b)x1(x(''U)1( 1
2
11
0
)b1)(b)x1(x(''U)1()a1)(a)x1(x(''U
)a)x1(x(''U)x1)(a1()a)x1(x('U
da
dx
2
11
2
11
111111
as the numerator is positive and the denominator negative.
We would expect that an increase in the probability of the unfavorable risky asset state a,
page-pf7
5.7 a.
)z
~
xzx(EUmax 2211
x,x 21
02211 Yxpxp .t.s
Since we assume (maintained assumption) U’( )>0,
02211 Yxpxp
, and
1
220
1p
xpY
x
The problem may thus be written :
221
1
220
xz
~
xz
p
xpY
EUmax
2
The necessary and sufficient F.O.C. (under the customary assumption) is :
0z
p
p
z
~
z
~
xz
p
xpY
'EU 1
1
2
2221
1
220
b. Suppose
AY
bea)y(U
; the above equation becomes :
0z
p
p
z
~
z
~
xz
p
xpY
expAbE 1
1
2
2221
1
220
equivalently,
0z
p
p
z
~
z
~
xz
p
xp
exp
p
zY
expAbE 1
1
2
2221
1
22
1
10
The first term which contains
0
Y
can be eliminated from the equation. The intuition for
5.8. The problem with linear mean variance utility is
22
02
1
~
)max( x
sxsEsy
~

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