Problem 1
Throughout this problem, note that conditional preference is well-de…ned, by P2.
(a) True. I show that B1 and B2 are null if and only if B1 [ B2 is null. The result
follows by induction on n.
Suppose that B1 and B2 are null. Then, for all f; g; h 2 F ,
fh
fjhB1 [B2 = fjBjB1 2
where the …rst
fh
gjBjB12 = fBg 2 nB1
gjhB
gh
gjBjB21 = gjhB1 [B2 ,
1
jB 2
follows because B1 is null and the second
follows because B2 is null. By
de…nition, this implies that B1 [ B2 is null.
Suppose that B1 [ B2 are null. Then, for all f; g; h 2 F ,
h
h
fjB
= fjB
1
1
h
h
h
gjB
1
jB1 [B2
jB1 [B2
= gjhB1 .
By de…nition, this implies that B1 is null. Similarly, B2 is null.
(b) True. Since C is …nite, there exists a …nite partition fD1 ; : : : ; Dn g of S such that,
if s; s0 2 Di for some i 2 f1; : : : ; ng, then f (s) = f (s0 ) and g (s) = g (s0 ). Let C 0 = ; and
S
let C i = ik=1 Dk . For any i 2 f1; : : : ; ng, let fx be the constant act that always yields
consequence x
consequence y
f (s) for some s 2 Di and let fy be the constant act that always yields
fg
fg
g (s) for some s 2 Di . If Di is null then (fx )jDjCi % (fy )jDjCi trivially, and if
fg
fg
i
i
Di is non-null then (fx )jDjCi % (fy )jDjCi by P3. Hence, for every i 2 f1; : : : ; ng,
i
g
fjC
i
i
fg
jC i
jDi
= (fx )
fg
g
% (fy )jDjCi = fjC
i 1.
i
g
Therefore, f = fjS
% fj;g = g.
(c) False. Let S = fs; tg, C = fx; x0 ; y; y 0 g. I denote an act f : S ! C by a pair (a; b),
with the interpretation that f (s) = a and f (t) = b.
Consider the preference relation %
given by
(x; x)
(x; x0 )
(x0 ; x)
(x0 ; x0 )
(x; y)
(x; y 0 )
(x0 ; y)
(x0 ; y 0 )
(y; x)
(y; x0 )
(y 0 ; x)
(y 0 ; x0 )
(y; y)
(y 0 ; y)
(y; y 0 )
(y 0 ; y 0 ) .
1
One can check that % satis…es P1-P3, but x
y 0 , ffx;x
sg
x0 , y
0
0
y;y
ffx;x
tg , and fftg
0
0
y;y
ffs
g .
Problem 2
(a) True.
_ is a qualitative probability.
By P1-P5, %
In what follows, “property 2” and
“property 3”refer to properties 2 and 3 of qualitative probability, on page 85 in the lecture
notes.
I …rst prove the result in the special case where B1 \ B2 = ;. Note that
_ B2 [ (A1 nB2 ) (by A2 %
_ B2 , A1 \ A2 = ;, and property 2)
A2 [ (A1 nB2 ) %
= A1 [ B2
= A1 [ (B2 nA1 )
_ B1 [ (B2 nA1 ) (by A1 %B
_ 1 , B1 \ B2 = ;, and property 2).
%
Since (A2 [ (A1 nB2 )) \ (A1 \ B2 ) = ; (by A1 \ A2 = ;) and (B1 [ (B2 nA1 )) \ (A1 \ B2 ) = ;
(by B1 \ B2 = ;), this implies that (using property 2 again)
_ (B1 [ (B2 nA1 )) [ (A1 \ B2 ) = B1 [ B2 .
A1 [ A2 = (A2 [ (A1 nB2 )) [ (A1 \ B2 ) %
_ 2 nB1 , because B1 \ B2 %
_ ; (by property
Now suppose that B1 \ B2 6= ;. Note that B2 %B
_ [ (B2 nB1 ) = (B2 nB1 ) (by property 2).
