VI.

Demand Relationships

D Cross-price effect
D The two-good case

• See Figure 6.1.

• When PY declines, if the substitution effect is small, consumption of X will increase:
Gross complements
• When PY declines, if the substitution effect is large, consumption of X will decrease:
Gross substitutes
D Slutsky equation for a change in PY on mX

• Differentiate both sides of H(PX , PY , U) ≡ X(PX , PY , E(PX , PY , U)) with respect to

PY .∂H
∂PY
∂X
= ∂PY
+ ∂X ∂E
∂I ∂PY
. By the Shephard lemma, ∂E
∂PY
= K = Y . Therefore, ∂X
∂PY
=
∂H ∂X
∂PY
− ∂I Y
• ∂H
∂PY
> 0 while − ∂X
∂I
< 0 assuming X is normal so that the combined effect is ambiguous
D Gross substitutes and gross complements

• ∂X
∂PY
> 0: X and Y are gross substitutes
• ∂X
∂PY
< 0: X and Y are gross complements
• Price effect needs not to be symmetric

– It is possible that ∂X
∂PY
> 0 and ∂Y
∂PX
<0
– Example: p.165
D Net substitutes and net complements


• define net substitutes and net complements with the sign of ∂H
∂PY
= ∂K
∂PX
or ∂X
∂PY  =
 U =const
∂Y 
∂PX 
U =const
∂2E
– By Shepherd’s lemma, ∂E
∂PX
= H and ∂E
∂PY
= K so that ∂PX ∂PY
= ∂H
∂PY
= ∂K
∂PX
by
Young’s Theorem.
– If there are only two goods, they must be net substitutes.
D Two important results of the theory of individual choice

• The own-substitute effect is nonpositive: ∂H

∂PX
≤ 0 and ∂K
∂PY
≤0
• The cross-substitution effects are symmetric: ∂H
∂PY
= ∂K
∂PX

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 • Example. Cobb-Douglas utility function with α = β = 12
t t
−3/2 −1/2 −1/2 −3/2
– Since H = U PPXY and K = U PPXY , ∂P ∂H
X
= − U2 PX PY ∂K
< 0, ∂PY
= − U2 PX PY <
∂H ∂K
0, ∂PY
= ∂PX
= U 2√P1X PY
• Composite Commodities
• Group goods into larger aggregates

– Food, clothing, shelter

– A specific good called X and ’’all other goods’’ called Y
D Consider the following problem:

max U(X, Y, Z) subject to PX X + PY Y + PZ Z = I

X,Y,Z

• Suppose the solution is given by X ∗ , Y ∗ and Z ∗ . The following two step maximization
generates the same solution.

(1)V (X, M) = max U(X, Y, Z) subject to PY Y + PZ Z = M

Y,Z

(2) max V (X, M) subject to PX X + M = I

X,M

– (In case you are interested) We first show that given X = X ∗ and M = M ∗ ≡
PY Y ∗ + PZ Z ∗ , (Y ∗ , Z ∗ ) is the solution to (1). In other words, V (X ∗ , M ∗ ) =
U (X ∗ , Y ∗ , Z ∗ ).

· First note that (Y ∗ , Z ∗ ) satisfies the constraint (1). Suppose that (Y , Z ) =

(Y ∗ , Z ∗ ) is a feasible solution to (1) and

U (X ∗ , Y , Z )
> U (X ∗ , Y ∗ , Z ∗ ).

Since

PX X ∗ + PY Y + PZ Z
= PX X ∗ + M ∗
= PX X ∗ + PY Y ∗ + PZ Z ∗
= I,

(X ∗ , Y , Z ) is feasible to the original problem, which is a contradiction.

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 – Next, we show that (X ∗ , M ∗ ) is the optimal solution to (2) using the above result.
e M)
· Suppose that (X, f = (X ∗ , M ∗ ) is a feasible solution to (2) such that

e M)
V (X, f > V (X ∗ , M ∗ ).

Let (Ye , M)
f be the optimal solution to (1) given (X,
e M).
f Then, since

e + PY Ye + PZ Ze
PX X
e +M
= PX X f

= I,
e Ye , Z)
(X, e is feasible to the original problem. But

e Ye , Z)
U (X, e = V (X,
e M)
f > V (X ∗ , M ∗ )

= U(X ∗ , Y ∗ , Z ∗ ),

which is a contradiction.
– Now, suppose that PY and PZ move together, that is, PY = tPY0 and PZ = tPZ0 for
some t and the initial prices, PY0 and PZ0 . If we define H = PY0 Y +PZ0 Y and therefore
M = tH, the above two-step maximization problem becomes

(1)V (X, H) = max U(X, Y, Z) subject to PY0 Y + PZ0 Z = H

Y,Z

(2) max V (X, H) subject to PX X + tH = I.

X,H

By inspecting (2), we can regard H as a (composite) good with price t, maximizing V with
respect to X and H.

• Example: p.169

– U (X, Y, Z) = − X1 − 1
Y
− 1
Z

– In the first step, maxY,Z − X1 − 1

Y
− 1
Z
subject to PY0 Y + PZ0 Z = H. L = − X1 −
1
− + λ(H − PY0 Y − PZ0 Z). LY = Y12 − λPY0 = 0 and LZ = Z12 − λPZ0 = 0.
1
Y Z t 0
PY
From this, Z = PZ0
Y. Using the constraint, we have Y = KHY and Z = KHZ
s s
where KY = PY0 + PY0 PZ0 and KZ = PZ0 + PY0 PZ0 . Next, maxX,H V (X, H) =
− X1 − KY +KZ
H
subject to PX X + tH = I. L = − X1 − KY H
+KZ
+ µ(I − PX X − tH)
s
LX = 1
X2
− µPX = 0 and LH = KYH+K 2
Z
− µt = 0 or H = (KY + KZ )PX /tX.
Substituting into the budget constraint, we have X = √ I √
PX + PX PY + PX PZ
. (You need

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 √ √ √
to make use of PY + PZ + 2 PY PZ = ( PY + PZ )2 .) Compare this with (6.32).

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