## 2009年7月21日火曜日

### 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.

## 2009年7月16日木曜日

### 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(.). "

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.

## 2009年7月15日水曜日

### 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...)

## 2009年7月10日金曜日

### 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.)

I will return them in today's TA session.
Afterwards, I will place them in front of 428.
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.
I gave Ino-san A+ for his excellent performance in A-D eqm.

## 2009年7月9日木曜日

### 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]).)

## 2009年7月3日金曜日

### 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*].