Comment on Problem 2 of hwk 8

Solution of problem 2 of hwk 8 is valid only for L=1.
When L>1, assume (to the contrary) that x_11 not equal to x_12.
Then from strict convexity of preference, we can say that
(x_11+x_12)/2 is strictly preferred to x_11 under type 1's preference.
(See MWG p.656 Prop. 18.B.2)

I thank Matsuda-san for pointing this out.

Answer to question on consumer choice

In general, if C({x,y})={x}, then x \succsim y & not「y \succsim x」 so that x \succ y holds. ...(*)

One student asked me that it seems contradicting to the statement in
the first example of p.6 (Section 2.3) in resume:
\succsim which rationalizes C(.) s.t. C({x,y})={x}, C({x,y,z})=x
is given by ① x \succsim y, x \succsim z, y \succsim z.

Using (*), you might think that it should be written as
"\succsim which satisfies 「 ①' x \succ y, x \succ z, y \succsim z」 rationalizes C(.). "

Here is Prof. Kamiya's reply:
In general, binary relation is represented using only \succsim
(and not \succ, \sim ...etc). Since ① does NOT contain
"y \succsim x", it implies that "y \succsim x" does NOT hold. Hence we can conclude that
「x \succsim y & not「y \succsim x」」 so that x \succ y.
Thus, representating \succsim by ① is correct.

New version of hwk 13

I made corrections to original solution of hwk 13 and new one is available at hp.
(I realized that (3-ii) was wrong.) Also, I wrote some comments by Prof. Kamiya regarding questions came up in yesterday's TA session.
(ex. BC inequality in Radner eqm, assuming symmetric price in answer sheet...)

New version of hwk 12's solution

New version of hwk 12's solution is now available
from coremicro 2009's homepage.
(Changes are written in blue.)

Hwk 11-12 Graded

I finished grading hwk 11-12.
I will return them in today's TA session.
Afterwards, I will place them in front of 428.
My grading policy:
C: incomplete
B: Many errors in basic problems or tragic error in Problem 3 of hwk 12
A: Most of the problem correct. solve problem 3 of hwk 12 correct.
If you have any comments, please contact me.
I gave Ino-san A+ for his excellent performance in A-D eqm.

Correction of Hwk11-Problem 4-ii

Hwk 11 (4-ii)'s solution is wrong.
I must show that for all F.
(New Answer: Fix arbitrary F. Then from Jensen's inequality,
E[u(x)] = E[-x^2 + 6x +1] = E[-x^2] + 6(E[x]) +1
< -(E[x])^2 + 6(E[x]) +1 = u(E[x]).)

Typo in Hwk 12

In Hwk 12 Problem (2-iii),
Prob[X* (smaller than or equal to) Y*] must be
Prob[X* (greater than or equal to) Y*].