Department of Economics Fall 2019 Problem Set 3 - Solutions

EC 501 MICROECONOMICS Bo˘gazi¸ciUniversity - Department of Economics Fall 2019 Problem Set 3 - Solutions

1. Derive the Walrasian demands and indirect utility function for each of the utility functions below.

α β (a) u(x1, x2) = x1 x2

(b) u(x1, x2) = ln(x1) + x2

Solution: (a) If either x1 or x2 is zero, utility is zero. But positive utility is possible. Thus this cannot

be optimal. That is both x1 and x2 must be strictly positive at the optimum. Using a Lagrangian

α β L = x1 x2 + λ[w − p1x1 − p2x2]

The FOC are α−1 β αx1 x2 − λp1 = 0

α β−1 βx1 x2 − λp2 = 0

And the budget constraint w = p1x1 − p2x2. We can solve the ﬁrst equation for λ and substitute it into the second to get

p1x1 = (α/β)p2x2

Substituting this into the budget constraint yields

p2x2[1 + (α/β)] = w

which gives us the Walrasian demand: αw x1 = p1[α + β] and βw x2 = p2[α + β] The indirect utility function is obtained by substituting these into the utility function. Thus we get

α β w v(p, w) = ( )α( )β( )α+β p1 p2 α + β

(b) Assume that we have an interior solution for a moment. The Lagrangian is

L = log(x1) + x2 + λ[w − p1x1 − p2x2]

The FOC are 1 − λp1 = 0 x1

1 1 − λp2 = 0

We get p2 x1 = p1 Substituting into the budget constraint we get

p2 + p2x2 = w that is w − p2 x2 = p2

Note that this solution only makes sense if w ≥ p2. If this condition does not hold, then the constraint x2 ≥ 0 is binding and we have a corner solution x2 = 0 and x1 = w/p1. Thus Walrasian demand is given by p w − p x(p, w) = ( 2 , 2 ) p1 p2 when w ≥ p2, and w x(p, w) = ( , 0) p1 when w < p2. The indirect utility function is then given by

p w − p v(p, w) = ln 2 ) + 2 p1 p2 when w ≥ p2, and w v(p, w) = ln( ) p1 when w < p2.

(c) Solve budget constraint for x2 and substitute it into the objective function. We get x2 = (w − p1x1)/p2 and substituting it, objective function becomes ax1 − (p1/p2)x1 + w/p2. This is linear in x1.

If a > p1/p2, then utility is strictly increasing in x1, so the optimal choice is x2 = 0 and x1 = w/p1.

If a < p1/p2, then utility is strictly decreasing in x1, so the optimal choice is x1 = 0 and x2 = w/p2.

And, if a = p1/p2, then any choice on the budget line is optimal. The indirect utility function is then w v(p, w) = min{p1/a, p2} √ (d) First note that we must have x1 = x2. Otherwise, for example, if x1 is larger, then a small reduction in the expenditure on x1 and a small increase in the expenditure on x2 must increase utility. √ 2 Hence we must have x1 = x2 at the optimum. So x2 = x1. Substituting this into the budget 2 constraint we get p1x1 + p2x1 = w. Solving for x1 we get

p 2 −p1 ± p1 + 4p2w x1 = 2p2

2 p 2 p 2 Since p1 + 4p2w > p1 = p1 we must have a +, so

p 2 −p1 + p1 + 4p2w x1(p, w) = 2p2 and p 2 −p1 + p1 + 4p2w 2 x2(p, w) = ( ) 2p2 Now the indirect utility function is

−p + pp2 + 4p w v(p, w) = 1 1 2 2p2

2. Suppose we have an agent who derives utility both from consumption and also wealth w, and she

2 α1 α2 1−α1−α2 chooses x ∈ R+ to maximize u(x, w) = x1 x2 w (subject to the same constraints as we had before) where αl > 0 for l = 1, 2, and α1 + α2 < 1. Derive the Walrasian demands and indirect utility function. Is the Lagrange multiplier still the marginal utility of wealth? Solution: The fact that w enters the utility function has no eﬀect on the demands. To see the point,

