Econ 4721 Money and Banking Midterm 1

Econ 4721 Money and Banking Midterm 1 - Midterm Practice...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Midterm Practice Problems- Answer Key Econ 4721: Money and Banking, Fall 2009 Problem 3 Government expenses are some number g > 0 per old person. These expenses do not directly affect the consumer in any way. The government keeps the money supply constant, z = 1, but uses a Lump-sum tax T goods on each young person, where T is a positive number. The economy’s parameters are: n = 1 y 1 = 20 β = 0 . 75 y 2 = 4 (a) Write the government’s budget constraint, equating revenue from taxes to expenses in period t . Answer: expenses N t - 1 g = revenues N t T Since n = 1 , we also have g = T. (b) Write consumer budget constraints in both periods of life. Answer: c 1 ,t + v t m t = y 1 - T c 2 ,t +1 = y 2 + v t +1 m t (c) Solve for the rate of return of money in a stationary equilibrium. The answer should be a number. Answer: v t m t = y 1 - c 1 - T v t = N t ( y 1 - c 1 - T ) M t v t +1 v t = N t +1 ( y 1 - c 1 - T ) M t +1 N t ( y 1 - c 1 - T ) M t = N t +1 N t M t M t +1 v t +1 v t = 1 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
v t +1 /v t = 1 because both the population and the money supply are constant. (d) Combine the consumer’s two budget constraints in part (b) into a single equation i.e. write the consumer lifetime budget constraint. Graph this equation on the ( c 1 ,c 2 ) plane and draw an indifference curve to illustrate the consumer’s optimal choice in a stationary equilibrium. Answer: c 1 + v t m t = y 1 - T = m t = 1 v t ( y 1 - T - c 1 ) Lifetime budget constraint: c 2 = y 2 + v t +1 v t ( y 1 - T - c 1 ) Since v t +1 /v t = 1 , c 2 = y 2 + y 1 - T - c 1 (e) Solve for the consumption allocation ( c 1 ,c 2 ) in a stationary equilibrium. The answers may depend on the lump-sum tax T . Answer: max c 1 ,c 2 ,m t ln ( c 1 ) + β ln ( c 2 ) subject to: c 2 = y 2 + y 1 - T - c 1 2
Background image of page 2
The solution to the problem can be found by setting all the partial derivatives of the following Lagrangian function to zero. L ( c 1 ,t ,c 2 ,t +1 t ) = ln ( c 1 ) + β ln ( c 2 ) - λ ( c 2 - y 2 - y 1 + T + c 1 ) First-order condition, setting derivative equal to zero: c 1 : 1 c 1 - λ = 0 c 2 : β 1 c 2 - λ = 0 λ 1 ,t : c 2 - y 2 - y 1 + T + c 1 = 0 which is the LBC From the first two 1 c 1 = β 1 c 2 or c 2 = βc 1 Substitute into
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 10

Econ 4721 Money and Banking Midterm 1 - Midterm Practice...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online