PS1 - PS1 solutions corrected by the professor. IBE 2019-2020 PDF

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PS1 solutions corrected by the professor. IBE 2019-2020...

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Problem Set #1: – Solutions:

April 6, 2020

Microeconomics I ¨ urk Fabrizio Germano, Tahir Ozt¨ To be solved for Seminar 1 (Week that starts on April 29)

Budget Constraints. 1. (∗ ) Elena has the following budget constraint p1 x1 + p2 x2 ≤ m where p1 = 2 and p2 = 3. (a) Elena’s budget line is 2x1 + 3x2 = 30. (b) The slope is −p1 /p2 = −2/3. This means that if Elena buys one extra unit of good 1 she has to give up 2/3 units of good 2 to remain with the same income. (c) If income increases by 20% then we can write m′ = 36. Elena’s budget line is now 2x1 + 3x2 = 36. be cause prices don’t change, the slope is unaffected. (d) We have m′ = 36 and p1′ = 3. and Elena’s budget line is now 3x1 + 3x2 = 36. The slope of the budget line is now −p′1 /p2 = −1.

(e) Between (a) and (c) Elena is clearly better off since she has 6 more units of income to spend goods 1 and 2. However, between (a) and (d) it is not so clear. While her income also increases by 6 units, the price of good 1 also increases. In particular, if Elena was buying more than 6 units of good 1 as part of her optimal bundle on the budget line in (a), then she can no longer buy that same bundle with the budget in (d).

2. (∗ ) Anna faces prices of pb = 1.80 crowns per kilo of bread, the pm = 1.20 crowns per litre of milk, and pp = 0.60 crowns per kilo potatoes. (a) If Anna’s yearly income is m = 900 crowns, her yearly budget set for food: 1.8xb + 1.2xm + 0.6xp ≤ 900. To draw the budget line, solve for xp to obtain: xp = 1500 − 3xb − 2xm . (b) With the euro currency and an exchange rate of 1 euro ≡ 1.5 crowns, ′ = 0.8, p′ = 0.4 and her income is prices in euros are now: p′b = 1.2, pm p

1

600 euros. Her budget constraint is therefore: Anna’s yearly budget set in euros is now: 2 2 (1.8xb + 1.2xm + 0.6xp ) ≤ 900 3 3 ⇐⇒ 1.8xb + 1.2xm + 0.6xp ≤ 900 (in euros).

1.2xb + 0.8xm + 0.4xp ≤ 600 ⇐⇒

(c) If we use one kilo of potatoes as the num´eraire, then prices expressed in ′′ kilos of potatos are: pb′′ = 3, pm = 2, pp′′ = 1 and her income is 1500 kilos of potato (=900/0.6). Anna’s budget set is: 1 1 900 (1.8xb + 1.2xm + 0.6xp ) ≤ 0.6 0.6 ⇐⇒ 1.8xb + 1.2xm + 0.6xp ≤ 900.

3xb + 2xm + 1xp ≤ 1500 ⇐⇒

(d) If we set Anna’s consumption of potatoes to zero in the above budget set, then Anna’s budget set for bread and milk becomes: 3xb + 2xm ≤ 1500 ⇐⇒ 1.8xb + 1.2xm ≤ 900. To draw the corresponding budget line, solve for xm to obtain: 3 xm = 750 − xb . 2 Preferences. 3. (∗ ) The preference relation of the group can be summarized as follows: Group : X ≻ Y, Y ≻ Z, Z ≻ X As such, we can say that: (a) It is complete. (b) It is not transitive. (c) Because it is not transitive, it cannot be represented by a utility function since u would have to satisfy both u(X) > u(Z) and u(Z) > u(X). On the other hand, if Bob changes his preferences, then the new preference relation of the group can be summarized as: Group : X ≻new Y, Z ≻new Y, Z ≻new X, As a result we can now say that: (d) The new preference relation is complete and transitive and can be represented by a utility function, for example, u(X) = 2, u(Y ) = 1, u(Z ) = 3. 2

4. No, two indifference curves of the same individual cannot intersect. Otherwise they would both belong to the same indifference curve since all bundles on the two indifference curves would be indifferent to the bundle at the intersection. 5. (∗ ) As mentioned in the hint, Alice’s preferences are lexicographic. To determine whether a car is preferred to another you go down a list of items as when looking for a word in a dictionary. (a) Let C1 = (s1 , e1 , c1 ) and C2 = (s2 , e2 , c2 ) be two cars, where si ∈ {4, 5}, ei ∈ {100, 120, 140} and ci ∈ { black, red } for i = 1, 2. Then Alice’s preference relation can be defined formally as follows: C1 ≻ C2 ⇐⇒ (s1 > s2 ) or (s1 = s2 and e1 > e2 ) or (s1 = s2 and e1 = e2 and c1 = black, c2 = red) C1 ∼ C2 ⇐⇒ s1 = s2 and e1 = e2 and c1 = c2 .