α1 α2 1−α1−α2 note that for a ﬁxed w, the bundle (x1, x2) which maximizes u(x, w) = x1 x2 w over the α1 α2 feasible set must also maximize x1 x2 over the feasible set. Hence the demands are just like the one in 8(a) above: αlw xl(p, w) = (α1 + α2)pl for l = 1, 2. On the other hand because w enters the utility function, the indirect utility function is now diﬀerent. We have α α v(p, w) = [ 1 ]α1 [ 2 ]α2 w (α1 + α2)p1 (α1 + α2)p2 So ∂v(p, w) α α = [ 1 ]α1 [ 2 ]α2 ∂w (α1 + α2)p1 (α1 + α2)p2 This is not equal to the Lagrange multiplier. To see this, recall that the Lagrange multiplier is the ratio of marginal utility to the price. Hence

α xα−1xα2 w1−α1−α2 λ(p, w) = 1 1 2 p1 α α w α w = 1 [ 1 ]α1 [ 2 ]α2 w1−α1−α2 p1 (α1 + α2)p1 (α1 + α2)p2 α α 1 α1 2 α2 = (α1 + α2)[ ] [ ] (α1 + α2)p1 (α1 + α2)p2

Note that λ 6= ∂v/∂w. For equality of the two, the argument was that λ should equal the derivative of the Lagrangian with respect to w since the first order effect of the change in w on the Lagrangian through the effect of w on x is zero. Now, though, we have a separate first order effect of a change in w on the Lagrangian through the fact that w enters the objective function.

2 3. Assuming X = R+, consider the lexicographic preferences with x = (x1, x2) y = (y1, y2) if x1 > y1 or if x1 = y1 and x2 > y2. Find the Walrasian demand. Is it continuous in (p, w). How about alternative

3 lexicographic preferences with x = (x1, x2) y = (y1, y2) if x1 > y1 or if x1 = y1 and x2 < y2?

Solution: The Walrasian demands are x1(p, w) = w/p1 and x2(p, w) = 0 for both type of lexicographic preferences. Even though the preferences are discontinuous, the demands are not only continuous but are also very simple. Indeed, the preferences are the same as one another and the same as those arising

from a utility function of u(x1, x2) = x1.

1/2 1/2 4. Consider the utility function u(x1, x2) = 2x1 + 4x2 for two goods. Assume an interior solution.

(a) Find the Hicksian demand function, h(p, u).

(b) Find the expenditure function, and verify that h(p, u) = Ope(p, u). (c) Find the indirect utility function. (d) Using Roy’s identity ﬁnd the Walrasian demand, x(p, w).

1/2 1/2 Solution: (a) Solve the following problem: minx1,x2 p1x1 +p2x2 subject to 2x1 +4x2 ≥ u assuming 1/2 1/2 x1 > 0, x2 > 0. So we can solve the following Lagrangian: L = p1x1 + p2x2 − λ[2x1 + 4x2 − u]. The FOC are 1 p − λ[ ] = 0 1 1/2 (x1) and 2 p − λ[ ] = 0 2 1/2 (x2) and the constraint 1/2 1/2 2x1 + 4x2 = u The FOC imply that (1/2) 2p1 1/2 x2 = x1 p2 Substitute this into the constraint and obtain

1/2 2p1 1/2 1/2 p1 2x1 + 4( x1 ) = 2x1 (1 + 4 ) = u p2 p2

which implies that 2 p2u h1(p, u) = 2(4p1 + p2)

and then solving for x2, we get 2 p1u h2(p, u) = 4p1 + p2

(b) Expenditure function is given by

e(p, u) = p1h1(p, u) + p2h2(p, u) 2 2 p2u p1u = p1 + p2 2(4p1 + p2) 4p1 + p2 u2p p = 1 2 4(4p1 + p2)

