Jaime Frade

ECO5282-Dr. Garriga Homework #3

1. Consider and economy with a representative consumer with preferences described by E_{0} ∑_{t=0}^{∞} β^{t} u(c_{t}) where u(c_{t}) = ln(c_{t} + γ) where γ ≥ 0 and c_{t} denotes consumption of the fruit in period t. The sole source of the single good is an everlasting tree that produces d_{t} units of the consumption good in period t. The dividend process d_{t} is Markov, with prob{d_{t+1} ≤ d′|d_{t} = d} = F (d′, d). Assume the conditional density f(d′, d) of F exists. There are competitive markets in the title of trees and in state-contingent claims. Let p_{t} be the price at t of a title to all future dividends from the tree.

(a) Prove that the equilibrium prices p_{t} satisfies

p_{t} = (d_{t} + γ) ∑_{j=1}^{∞} β^{t} E_{t} (\frac{d_{i+j}}{d_{i+j} + γ} )  (1)

For a given household, the comsumer solves

max E_{0} ∑_{t=0}^{∞} β^{t} u(c_{t})  (2)

Subject to the following budget constraint

A_{t+1} = R_{t}(A_{t} + y_{t} − c_{t})  (3)

From (3),

c_{t} ⟹ consumption of an agent at time t
y_{t} ⟹ agent’s labor income
A_{t} ⟹ amount of a single asset valued in the units of the

consumption good held at start of time t (Note: A_{0} is given)

R_{t} ⟹ real gross rate of return on asset between time t and t + 1

By using the Legrange multipliers, combining the first order conditions in (2) we obtain the Euler equation that is necessary condition of optimality for t = 0, 1, . . ..

u′(c_{t}) = E_{t}β R_{t}u′(c_{t+1})  (4)

However (4) does not applicable to models with complete general equilibrium. In the Lucas’s asset-pricing model, general equilibrium is used. This model assumes that the labor income is zero, y_{t} = 0, ∀ t. It also assumes that the only durable good in the economy is a set of trees, (which equal the number of people in economy). The tree also yields fruit (dividends) in the amount of d_{t} to its owner ar the beginning of time t. This fruit is also nonstorable. Each consumer starts a t = 0 with, one tree and initial dividend of good, (fruit).

Let p_{t} ⟹ price of a tree at time t
R_{t} = \frac{p_{t+1} + d_{t+1}}{p_{t}}  (5)

Using (5) subsitute into (4), will obtain the following:

1 = E_{t} β \frac{u′(c_{t+1})}{u′(c_{t})} (\frac{p_{t+1} + d_{t+1}}{p_{t}} )  (6)

In this economy, equilibrium must exist because all consumers are identical, the utility function is strictly increasing, and there is only one place for goods, period’s dividends. Therefore obtain:

c_{t} = d_{t}  (7)

Substituting (7) into (6) and solve for p_{t}, will obtain the following:

p_{t} = E_{t} β \frac{u′(d_{t+1})}{u′(d_{t})} (p_{t+1} + d_{t+1})  (8)

Given u(c_{t}) = ln(c_{t} + γ) where γ ≥ 0, and using (7) and differentiating to plug into (8) will obtain:

p_{t} = E_{t} β \frac{(d_{t} + γ)}{(d_{t+1} + γ)} (\underbrace{p_{t+1}}_{⇑} +d_{t+1})  (9)

Using recursions on (9) for the value p_{t+1}, will obtain:

p_{t} = E_{t} [β \frac{(d_{t} + γ)}{(d_{t+1} + γ)} (d_{t+1} + \underbrace{β \frac{(d_{t+1} + γ)}{(d_{t+2} + γ)} (p_{t+2} + d_{t+2})}_{=p_{t+1}})]

p_{t} = E_{t} [β \frac{(d_{t} + γ)}{(d_{t+1} + γ)} (d_{t+1} + β \frac{(d_{t+1} + γ)}{(d_{t+2} + γ)} (d_{t+2} + β \frac{(d_{t+2} + γ)}{(d_{t+3} + γ)} (d_{t+3} + p_{t+3})))]

p_{t} = E_{t} [β \frac{(d_{t} + γ)}{(d_{t+1} + γ)}d_{t+1} + β^{2} \frac{(d_{t} + γ)}{(d_{t+2} + γ)}d_{t+2} + β^{3} \frac{(d_{t} + γ)}{(d_{t+3} + γ)}d_{t+3} + ⋯]

