ECO362H1 Lecture Notes - Lecture 3: U.S. Route 30 In Pennsylvania, Budget Constraint
The household’s problem is given by
max
ca,cnφ(ca−¯c)−1
+ (1 −φ)c
−1
n
−1
subject to
paca+pncn=w
The Lagrangian is omitted for space. The first-order conditions are given by
0 = φc
−1
a+ (1 −φ)c
−1
n
−1
−1
φ(ca−¯c)−
1
−λpa
0 = φc
−1
a+ (1 −φ)c
−1
n
−1
−1
(1 −φ)c
−
1
n−λpn
plus the budget constraint. Putting the first two foc together:
pa(ca−¯c)
pncn
= φ
1−φ! pn
pa!−1
where we will define Γfor convenience. Rearranging we get
pncn= 1−φ
φ! pa
pn!−1
|{z }
=Γ
pa(ca−¯c)
We can plug the expression for pncninto the household’s budget constraint to get
paca+ Γpa(ca−¯c) = w
paca[1 + Γ] −Γpa¯c=w
ca=1
1+Γ"w
pa
+ Γ¯c#
ca=1
1+Γ
w
pa
+Γ
1+Γ¯c
1
Document Summary
The household"s problem is given by subject to max ca,cn (cid:20) (ca c) + (1 )c paca + pncn = w. 0 = (cid:20) c a + (1 )c a + (1 )c. Putting the rst two foc together: pa(ca c) pncn. 1 ! pn pa! 1 where we will de ne for convenience. Rearranging we get pncn = 1 . We can plug the expression for pncn into the household"s budget constraint to get pa(ca c) paca + pa(ca c) = w paca [1 + ] pa c = w ca = ca = We can then substitute this into the expression for cn to get pncn = pa(ca c) We can go through the above expression piece-by-piece. P1 a + (1 ) p1 n and. (cid:17) (cid:16) pa (1 ) p1 a + (1 ) p1 n n. 1+ together with the solutions for ca and cn, ca(pa, pn, w) = cn(pa, pn, w) =