4 To verify h(p, u) = Ope(p, u), we have u2p p ∂e(p, u) ∂[ 1 2 ] = 4(4p1+p2) ∂p1 ∂p1 2 p2u = = h1(p, u) 2(4p1 + p2)

u2p p ∂e(p, u) ∂[ 1 2 ] = 4(4p1+p2) ∂p2 ∂p2 2 p1u = = h2(p, u) 4p1 + p2

(v(p, w))2p p 1 2 = w 4(4p1 + p2)

solving for v(p, w) we get w 4w 1/2 v(p, w) = 2 + p1 p2

(d) To use Roy’s identity, ﬁrst we calculate the following from the indirect utilty function

∂v(p, w) w 4w 1 w w(4p1 + p2) 1 w − 2 − 2 = [ + ] [− 2 ] = [ ] [− 2 ] ∂p1 p1 p2 p1 p1p2 p1

∂v(p, w) w 4w 1 4w w(4p1 + p2) 1 4w − 2 − 2 = [ + ] [− 2 ] = [ ] [− 2 ] ∂p2 p1 p2 p2 p1p2 p2

∂v(p, w) w 4w − 1 1 4 w(4p1 + p2) − 1 4p1 + p2 = [ + ] 2 [ + ] = [ ] 2 [ ] ∂w p1 p2 p1 p2 p1p2 p1p2 Then Roy identity implies

w − 2 ∂v(p, w)/∂p1 p1 wp2 x1(p, w) = − = − 4p +p = 2 ∂v(p, w)/∂w 1 2 p1p2 + 4p p1p2 1 and 4w − 2 ∂v(p, w)/∂p2 p2 4wp1 x2(p, w) = − = − 4p +p = 2 ∂v(p, w)/∂w 1 2 4p1p2 + p p1p2 2

5. Given the utility function u(x1, x2) = ln(x1) + x2 for two goods, assume an interior solution.

(a) Find the Hicksian demand function h(p, u).

(b) Find the expenditure function e(p, u), and verify that h(p, u) = Ope(p, u). (c) Verify by substitution that the Hicksian demand functions you found can be obtained from the Walrasian demand functions you found in 8(b), using the expenditure function.

5 Solution: (a) When the solution is interior, the Lagrangian

L = p1x1 + p2x2 − λ[ln(x1) + x2 − u] yields the following FOC

p1 − λ(1/x1) = 0 and

p2 − λ = 0

These two together yield p2 x1 = p1 Plugging this into the constraint

ln(x1) + x2 = u we get x2 = u − ln(p2/p1). Thus, the Hicksian demand is

p2 h1(p, u) = p1

p2 h2(p, u) = u − ln p1

(b) From its deﬁnition, the expenditure function is

e(p, u) = p1h1(p, u) + p2h2(p, u) p2 p2 = p1 + p2 u − ln p1 p1 p2 = p2 1 + u − ln . p1

To verify h(p, u) = Ope(p, u), we have p2 ∂e(p, u) ∂[p2(1 + u − ln( ))] = p1 ∂p1 ∂p1 p2 ∂[−p2 ln( )] = p1 ∂p1 ∂[p ln(p )] = 2 1 ∂p1 p2 = = h1(p, u) p1

6 p2 ∂e(p, u) ∂[p2(1 + u − ln( ))] = p1 ∂p2 ∂p2 = 1 + u − ∂[p2 ln(p2) − p2 ln(p1)]/∂p2

= 1 + u − ∂[p2 ln(p2)]/∂p2 + ln(p1)

= 1 + u − (ln(p2) + 1) + ln(p1)

= u − ln(p2/p1) = h2(p, u)

p w − p x(p, w) = ( 2 , 2 ) p1 p2

The question asks us to verify by substitution, using the answers above, that xi(p, e(p, u) = hi(p, u). For i = 1 this is immediate, since

p2 x1(p, e(p, u)) = = h1(p, u). p1

For i = 2, we have

e(p, u) − p2 x2(p, e(p, u)) = p2

p2 p2 1 +u ¯ − ln p − p2 = 1 p2 p = u − ln 2 p1 = h2(p, u)

PL ρ 1/ρ 6. Suppose the consumer’s utility function is u(x) = ( l=1 xl ) where 0 6= ρ > 0.