From above equation, can see a pattern forming, obtain

p_{t} = (d_{t} + γ)E_{t} ∑_{j=0}^{∞} [β^{j} (\frac{d_{t+j}}{d_{t+j} + γ} )] + lim_{j→∞} E_{t} [ β^{j} (\frac{d_{t} + γ}{d_{t+j} + γ} ) p_{t+j}]

As j → ∞ lim → 0, will prove result for pricing equation.

p_{t} = (d_{t} + γ)E_{t} ∑_{j=0}^{∞} β^{j} (\frac{d_{t+j}}{d_{t+j} + γ} )  (10)

(b) Modify the Lucas Model to a market that consist of one- and two period perfectly safe loans, which bear gross rates of return R_{1t} and R_{2t}. R_{1t}^{−1} is the price of a perfectly sure claim to one unit of consumption at time (t + 1). R_{2t}^{−1} is the price of a perfectly sure claim to one unit of consumption at time (t + 2). The agent needs to solve the following maximum

max_{c_{t},L_{1t+1},L_{2t+1}} E_{0} ∑_{t=0}^{∞} β^{t} u(c_{t})  (11)

Subject to the following budget constraint, where L_{jt} is the amount lent for j period at time t.

c_{t} + L_{1t} + L_{2t} ≤ d_{t} + R_{1t−1}L_{1t−1} + R_{2t−2}L_{2t−2}  (12)

Solving (11), using Legrange Multiplier, where {λ_{t}} is a sequence of random Legrange Multipliers. Will obtain the following:

L = E_{0} ∑_{t=0}^{∞} β^{t} [u(c_{t}) + λ_{t}(d_{t} + L_{1t−1}R_{1t−1} + L_{2t−2}R_{2t−2} − c_{t} − L_{1t} − L_{2t})]  (13)

\underline{First Order Conditions (11)} with respect to \{c_{t}, L_{1t}, L_{2t}\}. Also E_{0} = E_{t} since the Markov property.

\begin{aligned}[c_{t}] : & E_{0}β^{t}(u′(c_{t}) − λ_{t}) = 0 & ⟹ λ_{t} = u′(c_{t}) \\ [L_{1t}] : & E_{0}β^{t}(βλ_{t+1}R_{1t} − λ_{t}) = 0 & ⟹ λ_{t} = E_{t}(βλ_{t+1}R_{1t}) \\ [L_{2t}] : & E_{0}β^{t}(β^{2}λ_{t+2}R_{2t} − λ_{t}) = 0 & ⟹ λ_{t} = E_{t}(β^{2}λ_{t+2}R_{2t}) \end{aligned}

Combining the first order conditions above will obtain the following Euler Equations:

E_{t} [β \frac{u′(c_{t+1})}{u′(c_{t})}R_{1t}] = 1  (14)
E_{t} [β^{2} \frac{u′(c_{t+2})}{u′(c_{t})}R_{2t}] = 1  (15)

Given that both assets are riskless, rewrite (14) and (15) as:

R_{1t}^{−1} = E_{t} [β \frac{u′(c_{t+1})}{u′(c_{t})}]  (16)
R_{2t}^{−1} = E_{t} [β^{2} \frac{u′(c_{t+2})}{u′(c_{t})}]  (17)

(i) Find a formula for the risk-free one-period interest rate R_{1t}.

From (16), using given u(c_{t}) = ln(c_{t} + γ) and (7) in equilibrium, will obtain a formula for th risk-free one-period interest rate R_{1t}.

R_{1t}^{−1} = E_{t} [β \frac{(d_{t} + γ)}{d_{t+1} + γ} ]  (18)

(ii) Prove that in the special case in which {d_{t}} is independently and identically distributed, R_{1t} is given by R_{1t}^{−1} = β k (d_{t} + γ), where k is a constant.

