Solutions to Assignment 1

Solutions to Assignment 1

Question 1

(a) We have to check if the condition C(t) £ Y(t) holds for periods t = 1, 2, 3, ...

Period 1: C(1) = c01(1)+c02(1)+c11(1)+c12(1) = 0.5+0.5+0.25+0.75 = 2 £ Y(1) = 2

Period 2: C(2) = c11(2)+c12(2)+c21(2)+c22(2) = 0.75+0.25+0.5+0.5 = 2 £ Y(2) = 2

Period 3: C(3) = c21(3)+c22(3)+c31(3)+c32(3) = 0.5+0.5+0.5+0.5 = 2 £ Y(3) = 2

Period 4, ... : Identical to period 3. Therefore, the allocation is feasible.



(b) We have to check if the condition C(t) = Y(t) holds for periods t = 1, 2, 3, ...

From the calculations in part (a) we see that the condition is satisfied for all periods. So the allocation is efficient.



(c) We have to find another (feasible) allocation where at least one person is better off and nobody is worse off. For members of generations t ³ 1, we have to check their utility levels. For members of generation 0, we have to check their period 1 consumption. Let's call the allocation given in the question allocation A. Here is one allocation, B, that is Pareto superior to A.

Generation 0: c01(1) = 0.5, c02(1) = 0.5.

Generations t=1,2,3,...:       ct1(t) = 0.5, ct2(t) = 0.5, ct1(t+1) = 0.5, c< i>t2(t+1) = 0.5.

Check members of generation 0

In B, they consume the same as in A so they are no better off and no worse off.

Check members of generation 1

In allocaton A, they have utility u1h = 0.250.5+0.750.5 = 1.366 for h = 1,2.

In allocaton B, they have utility u1h = 0.50.5+0.50.5 = 1.414 for h = 1,2.

So these two people are better off in allocation B.

Check members of generations 2, 3, ...

They have the same consumption under both allocations, so they have the same utility levels in both allocations. They are no better off and no worse off in B compared to A.

So in allocation B, members 1 and 2 of generation 1 are better off and nobody is worse off. Therefore, B is Pareto superior to A.

Question 2

(a)

young: cth(t)+lh(t) £ wth(t)

old: cth(t+1) £ wht(t+1)+r(t) lh(t)

lifetime: cht(t+1) £ wht(t+1)+ r(t) [wht(t)-cht(t)]



(b)

uht = ln(cht(t))+ 0.95 ln[wht(t+1)+ r(t) (wht(t)-cht(t))]

uht
cht(t)
 = 1
cht(t)
 + -0.95 r(t)
wht(t+1)+r(t)(wht(t)-cht(t))
 = 0
Solving for cht(t) yields the consumption function when young
cht(t) = wht(t)
1.95
 + wht(t+1)
1.95 r(t)
 º cht(r(t))
The savings function is then
sht(t) = wht(t)-cht(r(t)) = 0.95 wht(t)
1.95
 - wht(t+1)
1.95 r(t)



(c)

Time t good market clearing

N(t)
å
 h = 1 
 cht(t)+ N(t-1)
å
 h = 1 
 cht-1(t) = N(t)
å
 h = 1 
 wht(t)+ N(t-1)
å
 h = 1 
 wht-1(t)
Summing the time t old budget constraint yields
N(t-1)
å
 h = 1 
 cht-1(t) = N(t-1)
å
 h = 1 
 wht-1(t)+r(t-1) N(t-1)
å
 h = 1 
 lh(t-1)
Substituting in time t good market clearing we have
N(t)
å
 h = 1 
 cht(t)+ N(t-1)
å
 h = 1 
 wht-1(t)+r(t-1) N(t-1)
å
 h = 1 
 lh(t-1) = N(t)
å
 h = 1 
 wht(t)+ N(t-1)
å
 h = 1 
 wht-1(t)
Imposing market clearing on the time t-1 private borrowing and lending market,
åh = 1N(t-1) lh(t-1) = 0, we have
N(t)
å
 h = 1 
 cht(t) = N(t)
å
 h = 1 
 wht(t)
Using the utility maximization condition cht(t) = cht(r(t) we have
N(t)
å
 h = 1 
 (wht(t)-cht(r(t))) = St(r(t)) = 0



(d)

St(r(t))
= 20 æ
ç
è 		0.95
1.95
 - 1
1.95 r(t)
 ö
÷
ø 		+30 æ
ç
è 		0.95 (2)
1.95
 - 1
1.95 r(t)
 ö
÷
ø 		+50 æ
ç
è 		0.95 (3)
1.95
 - 1
1.95 r(t)
 ö
÷
ø
=
112.052- 51.282
r(t)



(e)

Imposing the competitive equilibrium condition, St(r(t)) = 0, we get the equilibrium interest rate r(t) = 0.458.

Time t consumption allocation

cht(t) = 1
1.95
 + 1
1.95 (0.458)
 = 1.633,       h = 1, ..., 20
cht(t) = 2
1.95
 + 1
1.95 (0.458)
 = 2.145,       h = 21, ..., 50
cht(t) = 3
1.95
 + 1
1.95 (0.458)
 = 2.658,       h = 51, ..., 100
Since all periods are identical, r(t) = r(t-1) = 0.458 and
cht-1(t) = cht(t+1) = 1+0.458 (1-1.633) = 0.710 ,       h = 1, ..., 20
cht-1(t) = cht(t+1) = 1+0.458 (2-2.145) = 0.934 ,       h = 21, ..., 50
cht-1(t) = cht(t+1) = 1+0.458 (3-2.648) = 1.161 ,       h = 51, ..., 100


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