(a) Find the Walrasian demand, x(p, w) and the indirect utility function, v(p, w) (b) Find the Hicksian demand, h(p, u) and the expenditure function, e(p, u). (c) Verify Roys Identity and the Slutsky equation.

Solution: (a) The FOC for the UMP can be written as

ρxρ−1 ρxρ−1 l = 1 pl p1

for each l = 1, ..., L. Rearranging this we get

1/(ρ−1) pl xl = x1 p1

7 Substituting into the budget constraint for all xls except x1, we get

1/(ρ−1) X pl p1x1 + plx1 = w p1 l6=1

Then get 1/(ρ−1) w wp1 x1(p, w) = = (ρ/(ρ−1)) P pr P pl l l p1 + l6=1 1/(ρ−1) p1 where r = ρ/(ρ − 1). Note that 1/(ρ − 1) = r − 1. Substituting for x1 into the FOC above we get

r−1 r−1 r−1 wp1 pl wpl xl(p, w) = P r = P r k pk p1 k pk

r−1 r−1 r−1 r−1 wp1 wp1 p2 wp2 For instance, when L = 2 we have x1(p, w) = pr+pr and x2(p, w) = pr+pr r−1 = pr+pr . Now, to get 1 2 1 2 p1 1 2 the indirect utility function, simply substitute for xls into the utility function and get

X r (−1/r) v(p, w) = w[ pk] k

(b) To get the Hicksian demands and expenditure function, you can go either of two ways. First, you can directly solve the expenditure minimization problem. Second, you can take advantage of the identities we have. Second approach is quicker. First use v(p, e(p, u)) = u to get e(p, u).

X r (−1/r) e(p, u)[ pl ] = u l which gives X r 1/r e(p, u) = u[ pl ] l

Now use h(p, u) = Ope(p, u) to get the Hicksian demands

!(1/r)−1 ∂e(p, u) r−1 X r hl(p, u) = = upl pk ∂pl k

!−1/r ∂v(p, w) X = pr ∂w k k and !(−1/r)−1 !(−1/r)−1 ∂v(p, w) w X r r−1 r−1 X r = − pk rpl = −wpl pk ∂pl r k k

8 Then Roy’s identity holds since

∂v(p,w) !−1 ∂p X − l = wpr−1 pr = x (p, w) ∂v(p,w) l k l ∂w k as we found above.

For the Slutsky equation, I will consider ∂hl(p, u)/∂pl, that is the entries in the diagonal. You can complete the rest yourself. We need

r−2 P r r−1 r−1 P r r ∂xl(p, w) (r − 1)pl ( k pk) − pl (rpl ) r−2 (r − 1)( k pk) − rpl = w P r 2 = wpl P r 2 ∂pl ( k pk) ( k pk) and r−1 ∂xl(p, w) pl = P r ∂w k pk Now, for Slutsky equation to hold we need

∂xl(p, w) ∂xl(p, w) (∂hl(p, u)/∂pl)|u=v(p,w) = + xl(p, w) ∂pl ∂w

The left hand side is

(r − 1)pr−2(P pr )1−(1/r) − pr−1(1 − (1/r))(P pr )−1/rrpr−1 (∂h (p, u)/∂p )| = v(p, w) l k k l k k l l l u=v(p,w) P r (2−2/r) ( k pk) r−2 P r 1−(1/r) r−1 P r −1/r r−1 X (r − 1)p ( p ) − p (r − 1)( p ) p = w( pr )(−1/r) l k k l k k l k (P pr )(2−2/r) k k k X X (P pr ) − pr = w( pr )(−1/r)(r − 1)pr−2( pr )−(1/r) k k l k l k (P pr )(2−2/r) k k k k P r r r−2 ( k pk) − pl = w(r − 1)pl P r 2 ( k pk)

