How Elegantly the Farkas’s Lemma Works:
My solution to the Homework 3-1 Fall 2001
Kan Takeuchi
October 9, 2001
Notation: Let x be a matrix. Then, x ≥ 0. ⇐⇒ all xi ≥ 0 , and ∃xj > 0.
1
The Strictly Dominated and Un-Rationalizable y
Proof. First of all, define some notations. Let s be the number of the possible states, and n be that
of the alternatives. And also


u(x1 , S1 ) − u(y, S1 ) u(x2 , S1 ) − u(y, S1 ) . . . u(xn , S1 ) − u(y, S1 )
u(x1 , S2 ) − u(y, S2 ) u(x2 , S2 ) − u(y, S2 ) . . . u(xn , S2 ) − u(y, S2 )


A := 
.
..
..
..
.
.
s×n


.
.
.
.
u(x1 , Ss ) − u(y, Ss )
u(x2 , Ss ) − u(y, Ss )
...
u(xn , Ss ) − u(y, Ss )
To apply the Farkas’s lemma, let us define the followings:


The System A: Aσ̃ = b and σ̃ ≥ 0.


The System B: P̃ T A ≤ 0 and P̃ T b > 0.
where σ̃ is an n × 1 matrix such that σ̃ = (σ̃1 . . . σ̃n )T , and P̃ is a s × 1 matrix. Note that we can
define b as an exogeneous valuable. Then, the powerful Farkas’s lemma implies that
The System B has no solution, for some b 0. ⇐⇒ The System A has a solution σ̃, for the b 0.
And next, please check that
σ strictly dominates y.
y is rationalizable.
y is not rationalizable.
⇐⇒ The System A has a solution σ̃ ∗ for some b 0.
∗
⇐⇒ The System B has a solution P̃ , ∀b 0.
⇐⇒ The System B has no solution, for some b 0.
(1.1)
(1.2)
(1.3)
P
1
We do not explicitly assume
p̃ = 1, but the System B still represents
the rationalizability
. If the
P
P
∗
System A has a solutionP
σ̃ for some b 0, then all σ̃i ≥ 0, and σ̃i u(xi , sk )− σ̃i u(y, sk ) = bk > 0
for all sk . Let σi = σ̃i / σ̃, then we can conclude that the σ = (σ1 . . . σn ) strictly dominates y.
Hence, the conclusion immediately follows.
1 First,
p˜i
note that P̃ = 0 cannot constitute any solution. Suppose the System B has a solution. Let pi = P , then
p̃
P̃ T A =
P
p̃Eu(x1 ) −
P
p̃Eu(y)
...
Thus, Eu(y) ≥ Eu(xi ) for all xi . That is, y is rationalizable.
1
P
p̃Eu(xn ) −
P
p̃Eu(y) ≤ 0.
2
The Weakly Dominated and Un-Rationalizable y by P > 0
Proof. A similar argument follows. Let us define the following two systems:


The System A: Aσ̃ = b and σ̃ ≥ 0.


The System B: P̃ T A ≤ 0 and P̃ T b > 0.
Please check that
σ weakly dominates y.
y is rationalizable by P 0.
y is not rationalizable by P 0.
⇐⇒ The System A has a solution σ̃ ∗ for some b ≥ 0.
∗
⇐⇒ The System B has a solution P̃ , ∀b ≥ 0.
⇐⇒ The System B has no solution, for some b ≥ 0.
(2.1)
(2.2)
(2.3)
Note that we can define b as an exogeneous valuable. Then, the powerful Farkas’s lemma implies
that
The System B has no solution, for some b ≥ 0. ⇐⇒ The System A has a solution σ̃, for the b ≥ 0.
The conclusion immediately follows.
2