Download Utility Maximisation in 2 Periods

Utility Maximisation in 2 Periods
Rob Pryce
www.robpryce.co.uk/teaching
1
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.
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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 20th 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
β
U(C1 ,C2 ) = C1α C2
(1)
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.
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2.1
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
C1 +C2 ≤ Y1 +Y2
(2)
where Y1 and Y2 are the incomes in period 1 and 2 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.
2.2
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
(1 + r)C1 +C2 ≤ (1 + r)Y1 +Y2
(3)
Firstly, note how if r = 0 (there is no real interest rate) the budget constraint is equal to
Equation 2. 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
Y2
C2
= Y1 +
(4)
C1 +
(1 + r)
(1 + r)
Either budget constraint is perfectly fine - they are the same (Equation 4 is found by dividing
Equation 3 by (1 + r)). The budget constraint is essentially setting the net present value of
consumption equal to the net present value of income.
2.3
Graphical Representation
Figure 1 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 1, then C1 =
Y2
. Likewise, if the individual saves everything until the second period, then C2 =
Y1 + (1+r)
Y1 (1 + r) +Y2 . Take a moment to work out what happens to the budget constraint if the interest
rate increases.
Figure 1: Graphical Representation
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General Problem - Worked Solution
β
Suppose the inidividual’s utility function is U = C1α C2 , the individual earns Y1 in the first period
and Y2 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
β
V = C1α C2 + λ ((1 + r)Y1 +Y2 − (1 + r)C1 −C2 )
(6)
Taking the differential with respect to C1 and C2 gives
δV
δC1
= C20.75 − λ (1 + r) = 0
(α−1) β
C2
= αC1
= λ (1 + r)
(7)
(α−1) β
=
αC1
C2
(1+r)
and
δV
δC2
= λ
β −1
= βC1α C2
=
−λ = 0
β −1
βC1α C2
= λ
(8)
meaning that
β −1
βC1α C2
C2 =
(α−1) β
αC1
C2
=
(1 + r)
β
(1 + r)C1
α
(9)
(10)
We then substitute this information back into the budget constraint to get
BC: (1 + r)C1 +C2 = (1 + r)Y1 +Y2
(1 + r)C1 +
β
(1 + r)C1 = (1 + r)Y1 +Y2
α
α +β
(1 + r)C1 = (1 + r)Y1 +Y2
α
C1 =
α (1 + r)Y1 +Y2
α +β
(1 + r)
3
(11)
(12)
(13)
(14)
C2 is found as
β
(1 + r)C1
α
(15)
α (1 + r)Y1 +Y2
β
(1 + r)
α
α +β
(1 + r)
(16)
β
(1 + r)Y1 +Y2
α +β
(17)
C2 =
C2 =
C2 =
3.1
What Happens when Parameters Change?
Using equation (10) we can see that
δC2
δ α <0
δC2
δ β >0
δC2
δ r >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.
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Numerical Problem - Worked Solution
Suppose U = C10.6C20.4 , income is given as Y1 = 20 and Y2 = 30. The real interest rate, r is 5%.
4.1
Step 1 - Write out the Lagrangian
V = C10.6C20.4 + λ (1.05Y1 +Y2 − 1.05C1 −C2 )
4.2
(18)
Take the Differentials
Taking the differentials and setting equal to zero gives
δV
= 0.6C1−0.4C20.4 − 1.05λ = 0
δC1
(19)
δV
= 0.6C1−0.4C20.4 = 1.05λ
δC1
(20)
4
1
δV
=
0.6C1−0.4C20.4 = λ
δC1 1.05
(21)
δV
= 0.4C10.6C2−0.6 − λ = 0
δC2
(22)
δV
= 0.4C10.6C2−0.6 = λ
δC2
(23)
and
4.3
Set Lambdas Equal
Using equations (18) and (20) we get
1
0.6C1−0.4C20.4 = 0.4C10.6C2−0.6
1.05
(24)
which we can re-arrange to get
C2 = (1.05)
4.4
0.4
C1
0.6
(25)
Substitute into Budget Constraint
BC: 1.05C1 +C2 = 1.05Y1 +Y2
(26)
0.4
Substituting Y1 = 20, Y2 = 30, and C2 = (1.05) 0.6
C1 gives
1.05C1 + (1.05)
0.4
C1 = 1.05(20) + 30
0.6
(27)
5
1.05(C1 ) = 51
3
(28)
C1 = 29.143
(29)
C2 = (1.05)
0.4
C1 = 20.4
0.6
(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.
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5
Practice Exercises
Question 1
Question 1 will change only one variable at a time.
Question 1a
U = C10.8C20.2 , Y1 = 100, Y2 = 100, r = 5%
Question 1b
U = C10.8C20.2 , Y1 = 200, Y2 = 100, r = 5%
Question 1c
U = C10.8C20.2 , Y1 = 100, Y2 = 0, r = 5%
Question 1d
U = C1C20.2 , Y1 = 100, Y2 = 100, r = 5%
Question 1e
U = C10.8C20.8 , Y1 = 100, Y2 = 100, r = 5%
Question 1f
U = C10.8C20.2 , Y1 = 100, Y2 = 100, r = 10%
Question 1g
U = C10.8C20.2 , Y1 = 100, Y2 = 100, r = 1%
Question 2
You are a student. You live for 2 periods, your ‘student’ year and your ‘working’ year. You
don’t earn any money in your student year, so Y1 = 0. Your ‘working’ year consists of your
graduate salary Y2 = 20 (where Y is measured in thousands of pounds). Your utility function is
U = C1C20.95 and the real interest rate is 1%. How much do you borrow in your student period?
Question 3
U = (C1 − 50)0.8 (C2 − 50)0.2 , Y1 = 100, Y2 = 100, r = 5%
(notice how this is similar to question 1a, but the individual has decided that he needs C ≥ 50
in either period to survive).
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