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Microeconomics
Problem Set 1
NCKU Economics
𝑤
1. Suppose 𝑥𝑘 (𝑝, 𝑤) = 𝐿𝑝 for 𝑘 = 1, ⋯ , 𝐿.
𝑘
(a.) Show that 𝑥(𝑝, 𝑤) is homogeneous of degree zero (hd0) in (𝑝, 𝑤)
𝛼𝑤
𝑤
𝑥𝑘 (𝛼𝑝, 𝛼𝑤) =
=
= 𝑥𝑘 (𝑝, 𝑤)
𝐿𝛼𝑝𝑘 𝐿𝑝𝑘
(b.) Show that 𝑥(𝑝, 𝑤) satisfies Walras’ law (WL).
𝐿
𝑝 ∙ 𝑥(𝑝, 𝑤) = ∑
𝐿
𝑝𝑘 𝑥𝑘 (𝑝, 𝑤) = ∑
𝑘=1
𝑘=1
𝑝𝑘
𝐿
𝑤
𝑤
=∑
=𝑤
𝐿𝑝𝑘
𝑘=1 𝐿
(c.) Let 𝐿 = 2. Derive its Slutsky substitution matrix.
−𝑤
4𝑝2
𝑆(𝑝, 𝑤) = [ 𝑤1
4𝑝1 𝑝2
𝑤
4𝑝1 𝑝2
−𝑤 ]
4𝑝22
2. For computing the change in consumption quantities between two periods,
Laspeyres quantity index ( 𝐿𝑄 = 𝑝0 ∙ 𝑥1 /𝑝0 ∙ 𝑥0 ) uses period 0 prices as
weights while Paasche quantity index (𝑃𝑄 = 𝑝1 ∙ 𝑥1 /𝑝1 ∙ 𝑥0 PQ=p1x1/p1x0) uses
period 1 prices as weights. Show that
(a.) If 𝐿𝑄 < 1, then the consumer has a revealed preference for 𝑥0 over 𝑥1 .
If 𝐿𝑄 < 1, 𝑝0 ∙ 𝑥1 < 𝑝0 ∙ 𝑥0 . When one can afford both 𝑥0 and 𝑥1 , and 𝑥0
is chosen, 𝑥0 is preferred to 𝑥1 .
(b.) If 𝑃𝑄 > 1, then the consumer has a revealed preference for 𝑥1 over 𝑥0 .
If 𝑃𝑄 > 1, 𝑝1 ∙ 𝑥1 > 𝑝1 ∙ 𝑥0 . When one can afford both 𝑥0 and 𝑥1 , and 𝑥1
is chosen, 𝑥1 is preferred to 𝑥0 .
3. Verify if one can derive 𝐷𝑝 𝑥(𝑝, 𝑤)𝑝 + 𝐷𝑤 𝑥(𝑝, 𝑤)𝑤 = 0 simply from
𝑝𝑇 𝐷𝑝 𝑥(𝑝, 𝑤) + 𝑥(𝑝, 𝑤)𝑇 = 0 and 𝑝𝑇 𝐷𝑤 𝑥(𝑝, 𝑤) = 1 . If not, discuss the
necessary condition for the property to hold.
𝑝𝑇 𝐷𝑝 𝑥(𝑝, 𝑤)𝑝 + 𝑥(𝑝, 𝑤)𝑇 𝑝 = 0𝑇 𝑝;
𝑝𝑇 𝐷𝑝 𝑥(𝑝, 𝑤)𝑝 + 𝑤 = 0;
𝐷𝑤 𝑥(𝑝, 𝑤)𝑝𝑇 𝐷𝑝 𝑥(𝑝, 𝑤)𝑝 + 𝐷𝑤 𝑥(𝑝, 𝑤)𝑤 = 0 is different from 𝐷𝑝 𝑥(𝑝, 𝑤)𝑝 +
𝐷𝑤 𝑥(𝑝, 𝑤)𝑤 = 0.
In order to derive 𝐷𝑝 𝑥(𝑝, 𝑤)𝑝 + 𝐷𝑤 𝑥(𝑝, 𝑤)𝑤 = 0. We need 𝑥(𝑝, 𝑤) to be
hd0, i.e. 𝑥(𝛼𝑦) = 𝑥(𝑦).
𝑝
Let 𝑦 = ( ). Differentiate both sides of 𝑥(𝛼𝑦) ≡ 𝑥(𝑦) with respect to 𝛼.