(b) Yes, you can check that Alice’s preferences are both complete and transitive. (c) If the set of alternatives is finite and the preference relation is complete and transitive, then a utility function always exists that represents the preferences. Here’s an example in this case: u(4, 100, r) = 0, u(4, 100, b) = 0.001, u(4, 120, r) = 0.010, u(4, 120, b) = 0.011, u(4, 140, r) = 0.020, u(4, 140, b) = 0.021 u(5, 100, r) = 0.1, u(5, 100, b) = 0.101, u(5, 120, r) = 0.110, u(5, 120, b) = 0.111, u(5, 140, r) = 0.120, u(5, 140, b) = 0.121.

Utility Functions. √ 6. The utility functions that represent the same preferences as U (x1 , x2 ) = x1 x2 (where x1 > 0 y x2 > 0) are (c), (d), (e), (f ). It is easy to see that (a) and (b) have different indifference curves and in general different MRS than U . The positive monotonic transformations that show that (c)–(f ) represent the same preferences are: (c) f (U ) = ln U , (d) f (U ) = 2 ln U , (e) f (U ) = 32 + U 2 , (f ) f (U ) = 32 + 2U . Note that all these functions are strictly increasing in U , for U > 0. 7. (∗ ) See Figure 1 for a sketch of the indifference curves. Moreover: (a) u(x1 , x2 ) = x12 + 3 — preferences are monotone and convex, but neither strictly monotone nor strictly convex; good 2 is a neutral good. (b) u(x1 , x2 ) = −x1 + 4x2 — preferences are not monotone and are convex but not strictly convex; good 1 is a bad. (c) u(x1 , x2 ) = x12 +x2 — preferences are strictly monotone and are not convex (they are actually concave). (d) u(x1 , x2 ) = x12+x22 — preferences are strictly monotone and are not convex (they are actually strictly concave). 3

(e) u(x1 , x2 ) = (x1 + 4x2 )2 — preferences are strictly monotone and convex, but not strictly convex; goods 1 and 2 are perfect substitutes. (f ) u(x1 , x2 ) = −(x1 − 2)2 − (x2 − 3)2 — preferences are not monotone and are strictly convex; bundle (2, 3) is a satiation point. (g) u(x1 , x2 ) = min{x1 + 2x2 , 2x1 + x2 } — preferences are strictly monotone and convex, but not strictly convex. (h) u(x1 , x2 ) = max{x1 + 2x2 , 2x1 + x2 } — preferences are strictly monotone, but not convex. 8. (∗ ) Again, see Figure 1 for a sketch of the indifference curves. Moreover: (a) u(x1 , x2 ) = x12 +3 — MU1 (x1 , x2 ) = 2x1 , MU2 (x1 , x2 ) = 0, MRS(x1 , x2 ) = −∞, MRS(4, 2) = −∞. Consumer does not care for good 2 but likes to have more of good 1 and hence is willing to give up “infinitely” many units of good 2 for an extra unit of good 1. (b) u(x1 , x2 ) = −x1 +4x2 — MU1 (x1 , x2 ) = −1, MU2 (x1 , x2 ) = 4, MRS(x1 , x2 ) = 1/4, MRS(4, 2) = 1/4. Consumer always willing to receive 1/4 units of good 2 for an extra unit of good 1 (remember good 1 is a bad). (c) u(x1 , x2 ) = x12 +x2 — MU1 (x1 , x2 ) = 2x1 , MU2 (x1 , x2 ) = 1, MRS(x1 , x2 ) = −2x1 , MRS(4, 2) = −8. Consumer willing to give up 2x1 units of good 2 for an extra unit of good 1. In particular, willing to give up 8 units of good 2 for an extra unit of good 1 at bundle (4, 2). (d) u(x1 , x2 ) = x12+x22 — MU1 (x1 , x2 ) = 2x1 , MU2 (x1 , x2 ) = 2x2 , MRS(x1 , x2 ) = −x1 /x2 , MRS (4, 2) = −2. Consumer willing to give up x1 /x2 units of good 2 for an extra unit of good 1. In particular, willing to give up 2 units of good 2 for an extra unit of good 1 at bundle (4, 2). (e) u(x1 , x2 ) = (x1 + 4x2 )2 — MU1 (x1 , x2 ) = 2x1 + 8x2 , MU2 (x1 , x2 ) = 8x1 + 32x2 , MRS(x1 , x2 ) = −1/4, MRS (4, 2) = −1/4. Consumer always willing to give up 1/4 units of good 2 for an extra unit of good 1. (f ) u(x1 , x2 ) = −(x1 −2)2 −(x2 − 3)2 — MU1 (x1 , x2 ) = 4−2x1 , MU2 (x1 , x2 ) = −2x1 6 − 2x2 , MRS(x1 , x2 ) = − 46−2x , MRS (4, 2) = −(−4)/2 = 2. The MRS 2 can be positive or negative depending on x1 and x2 . At (4, 2) consumer is willing to receive 2 units of good 2 for an extra unit of good 1 (remember the consumer is satiated with good one after x1 = 2). 9. The utility functions: (a) u(x, y, x) = x(y + z), (c) u(x, y, x) =