If assume that {d_{t}} is i.i.d, then (18) can be rewritten:

R_{1t}^{−1} = (d_{t} + γ)β E_{t} [\frac{1}{d_{t+1} + γ} ]
R_{1t}^{−1} = (d_{t} + γ)β E_{t} [\frac{1}{d_{t+1} + γ} ] ⟹ Let k = E_{t} \frac{1}{d_{t+1} + γ}
R_{1t}^{−1} = (d_{t} + γ)β k  (19)

(iii) Give a formula for k. From (19), and assumption that {d_{t}} is i.i.d, obtain a formula for k, which is not a function of time:

k = E_{t} [\frac{1}{d_{t+1} + γ} ]  (20)

(c) (i) Find a formula for the risk-free two-period interest rate R_{2t}. From (17), using given u(c_{t}) = ln(c_{t} + γ) and (7) in equilibrium, will obtain a formula for th risk-free one-period interest rate R_{2t}.

R_{2t}^{−1} = E_{t} [β^{2} \frac{(d_{t} + γ)}{d_{t+2} + γ} ]  (21)

(ii) Prove that in the special case in which {d_{t}} is independently and identically distributed, R_{2t} is given by R_{2t}^{−1} = β^{2} k (d_{t} + γ), where k is the same constant you found in part (b). If assume that {d_{t}} is i.i.d, then (18) can be rewritten:

R_{2t}^{−1} = (d_{t} + γ)β^{2} E_{t} [\frac{1}{d_{t+2} + γ} ]
R_{2t}^{−1} = (d_{t} + γ)β^{2} E_{t} [\frac{1}{d_{t+2} + γ} ] ⟹ Let j = E_{t} \frac{1}{d_{t+2} + γ}
R_{2t}^{−1} = (d_{t} + γ)β^{2} j  (22)

• Prove that j from (22) is same k from (20). From {d_{t}} is i.i.d and given that the dividend process d_{t} is Markov, with prob{d_{t+1} ≤ d′|d_{t} = d} = F (d′, d).

j = E_{t} [\frac{1}{d_{t+2} + γ} ]
= E_{t+1} [\frac{1}{d_{t+2} + γ} ]
= k

2. Consider the following version of the Lucas’s tree economy. There are two kinds of trees. The first kind is ugly and gives no direct utility to consumers, but yields a stream of fruit {d_{1t}}, where d_{1t} denotes a positive random process obeying a first-order Markov process. The second tree is beautiful and yields utilty on itself. This tree also yields a stream of the same kind of fruit d_{2t}, where it happens that d_{2t} = d_{1t} = (\frac{1}{2}) d_{t} ∀ t, so that the physical yields of the two kinds of trees are equal. There is one of each tree for each N individuals in the economy. Trees last forever, but the fruit is not storable. Trees are the only source of fruit.

Each of the N individuals in the economy has preferences described by

E_{0} ∑_{t=0}^{∞} β^{t} u(c_{t}, s_{2t})  (23)

where u(c_{t}, s_{2t}) = ln c_{t} + γ ln(s_{2t}) where γ ≥ 0, c_{t} denotes consumption of the fruit in period t and s2t is the stock of beautiful trees owned at the beginning of the period t. The owner of a tree of either kind i at the start of the period receives the fruit d_{it} produced by the tree during that period.

Let p_{it} be the price of a tree of type i (where i = 1, 2) during period t. Let R_{it} be the gross rate of returns of tree i during that period held from period t to t + 1.

(a) Write down the consumer optimization problem in sequential and recursive form.

Consumer optimization problem in \underline{sequential form}

max_{{c_{t},s_{1t+1},s_{2t+1}}} E_{0} ∑_{t=0}^{∞} β^{t} (ln(c_{t}) + γ ln(s_{2t}))  (24)

subject to c_{t} + p_{1t}s_{1t+1} + p_{2t}s_{2t+1} ≤ (d_{1t} + p_{1t})s_{1t} + (d_{2t} + p_{2t})s_{2t}.  (25)

Consumer optimization problem in \underline{recursive form}

The Bellman’s equation is given by

v(d_{t}, s_{1t}, s_{2t}) = max_{{c_{t},s_{1t+1},s_{2t+1}}} (ln(c_{t}) + γ ln(s_{2t}) + E_{t} β v(d_{t}, s_{1t+1}, s_{2t+1}))  (26)

subject to c_{t} + p_{1t}s_{1t+1} + p_{2t}s_{2t+1} ≤ (d_{1t} + p_{1t})s_{1t} + (d_{2t} + p_{2t})s_{2t}.  (27)

(b) Define a rational expectations equilibrium.