𝑤
By chain rule, 𝐷𝛼𝑦 𝑥(𝛼𝑦)𝑦 = 0 and 𝐷𝑦 𝑥(𝑦)𝑦 = 0 when evaluated at α = 1.
𝐷𝑦 𝑥(𝑦)𝑦 = (𝐷𝑝 𝑥(𝑝, 𝑤)
𝑝
𝐷𝑤 𝑥(𝑝, 𝑤)) ( ) = 𝐷𝑝 𝑥(𝑝, 𝑤)𝑝 + 𝐷𝑤 𝑥(𝑝, 𝑤)𝑤.
𝑤
Microeconomics
Problem Set 1
NCKU Economics
4. Suppose 𝑥(𝑝, 𝑤) satisfies WL and hd0 in (𝑝, 𝑤). Show that WA holds if and
only if it holds for any compensated price change. (Refer to p.31 of the
textbook)
a. WA holds → WA for compensated price change holds.
Since WA holds for any change in price and income pair, certainly it will hold
for any compensated price change.
b. WA doesn’t hold → WA for compensated price change doesn’t hold.
If WA doesn’t hold, for any change from (𝑝′ , 𝑤′) to (𝑝′′ , 𝑤 ′′ ) such that
𝑥(𝑝′ , 𝑤′) ≠ 𝑥(𝑝′′ , 𝑤 ′′ ), we have 𝑝′ ∙ 𝑥(𝑝′′ , 𝑤 ′′ ) ≤ 𝑤′ and 𝑝′′ ∙ 𝑥(𝑝′ , 𝑤 ′ ) ≤ 𝑤′′.
If one of the two weak inequalities holds with equality, it is indeed a violation of
WA for compensated price change.
Next focus on the case with 𝑝′ ∙ 𝑥(𝑝′′ , 𝑤 ′′ ) < 𝑤′ and 𝑝′′ ∙ 𝑥(𝑝′ , 𝑤 ′ ) < 𝑤′′.
First, by above inequalities, there exists α ∈ (0,1) such that (α𝑝′ + (1 − 𝛼)𝑝′′ ) ∙
𝑥(𝑝′ , 𝑤 ′ ) = (α𝑝′ + (1 − 𝛼)𝑝′′ ) ∙ 𝑥(𝑝′′ , 𝑤 ′′ ) . (Note when 𝛼 = 0 , (α𝑝′ + (1 −
𝛼)𝑝′′ ) ∙ 𝑥(𝑝′ , 𝑤 ′ ) < (α𝑝′ + (1 − 𝛼)𝑝′′ ) ∙ 𝑥(𝑝′′ , 𝑤 ′′ ) and when 𝛼 = 1 , (α𝑝′ +
(1 − 𝛼)𝑝′′ ) ∙ 𝑥(𝑝′ , 𝑤 ′ ) > (α𝑝′ + (1 − 𝛼)𝑝′′ ) ∙ 𝑥(𝑝′′ , 𝑤 ′′ ))
Let 𝑝′′′ = (α𝑝′ + (1 − 𝛼)𝑝′′ ), 𝑤 ′′′ = 𝑝′′′ ∙ 𝑥(𝑝′ , 𝑤 ′ ) = 𝑝′′′ ∙ 𝑥(𝑝′′ , 𝑤 ′′ ).
α𝑤 ′ + (1 − 𝛼)𝒘′′ > α𝑝′ ∙ 𝑥(𝑝′ , 𝑤 ′ ) + (1 − 𝛼)𝒑′′ ∙ 𝒙(𝒑′ , 𝒘′ )
= 𝑝′′′ ∙ 𝑥(𝑝′ , 𝑤 ′ )
= 𝑤 ′′′
= 𝑝′′′ ∙ 𝑥(𝑝′′′ , 𝑤 ′′′ )
= α𝑝′ ∙ 𝑥(𝑝′′′ , 𝑤 ′′′ ) + (1 − 𝛼)𝑝′′ ∙ 𝑥(𝑝′′′ , 𝑤 ′′′ )
When 𝑥(𝑝′′′ , 𝑤 ′′′ ) ≠ 𝑥(𝑝′ , 𝑤 ′ ) and 𝑥(𝑝′′′ , 𝑤 ′′′ ) ≠ 𝑥(𝑝′′ , 𝑤 ′′ ) either 𝑤 ′ > 𝑝′ ∙
𝑥(𝑝′′′ , 𝑤 ′′′ ) or 𝑤 ′′ > 𝑝′′ ∙ 𝑥(𝑝′′′ , 𝑤 ′′′ ) implies the violation of WA for
compensated price change since 𝑤 ′′′ = 𝑝′′′ ∙ 𝑥(𝑝′ , 𝑤 ′ ) = 𝑝′′′ ∙ 𝑥(𝑝′′ , 𝑤 ′′ ).