1 xy 3

+ 31 xz, (d) u(x, y, x) =

√ xy + xz

represent the same √ for (a) and √ preferences. To see this, use f (u) = u/3 to show (c), use f (u) = u to show for (a) and (d), and use f (u) = 3u to show for (c) and (d). Also, the utility functions: 4

Figure 1: Indifference curves for Questions 4 and 5 5

1

1

1

(b) u(x, y, x) = xyz, (e) u(x, y, x) = ln(x) + ln(y) + ln(z), (g) u(x, y, x) = x 3 y 3 z 3

represent the same preferences. To see this, use f (u) = ln u to√show for (b) and √ 3 (e), use f (u) = 3 u to show for (b) and (g), and use f (u) = e u to show for (e) and (g). Finally, no other utility function represents the same preferences as (f ) u(x, y, x) = x2 y 2 + 2xyz + x2 z 2 . √ 10. (∗ ) Bob’s utility function u(x, y) = ln(x) + y for x, y > 0. beginenumerate √ y+10

(a) To see why v(x, y) = eln(x)+ that the function

represents the same preferences, notice

f (u) = eu+10 is a positive monotonic transformation that maps u to v. That is f is strictly increasing (since (f ′ (u) = eu > 0 for all u) and f (u(x, y)) = v(x, y ) for all x, y > 0. This implies that u and v have the same indifference curves. (b) Fix u and (x0 , y0 ). Then MUx (x0 , y0 ) =

∂u(x0 , y0 ) ∂u(x0 , y0 ) 1 1 = √ = , MUy (x0 , y0 ) = 2 y0 x0 ∂x ∂x

(c) Fix now v and (x0 , y0 ). Then √

√ ∂v(x0 , y0 ) ∂v(x0 , y0 ) x0 · e y0 +10 = e y0 +10 , MVy (x0 , y0 ) = MVx (x0 , y0 ) = = √ ∂x ∂x 2 y0

The marginal utilities are not the same. We have MVx = f ′ (u(x0 , y0 )) · MUx (x0 , y0 ), MVy = f ′ (u(x0 , y0 )) · MUy (x0 , y0 ).

(d) Using u we have:

√ 2 y0 MUx (x0 , y0 ) 1/x0 MRS(x0 , y0 ) = − . =− √ =− 1/2 y0 x0 MUy (x0 , y0 ) (e) Using v we have: √ √ 2 y0 MVx (x0 , y0 ) e y0 +10 √ . MRS(x0 , y0 ) = − =− √ =− x0 MVy (x0 , y0 ) x0 · e y0 +10 /2 y0

The two utility functions have the same MRS at (x0 , y0 ). Since MVx = f ′ (u(x0 , y0 )) · MUx (x0 , y0 ) and MVy = f ′ (u(x0 , y0 )) · MUy (x0 , y0 ), we have MRS(x0 , y0 ) = −

MUx (x0 , y0 ) f ′ (u(x0 , y0 )) · MUx (x0 , y0 ) =− . ′ f (u(x0 , y0 )) · MUy (x0 , y0 ) MUy (x0 , y0 )

And because (x0 , y0 ) was arbitrary, we deduce that u and v have the same MRS everywhere.

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