Defintion: A \underline{rational expectations equilibrium} is an allocation \{\{c_{t}^{i}\}_{t=0}^{∞}\}_{i=1}^{I}, \{\{s_{1t}^{i}, s_{2t}^{i}\}_{t=0}^{∞}\}_{i=1}^{I}, and a price system, {p_{1t}, p_{2t}}_{t=0}^{∞}, such it solves each household problem and satisfies the market clear condition.

Defintion: For the \underline{market clearing condition} to exist,

∑_{i=1}^{I} c_{t}^{i} = ∑_{i=1}^{I} d_{t}^{i} , ∑_{i=1}^{I} s_{1t}^{i} = ∑_{i=1}^{I} s_{10}^{i} = I, ∑_{i=1}^{I} s_{2t}^{i} = ∑_{i=1}^{I} s_{20}^{i} = I,

where s_{10}^{i} and s_{20}^{i} are each agent’s number of trees at initial time.

The \underline{household problem} is defined sequentially by each agent’s utility optimization problem:

max_{{c_{t},s_{1t+1},s_{2t+1}}} E_{0} ∑_{t=0}^{∞} β^{t} (ln(c_{t}) + γ ln(s_{2t}))  (28)

subject to c_{t} + p_{1t}s_{1t+1} + p_{2t}s_{2t+1} ≤ (d_{1t} + p_{1t})s_{1t} + (d_{2t} + p_{2t})s_{2t}  (29)

(c) Find the pricing functions mapping the state of the ecomony at t unto p_{1t} and p_{2t} (give precise formulas). [Hint: You should be able to directly derive p_{1t} from the example seen in class, then since pricing function have to be linear you can guess a pricing function p_{2t} = kd_{t} and solve for k parameter using Euler equation of the second stock.]

From (28), using Legrange multipilers with the given budget constraints, (29), we obtain the following:

L = E_{0} ∑_{t=0}^{∞} β^{t} ((ln(c_{t}) + γ ln(s_{2t}))
+ λ_{t}((d_{1t} + p_{1t})s_{1t} + (d_{2t} + p_{2t})s_{2t} − c_{t} − p_{1t}s_{1t+1} − p_{2t}s_{2t+1}))  (30)

\underline{First Order Conditions (30)} with respect to \{c_{t}, s_{1t+1}, s_{2t+1}\}. Also using E_{0} = E_{t} since the Markov property.

[c_{t}] :  E_{0} β^{t} (\frac{1}{c_{t}} − λ_{t}) = 0  ⟹ \frac{1}{c_{t}} = λ_{t}
[s_{1t+1}] :  E_{0} β^{t} (βλ_{t+1}(d_{1t+1} + p_{1t+1}) − λ_{t}p_{1t}) = 0  ⟹ p_{1t} = E_{t}β \begin{matrix}λ & t+1 \\ λ & t \end{matrix} (d_{1t+1} + p_{1t+1})
[s_{2t+1}] :  E_{0} β^{t} (βλ_{t+1}(d_{2t+1} + p_{2t+1}) − λ_{t}p_{2t} + β \frac{γ}{s_{2t+1}} ) = 0

⟹ p_{2t} = E_{t}β \begin{matrix}λ & t+1 \\ λ & t \end{matrix} (d_{2t+1} + p_{2t+1}) + β \frac{γ}{s_{2t+1}λ_{t}}

Combining the first order conditions above will obtain the following pricing formula:

p_{1t} = E_{t} β \frac{c_{t}}{c_{t+1}} (d_{1t+1} + p_{1t+1})  (31)
p_{2t} = E_{t} β \frac{c_{t}}{c_{t+1}} (d_{2t+1} + p_{2t+1}) + β \frac{γc_{t}}{s_{2t+1}}  (32)

In this economy, for general equilibrium to exist:

c_{t} = d_{t}  (33)

From (33), substitute into the pricing formulas in (31) and (32), obtain:

p_{1t} = E_{t} β \frac{d_{t}}{d_{t+1}} (d_{1t+1} + p_{1t+1})  (34)
p_{2t} = E_{t} β \frac{d_{t}}{d_{t+1}} (d_{2t+1} + p_{2t+1}) + β \frac{γd_{t}}{s_{2t+1}}  (35)