3), and therefore B2 = (B1 \ B2 ) [ (B2 nB1 ) %;
_ 2 nB1 , so the fact that the result holds in the special case where B1 \ B2 = ;
Therefore, A2 %B
_ 1 [ (B2 nB1 ) = B1 [ B2 .
implies that A1 [ A2 %B
_ for
(b) False. For the simplest counterexample, let D = ;, in which case clearly A%B
all A; B
_ S, which contradicts property 3 of qualitative probability.
S, and in particular ;%
_ given
One can derive a similar contradiction for any null event D, and one can show that %
D is indeed a qualitative probability if D is non-null.
(c) True. Note that, by property 1 of a probability measure (also on page 85 in the
lecture notes), p (;) = 0 (because p (;) = p (; [ ;) = 2p (;)).
If A
S is null, then
0
0
0
fAx;x = (fx )xjA
0
(fx0 )xjA = f;x;x ,
2
_ , it follows that p (A) = p (;) = 0.
and therefore A _ ;. Hence, if p represents %
_ then A _ ;. Therefore,
Conversely, if p (A) = 0, then p (A) = p (;), so if p represents %
0
0
0
0
0
(fx0 )xjA = f;x;x . But if A were not null, then P3 would imply that (fx )xjA
fAx;x = (fx )xjA
0
(fx0 )xjA , so it must be that A is null.
Problem 3
(Thanks to Hongkai Zhang, whose solution I used as a model for this)
As per Muhamet’s hint, each player has a well-de…ned continuation value at every history,
and she accepts her opponent’s proposal if and only if it gives her expected utility at least
as great as her continuation value.
Therefore, at every history the proposer proposes a
Pareto e¢ cient allocation that gives her opponent exactly her continuation value. Due to
CARA utility and normally distributed payo¤s, the formula for e¢ cient risk-sharing derived
in lecture implies that player i’s time-t proposal is
xti
=
i
i
i+
i
=
1=
1=
+
i s=t
T 1
X
i
xt i =
(noting that
T 1
X
i
i+
i
i +1=
i
(xi;s + x
(xi;s + x
i;s )
+
i;s )
+
i s=t
t 1
X
s=0
t 1
X
xi;s +
x
t
t
i;s
s=0
).
It remains only to determine
t
(and to verity that the …xed component of xti does indeed
depend on the history only through t). Note that
t
is determined by the condition that
player i is indi¤erent between the assets xt i and xt+1
at time t. Using certainty equivalents,
i
this condition is
CE
i
i
i
+
T 1
X
(xi;s + x
i;s )
i s=t
+
t 1
X
x
i;s
t
s=0
= CE
T 1
X
i
i
i
+
i
i
i+
(xi;t + x
i;t )
i
Recall the following two facts:
3
(xi;s + x
i;s )
i s=t+1
or equivalently
CE
!
t
= CE
+
t
X
s=0
i
(x
i;t
+
t+1 ) .
x
i;s
+
t+1
!
,
(1)
2
1. If X and Y are iid random variables distributed N (0;
N (0; 2k 2
2
), then k (X + Y ) is distributed
).
2
2. With CARA utility, if X is distributed N ( ;
), then CE (X) =
2
2
.
Using these facts, (1) may be rewritten as
2
i
i
2
i
+
t
Noting that
T
=
2
1
2
i
= 0, (2) determines
2
2
i
or equivalently
t
i
=
!
i
2
i
+
t
for all t 2 f0; : : : ; T
i
t+1 ,
+
t+1 .
(2)
1g (where i = 1 if T is odd,
and i = 2 if T is even). (2) can be written more concisely as follows:
!
2
1
i
2
t =
i
t+1 if t is even
2
+
i
i
t
=
T
t+1
2
1
+
2 (
i
i
+
4
i
2
i)
(
i
i)
2
if t is odd.
MIT OpenCourseWare
http://ocw.mit.edu
14.123 Microeconomic Theory III
Spring 2015
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