Utility Maximisation in 2 Periods

Transcription

1 Utility Maximisation in Periods Rob Pryce Introduction This short chapter gives an overview of inter-temporal utility maximisation. It begins with a brief discussion of the theory, to give us some priors we can use when checking the answers we get for problems. It provides a full solution to a general algebraic problem, before a numerical inter-temporal problem. It concludes with several problems for you to work through. Feel free to go through the numerical example before the algebraic if this makes you feel more comfortable. Theory A simple utility maximisation problem normally focuses on two (or more) goods, where the individual makes a consumption decision in one period. The individual is then assumed to make other, separate choices in future periods. In the latter half of the 0th century, economists developed models which dealt with inter-temporal consumption choice, where individuals make forward-looking consumption decisions. This makes sense in the real world - we don t typically spend all of our current income within the current period; we save for, or borrow from, our future selves. Friedman s permanent income hypothesis is an example of work that attempted to explain why and how individuals make consumption decisions. The inter-temporal utility function is not too different from the standard, two-good static utility function. The individual will have preferences with regards to current and future consumption, just as he has preferences with regards to good A and good B. The fact that the individual typically prefers consumption today compared to tomorrow is called discounting. An example of an inter-temporal, Cobb-Douglas utility function is U (C,C ) = C α Cβ () although remember that there is no reason why it should be Cobb-Douglas - this is just a simplification. Note that discounting would mean that α is greater than β. The main difference between inter-temporal utility maximisation and the static, two-good utility maximisation is the budget constraint. This is because the idea of price is perhaps harder to understand. In inter-temporal problems, the price element arises because of interest and inflation - the combination of the two forming the real interest rate r.

2 . No Real Interest Rate It makes sense to start from basics, where there is no real interest rate - imagine that there is no inflation and no interest rate available. In this case the budget constraint is simply C +C Y +Y () where Y and Y are the incomes in period and respectively. In an even simpler case, with no discounting (ie. the individual gets as much utility from tomorrow as he does today) and no difference in income between the two periods, the individual will divide consumption equally between the two periods.. Adding a Real Interest Rate If we add in a real interest rate, then any consumption today comes at the opportunity cost of consumption tomorrow. We can now write the budget constraint as ( + r)c +C ( + r)y +Y (3) Firstly, note how if r = 0 (there is no real interest rate) the budget constraint is equal to Equation. Secondly, note that income we get in the first period can be saved and earn r interest. Another important note is that the budget constraint is sometimes written differently as C + C ( + r) = Y + Y (4) ( + r) Either budget constraint is perfectly fine - they are the same (Equation 4 is found by dividing Equation 3 by ( + r)). The budget constraint is essentially setting the net present value of consumption equal to the net present value of income..3 Graphical Representation Figure shows the inter-temporal problem graphically. Notice that it looks very similar to the static, two-good problem. If the individual consumes everything in period, then C = Y + Y (+r). Likewise, if the individual saves everything until the second period, then C = Y ( + r) +Y. Take a moment to work out what happens to the budget constraint if the interest rate increases. Figure : Graphical Representation

3 3 General Problem - Worked Solution Suppose the inidividual s utility function is U = C α Cβ, the individual earns Y in the first period and Y in the second period, and the real interest rate r. To solve this problem, we can use the Lagrange multiplier method like we do for the static, two-good problem. Remember that the Lagrange multiplier method is set out as V = U +λ (BC) whatever is being written so that the part in the maxmised/minimised brackets is equal to zero (5) so in this case, the Lagrangian is written V = C α Cβ + λ(( + r)y +Y ( + r)c C ) (6) Taking the differential with respect to C and C gives δc = C 0.75 λ( + r) = 0 = αc (α ) C β = λ( + r) (7) and meaning that = αc(α ) C β (+r) = λ = βc αcβ λ = 0 = βc αcβ = λ βc α Cβ = αc(α ) C β ( + r) (8) (9) C = β α ( + r)c (0) We then substitute this information back into the budget constraint to get BC: ( + r)c +C = ( + r)y +Y () ( + r)c + β α ( + r)c = ( + r)y +Y () α + β α ( + r)c = ( + r)y +Y (3) C = α ( + r)y +Y α + β ( + r) (4) 3

4 C is found as C = β α ( + r)c (5) C = β α ( + r) α α + β C = ( + r)y +Y ( + r) 3. What Happens when Parameters Change? Using equation (0) we can see that (6) β α + β ( + r)y +Y (7) δα <0 δβ >0 δr >0 so that, holding all else constant, if the individual is a heavier discounter (α increases relative to β) then consumption goes up more in the first period relative to the second period s consumption. If the interest rate increases, the individual saves more in the first period and consumes more in the second. 4 Numerical Problem - Worked Solution Suppose U = C 0.6 C0.4, income is given as Y = 0 and Y = 30. The real interest rate, r is 5%. 4. Step - Write out the Lagrangian 4. Take the Differentials V = C 0.6 C0.4 + λ(.05y +Y.05C C ) (8) Taking the differentials and setting equal to zero gives δc = 0.6C 0.4 C λ = 0 (9) δc = 0.6C 0.4 C 0.4 =.05λ (0) 4

5 and = δc C 0.4 C 0.4 = λ () = 0.4C 0.6 C 0.6 λ = 0 () = 0.4C 0.6 C 0.6 = λ (3) 4.3 Set Lambdas Equal Using equations (8) and (0) we get which we can re-arrange to get 4.4 Substitute into Budget Constraint Substituting Y = 0, Y = 30, and C = (.05) C gives C 0.4 C 0.4 = 0.4C 0.6 C 0.6 (4) C = (.05) C (5) BC:.05C +C =.05Y +Y (6).05C + (.05) C =.05(0) + 30 (7) (C ) = 5 (8) C = 9.43 (9) C = (.05) C = 0.4 (30) It is possible to use the algebraic solution to solve this numerical example - you will get the same answers. The next page has several problems for you to work through. The first question changes each variable one at a time to allow comparison. The second provides a real-life example. The third uses a slightly different (so-called Stone-Geary) utility function. 5

6 5 Practice Exercises Question Question will change only one variable at a time. Question a C0., Y = 00, Y = 00, r = 5% Question b C0., Y = 00, Y = 00, r = 5% Question c C0., Y = 00, Y = 0, r = 5% Question d U = C C 0., Y = 00, Y = 00, r = 5% Question e C0.8, Y = 00, Y = 00, r = 5% Question f C0., Y = 00, Y = 00, r = 0% Question g C0., Y = 00, Y = 00, r = % Question You are a student. You live for periods, your student year and your working year. You don t earn any money in your student year, so Y = 0. Your working year consists of your graduate salary Y = 0 (where Y is measured in thousands of pounds). Your utility function is U = C C 0.95 and the real interest rate is %. How much do you borrow in your student period? Question 3 U = (C 50) 0.8 (C 50) 0., Y = 00, Y = 00, r = 5% (notice how this is similar to question a, but the individual has decided that he needs C 50 in either period to survive). 6