The right hand side is

P r r r−1 ∂xl(p, w) ∂xl(p, w) r−2 (r − 1)( k pk) − rpl pl + xl(p, w) = wpl P r 2 + P r xl(p, w) ∂pl ∂w ( k pk) k pk P r r r−1 r−1 r−2 (r − 1)( k pk) − rpl pl wpl = wpl P r 2 + P r P r ( k pk) k pk k pk P r r r 1 r r−2 ( k pk) − r−1 pl + r−1 pl = w(r − 1)pl P r 2 ( k pk) P r r r−2 ( k pk) − pl = w(r − 1)pl P r 2 ( k pk)

9 The two sides are equivalent. Thus, Slutsky equation is established (on the diagonal).

7. (a) Find the Hicksian demands and the expenditure function implied by the indirect utility function given by v(x , x , w) = w + w . 1 2 p1 p2 (b) Now suppose that you only know the expenditure function you have found in part (a) above. Find the Walrasian demands and the indirect utility function implied by this expenditure function. (c) Verify that the Slutsky equation holds for good 1. Solution: (a) The expenditure function can be obtained through the identity v(p, e(p, u)) = u, that is,

e(p, u)/p1 + e(p, u)/p2 = u

which implies p p u e(p, u) = 1 2 p1 + p2

Finally, to get the Hicksian demands, we can diﬀerentiate the expenditure function.

2 ∂e(p, u) p2u p1p2u p2u h1(p, u) = = − 2 = 2 ∂p1 p1 + p2 (p1 + p2) (p1 + p2)

2 ∂e(p, u) p1u p1p2u p1u h2(p, u) = = − 2 = 2 ∂p2 p1 + p2 (p1 + p2) (p1 + p2)

(b) Using e(p, v(p, w)) = w we can get the indirect utility function.

p p v(p, w) 1 2 = w p1 + p2

implying (p + p )w w w v(p, w) = 1 2 = + p1p2 p1 p2

Now using Roy’s identity, we can get the Walrasian demands

2 ∂v(p, w)/∂p1 w/(p1) wp2 x1(p, w) = − = = ∂v(p, w)/∂w (1/p1) + (1/p2) p1(p1 + p2)

and 2 ∂v(p, w)/∂p2 w/(p2) wp1 x2(p, w) = − = = ∂v(p, w)/∂w (1/p1) + (1/p2) p2(p1 + p2)

2 ∂h1(p, u) 2p2u = − 3 ∂p1 (p1 + p2)

and ∂h1(p, u) 2p1p2u = 3 ∂p2 (p1 + p2)

10 We evaluate these at u = v(p, w) to able to compare to the Walrasian derivatives. Plugging v(p, w) for u, we get ∂h1(p, v(p, w)) 2p2w = − 2 ∂p1 p1(p1 + p2) and ∂h1(p, v(p, w)) 2w = 2 ∂p2 (p1 + p2) For the Walrasian demands we get

2 ∂x1(p, w) ∂x1(p, w) p2(2p1 + p2)w p2w 2p2w + x1(p, w) = − 2 2 + 2 2 = − 2 ∂p1 ∂w p1(p1 + p2) p1(p1 + p2) p1(p1 + p2) and ∂x1(p, w) ∂x1(p, w) w w 2w + x2(p, w) = 2 + 2 = 2 ∂p2 ∂w (p1 + p2) (p1 + p2) (p1 + p2) Thus, Slutsky equation holds for good 1 (you can show that it also holds for good 2, analogously).