5. Show that if 𝑥(𝑝, 𝑤) satisfies WA, it must be hd0 in (𝑝, 𝑤).
𝑝 ∙ 𝑥(𝑝, 𝑤) ≤ 𝑤 and 𝛼𝑝 ∙ 𝑥(𝛼𝑝, 𝛼𝑤) ≤ 𝛼𝑤 because 𝑥(𝑝, 𝑤) is affordable.
Equivalently we have 𝛼𝑝 ∙ 𝑥(𝑝, 𝑤) ≤ 𝛼𝑤 and 𝑝 ∙ 𝑥(𝛼𝑝, 𝛼𝑤) ≤ 𝑤. If WA
holds, then it must be that 𝑥(𝑝, 𝑤) = 𝑥(𝛼𝑝, 𝛼𝑤) so that 𝑥(𝑝, 𝑤) is hd0.
6. Show that for L=2, the Slutsky matrix, 𝑆(𝑝, 𝑤), is always symmetric.
Since 𝑆(𝑝, 𝑤)𝑝 = 0 and 𝑝𝑇 𝑆(𝑝, 𝑤) = 0, we have
[
𝑆11
𝑆21
[𝑝1
𝑆 𝑝 +𝑆 𝑝
𝑆12 𝑝1
] [ ]=[ 11 1 12 2 ]=0
𝑆22 𝑝2 𝑆21 𝑝1 + 𝑆22 𝑝2
𝑝2 ] [𝑆11
𝑆21
𝑆12
]=[𝑆11 𝑝1 + 𝑆21 𝑝2
𝑆22
𝑆12 𝑝1 + 𝑆22 𝑝2 ]=0
Since the above two both are true, we must have S21=S12.
Microeconomics
Problem Set 1
NCKU Economics
7. Walrasian demand 𝑥(𝑝, 𝑤) is said to satisfy the uncompensated law of
demand (ULD) if for any two prices, 𝑝′ , 𝑝′′ and the given income, 𝑤 ′ , we
have (𝑝′′ − 𝑝′ ) ∙ (𝑥(𝑝′′ , 𝑤 ′ ) − 𝑥(𝑝′ , 𝑤 ′ )) ≤ 0 , with strict inequality if
𝑥(𝑝′ , 𝑤 ′ ) ≠ 𝑥(𝑝′′ , 𝑤 ′ ). Show that if 𝑥(𝑝, 𝑤) satisfies WL, hd0, and ULD for
any 𝑝′ , 𝑝′′ , and 𝑤 ′ , it will also satisfy WA. (Refer to p.112 of the textbook)
Take any (𝑝′ , 𝑤 ′ ) and (𝑝′′ , 𝑤 ′′ ) with 𝑥(𝑝′ , 𝑤 ′ ) ≠ 𝑥(𝑝′′ , 𝑤 ′′ ) and 𝑝′ ∙
𝑥(𝑝′′ , 𝑤 ′′ ) ≤ 𝑤 ′ . To show WA, we have to show 𝑝′′ ∙ 𝑥(𝑝′ , 𝑤 ′ ) > 𝑤 ′′ .
𝑤′
𝑤′
𝑤′
Let 𝑝′′′ = 𝑤′′ 𝑝′′ , first note that 𝑥(𝑝′′′ , 𝑤 ′ ) = 𝑥 (𝑤′′ 𝑝′′ , 𝑤′′ 𝑤 ′′ ) = 𝑥(𝑝′′ , 𝑤 ′′ )
by hd0. By ULD, (𝑝′′′ − 𝑝′ ) ∙ (𝑥(𝑝′′′ , 𝑤 ′ ) − 𝑥(𝑝′ , 𝑤 ′ )) ≤ 0, or equivalently 𝑝′′′ ∙
𝑥(𝑝′′′ , 𝑤 ′ ) − 𝑝′′′ ∙ 𝑥(𝑝′ , 𝑤 ′ ) − 𝑝′ ∙ 𝑥(𝑝′′′ , 𝑤 ′ ) + 𝑝′ ∙ 𝑥(𝑝′ , 𝑤 ′ ) ≤ 0.
