14.123
Microeconomic
Theory
III
Final
Make
Up
Exam
(80
Minutes)
1. (30
points)
This
question
assesses
your
understanding
of
expected
utility
theory.
(a)
Show
that
there
exists
a
preference
relation
on
preferences
that
satis
fi
es
the
in
dependence
axiom
but
is
discontinuous.
(For
an
example,
you
can
take
the
set
of
consequences
as
{
x, y, z
}
and
consider
lexicographic
preferences.)
Answer:
Denote
the
lotteries
by
(
p
x
, p
y
,
1
−
p
x
−
p
y
)
.
Consider
the
lexicographic
relation
p
º
q
[(
p
x
> q
x
)
or
[
p
x
=
q
x
and
p
y
> q
y
]]
.
⇐⇒
This
is
a
discontinuous
preference
relation
because
the
upper
counter
set
for
p
is
{
q

q
x
> p
x
}
∪
{
q

q
x
=
p
x
, q
y
≥
p
y
}
,
which
is
clearly
not
a
closed
set
when
p
is
in
the
interior.
To
check
that
it
satis
fi
es
the
independence
axiom,
take
any
p, q, r
∈
P
and
a
∈
(0
,
1]
. If
p
∼
q
, then
p
=
q
(as
the
indi
ff
erence
sets
are
singletons),
and
hence
ap
+(1
−
a
)
r
=
aq
+(1
−
a
)
r
,
showing
that
ap
+(1
−
a
)
r
∼
aq
+(1
−
a
)
r
.
If
p
Â
q
, then
either
p
x
> q
x
,
in
which
case
ap
x
+ (1
−
a
)
r
x
> aq
x
+ (1
−
a
)
r
x
,
showing
that
ap
+ (1
−
a
)
r
Â
aq
+ (1
−
a
)
r
,
or
p
x
=
q
x
and
p
y
> q
y
, in
which
case
ap
x
+ (1
−
a
)
r
x
=
aq
x
+ (1
−
a
)
r
x
and
ap
y
+ (1
−
a
)
r
y
> aq
y
+ (1
−
a
)
r
y
,
showing
once
again
that
ap
+ (1
−
a
)
r
Â
aq
+ (1
−
a
)
r
.
This
shows
that
if
p
º
q
=
⇒
ap
+ (1
−
a
)
r
º
aq
+ (1
−
a
)
r
. Conversely,
if
p
º
6
q
, then
q
Â
p
(by
completeness)
and
as
we
just
shows
this
imples
that
aq
+(1
−
a
)
r
Â
ap
+(1
−
a
)
r
,
showing
that
ap
+ (1
−
a
)
r
º
6
aq
+ (1
−
a
)
r
.
(b)
Under
Postulates
P15
of
Savage,
consider
the
as
likely
as
relation
∼
˙
between
events,
derived
from
betting
preferences
as
in
the
class.
Consider
events
A
and
B
such
that
A
∼
˙
S
\
A
and
B
∼
˙
S
\
B
, where
S
is
the
state
space.
Show
that
A
∼
˙
B
.
Answer:
Recall
from
the
class
that
under
P15,
º
˙
is
a
qualitative
probability.
In
particular,
if
C, D, E
are
disjoint
events,
C
Â
˙
D
⇐⇒
C
∪
E
Â
˙
D
∪
E.
(1)
Now,
for
simplicity
of
notation,
let
W
=
A
∩
B
,
X
=
B
\
C
,
Y
=
A
\
B
, and
Z
=
S
\
(
A
∪
B
)
.
Suppose
for
a
contradiction
that
A
Â
˙
B
.
Then,
by
(1),
Y
Â
˙
X
.
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 Spring '10
 Yildiz
 Game Theory, player, 80 minutes

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