For p_{1t}, if assuming a the pricing function have to be of linear form obtain:

p_{1t} = k_{1t}d_{t}  (36)

From (36), substitute this into (??) to obtain:

k_{1t}d_{t} = E_{t} β \frac{d_{t}}{d_{t+1}} (d_{1t+1} + k_{1t+1}d_{t+1})
k_{1t} = E_{t} β \underbrace{k_{1t+1}}_{⇑} +E_{t} β \frac{d_{1t+1}}{d_{t+1}}  (37)

⟹ apply recursions onk_{1t+1}

k_{1t} = E_{t} ∑_{j=1}^{∞} β^{j} \frac{d_{1t+j}}{d_{t+j}}  (38)

From d_{2t} = d_{1t} = (\frac{1}{2}) d_{t} ∀ t, rewrite (38) to obtain:

k_{1t} = \frac{1}{2} E_{t} ∑_{j=1}^{∞} β^{j}

= \frac{β}{2(1 − β)}  (39)

For p_{2t}, if assuming a the pricing function have to be of linear form obtain:

p_{2t} = k_{2t}d_{t}  (40)

From (40), substitute this into (??) to obtain:

k_{2t}d_{t} = E_{t} β \frac{d_{t}}{d_{t+1}} (d_{2t+1} + k_{2t+1}d_{t+1}) + β \frac{γd_{t}}{s_{2t+1}}
k_{2t} = E_{t} β \underbrace{k_{2t+1}} +E_{t} β (\frac{d_{2t+1}}{d_{t+1}} + \frac{γ}{s_{2t+1}} )  (41)

⟹ apply recursions onk_{2t+1}

k_{2t} = E_{t} ∑_{j=1}^{∞} β^{j} (\frac{d_{2t+j}}{d_{t+j}} + \frac{γ}{s_{2t+1}} )  (42)

From the problem, d_{2t} = \frac{1}{2}d_{t} and s_{2t+1} = 1 in equilibrium, rewrite (42) to obtain:

k_{2t} = E_{t} ∑_{j=1}^{∞} β^{j} (\frac{1}{2} + γ)

= (\frac{1}{2} + γ) \frac{β}{1 − β}  (43)

Subsititing (39) and (43) into (34) and (34), will obtain the pricing functions:

p_{1t} = (\frac{β}{2(1 − β)}) d_{t}  (44)
p_{2t} = ((\frac{1}{2} + γ) \frac{β}{1 − β} ) d_{t}  (45)

(d) Prove that if γ > 0, then R_{1t} > R_{2t} ∀ t

From (5), derive the following definitions for R_{1t} and R_{2t}.

R_{1t} = \frac{p_{1t+1} + d_{1t+1}}{p_{1t}}  (46)
R_{2t} = \frac{p_{2t+1} + d_{2t+1}}{p_{2t}}  (47)

Substiting (44) and (45) the pricing functions, into (46) and (46) can rewrite equation for R_{1t} and R_{2t} as:

R_{1t} = \frac{\frac{1}{2} (\frac{β}{1 − β} ) d_{t+1} + \frac{1}{2} d_{t+1}}{\frac{1}{2} (\frac{β}{1 − β} ) d_{t}}  (48)

R_{2t} = \frac{(\frac{1}{2} + γ) (\frac{β}{1 − β} ) d_{t+1} + \frac{1}{2} d_{t+1}}{(\frac{1}{2} + γ) (\frac{β}{1 − β} ) d_{t}}  (49)

Analyzing difference in the returns, R_{1t} and R_{2t}, is:

R_{1t} − R_{2t} = [ \frac{1}{\frac{1}{2} \frac{β}{1−β}} − \frac{1}{(γ + \frac{1}{2}) \frac{β}{1−β}} ] \frac{d_{t+1}}{d_{t}}

= \frac{1 − β}{β} (1 − \frac{1}{1 + 2γ} ) \frac{d_{t+1}}{d_{t}}  (50)

From (50), if let γ > 0, ⟹ R_{1t} − R_{2t} > 0. Able to obtain:

R_{1t} > R_{2t}  (51)