Since 𝑝′ ∙ 𝑥(𝑝′′ , 𝑤 ′′ ) ≤ 𝑤 ′ or equivalently −𝑝′ ∙ 𝑥(𝑝′′′ , 𝑤 ′ ) + 𝑝′ ∙ 𝑥(𝑝′ , 𝑤 ′ ) ≥ 0,
𝑤′
we must have 𝑝′′′ ∙ 𝑥(𝑝′′′ , 𝑤 ′ ) − 𝑝′′′ ∙ 𝑥(𝑝′ , 𝑤 ′ ) < 0, thus
𝑤′
𝑤 ′′
𝑝′′ ∙ 𝑥(𝑝′′ , 𝑤 ′′ ) −
𝑝′′ ∙ 𝑥(𝑝′ , 𝑤 ′ ) < 0 or 𝑝′′ ∙ 𝑥(𝑝′ , 𝑤 ′ ) > 𝑤 ′′ .
8. Let
[
𝑤 ′′
𝑥1
𝑥
𝑥 = [ 2] ,
𝑥3
𝑔1 (𝑥)
𝑥1 + 𝑥2
𝑔2 (𝑥)
𝑥 +𝑥
𝑔(𝑥) =
= [ 𝑥2 + 𝑥3 ] ,
𝑔3 (𝑥)
1
3
𝑥1 + 𝑥2 + 𝑥3
[ 𝑔4(𝑥) ]
𝑓(𝑔) = [
𝑓1 (𝑔)
]=
𝑓2 (𝑔)
−𝑔1 − 𝑔2 − 𝑔3 + 2𝑔4
]. Show that 𝐷𝑥 𝑓(𝑔(𝑥)) = 𝐷𝑔 𝑓(𝑔)𝐷𝑥 𝑔(𝑥).
𝑔1 + 𝑔2 + 𝑔3 + 𝑔4
𝑓(𝑥) = [
0
0
]; 𝐷𝑥 𝑓(𝑔(𝑥)) = [
3𝑥1 + 3𝑥2 + 3𝑥3
3
−1 −1
𝐷𝑔 𝑓(𝑔)𝐷𝑥 𝑔(𝑥) = [
1
1
1
−1 2 0
][
1 1 1
1
1
1
0
1
0 0
];
3 3
0
0
1
]=[
3
1
1
𝑥1
9. Let 𝑥 = [𝑥 ], and 𝑔(𝑥) = 𝑥1 + 𝑥2 + 𝑥1 𝑥2 . Show that
2
0 0
];;
3 3
𝜕𝑔(𝛼𝑥)
𝜕𝛼
= 𝐷𝛼𝑥 𝑔(𝛼𝑥)𝑥 ,
where 𝐷𝛼𝑥 𝑔(𝛼𝑥) is the derivative of 𝑔(𝛼𝑥) with respect 𝛼𝑥 evaluated at
𝛼𝑥 . Note that 𝑔(𝛼𝑥) and 𝑔(𝑥) are evaluating the same function at
different values.
𝑔(𝛼𝑥) = 𝛼𝑥1 + 𝛼𝑥2 + 𝛼 2 𝑥1 𝑥2 ;
𝜕𝑔(𝛼𝑥)
𝜕𝛼
= 𝑥1 + 𝑥2 + 2𝛼𝑥1 𝑥2 ;
Microeconomics
Problem Set 1
NCKU Economics
𝑦1
Let 𝑦 = [𝑦 ] = 𝛼𝑥,
2
𝐷𝛼𝑥 𝑔(𝛼𝑥) = 𝐷𝑦 𝑔(𝑦) = [1 + 𝑦2
𝐷𝛼𝑥 𝑔(𝛼𝑥)𝑥=[1 + 𝛼𝑥2
1 + 𝑦1 ] = [1 + 𝛼𝑥2
1 + 𝛼𝑥1 ];
𝑥
1 + 𝛼𝑥1 ] [𝑥1 ]=𝑥1 + 𝑥2 + 2𝛼𝑥1 𝑥2
2