Jaime Frade

ECO5282-Dr. Garriga Homework #2

1. Consider an economy with a single consumer. There is one good in the economy, which arrives in the form of an exogenous endowement obeying, y_{t+1} = λ_{t+1} y_{t}, where y_{t} is the endowment at time t and {λ_{t+1}} is

governed by a two-state Markov chain with transition matrix, P = [ \begin{matrix}p_{11} & 1 − p_{11} \\ 1 − p_{22} & p_{22} \end{matrix} ] , and intial distribution

π_{λ} = [π_{0}, 1 − π_{0}]. The value of λ_{t} is given by \overline{λ_{1}} = 0.98 in state 1 and \overline{λ_{2}} = 1.03 in state 2. Assume that the history of y_{s} and λ_{s}, up to t is observed at time t. The consumer has endowment process {y_{t}} and has

preferences over consumption streams that are ordered by E_{0} ∑_{t=0}^{∞} β^{t} \frac{c^{1−γ}}{1 − γ}, here β ∈ (0, 1) and γ ≥ 1.

(a) Define a competitive equilibrium.

Definition A allocation is said to be a \underline{feasible allocation} if it satisfies

∑_{i=1}^{2} c_{t}^{i}(s^{t}) = ∑_{i=1}^{2} y_{t}^{i}(s_{t})  (1)
Since one agent → c_{t}(s^{t}) = y_{t}(s_{t})  (2)

Definition A \underline{competitive equilibrium} is a feasible allocation, \{c^{i}\}_{i=1}^{2} = \{\{c_{t}^{i}(s^{t})\}_{t=0}^{∞}\}_{i=1}^{2}, and a price system, {p_{t}^{0}(s^{t})}_{t=0}^{∞}, such that the allocation solves each household problem.

For a given household i solves

U(c^{i}) = max_{{c_{t}^{T} (s^{t})}} E_{0} ∑_{t=0}^{∞} β^{t} u(c_{t}^{i})

U(c^{i}) = max_{{c_{t}^{T} (s^{t})}} ∑_{t=0}^{∞} ∑_{s^{t}} β^{t} π(s^{t}|s_{0}) u(c_{t}^{i}(s^{t}))  (3)

subject to ∑_{t=0}^{∞} ∑_{s^{t}} p_{t}^{0}(s^{t})c_{t}^{i}(s^{t}) = ∑_{t=0}^{∞} ∑_{s^{t}} p_{t}^{0}(s^{t})y_{t}^{i}(s_{t}), and c_{t}^{1}(s^{t}), c_{t}^{2}(s^{t}) ≥ 0  (4)

(b) From (2), consumption = endowment, since only one agent obtain the following:

c_{t} = y_{t}  (5)

In (5), y is defined as y_{t+1} = λ_{t+1} y_{t}, where λ_{t+1} is defined as a two state Markov Chain. Using the properites of Markov process,

y_{t+1} = ∏_{i=0}^{t} λ_{i} y_{0}  (6)

Substituting (5) into the first order conditions of the household problem, into (??), will obtain an equation that will solve the price system for any asset that will lead to competitive equilibrium.

p_{t}^{0}(s^{t}) = β^{t} π(s^{t}|s_{0}) \frac{u′(c_{t}^{i}(s^{t}))}{u′(c_{0}^{i}(s^{0}))}  from (??)  (7)

p_{t}^{0}(s^{t}) = β^{t} π(s^{t}|s_{0}) \frac{u′(y_{t})}{u′(y_{0})}  using (5)  (8)

Suppose p_{11} = .8, p_{22} = .85, π_{0} = .5, β = .96, and γ = 2. Suppose economy starts off with λ_{0} = .98 and y_{0} = 1.

Also, P = [ \begin{matrix}p_{11} & 1 − p_{11} \\ 1 − p_{22} & p_{22} \end{matrix} ] = [ \begin{matrix}.8 & .2 \\ .15 & .85 \end{matrix} ] and intial distribution π_{λ} = [π_{0}, 1 − π_{0}] = [0.5, .5]. The

value of λ_{t} is given by \overline{λ_{1}} = 0.98 in state 1 and \overline{λ_{2}} = 1.03 in state 2.

(c) I will use the given conditions in the setup of the problem, stated above, and using the properties of a Markov chain, (that is, as time → ∞, the probability of each state occuring will converge to a stationary distrubition) to solve for the average growth rate of consumption. Goal: Find a stationary distribution such that,

∑ a_{i} = 1 and a = aP, where a = [α, 1 − α]  (9)

Using P, setup a system of equations to solve for α.

a = aP ⟶ [α, 1 − α] = [α, 1 − α] [ \begin{matrix}p_{11} & 1 − p_{11} \\ 1 − p_{22} & p_{22} \end{matrix} ]

From top equation, (Eq1)  From bottom equation, (Eq2)
α = α p_{11} + (1 − α)(1 − p_{22})  and (1 − α) = α (1 − p_{11}) + (1 − α)p_{22}
α = α p_{11} + 1 − α − p_{22} + α p_{22}  (1 − α) + (1 − α)p_{22} = α (1 − p_{11})
α + α − α p_{11} − α p_{22} = 1 − p_{22}  (1 − α)(1 − p_{22}) = α (1 − p_{11})
α(2 − p_{11} − p_{22}) = 1 − p_{22}  \frac{(1 − α)}{α} = \frac{(1 − p_{11})}{(1 − p_{22})}
α = \frac{1 − p_{22}}{(2 − p_{11} − p_{22})}  \frac{1}{α} = \frac{(1 − p_{11})}{(1 − p_{22})} + 1 = \frac{(1 − p_{11} + 1 − p_{22})}{(1 − p_{22})}

Solving in both equations for α obtain equalvalent realtions, which is suppose to happen since it is a stationary distribution.

From Eq1: α = \frac{1 − p_{22}}{2 − p_{11} − p_{22}}  From Eq2: α = \frac{1 − p_{22}}{2 − p_{11} − p_{22}}  (10)

Therefore subsituting the values for p_{11} = .8 and p_{22} = .85 in (10) will obtain:

α = \frac{1 − p_{22}}{2 − p_{11} − p_{22}} = \frac{1 − .85}{2 − .80 − .85} = \frac{1 − .85}{2 − .80 − .85} = \frac{3}{7}

a = [α, 1 − α] = [\frac{3}{7} , \frac{4}{7}]  (11)

By properties of Markov process, as time → ∞, the probability of the each state occuring will converge to the corresponding probabilities in the stationary distribution. Hence:

P (λ_{t} = \overline{λ_{i}}) = a where i=1,2  (12)

Therefore will obtain

P (λ_{t} = \overline{λ_{1}} ) = \frac{3}{7} and P (λ_{t} = \overline{λ_{2}} ) = \frac{4}{7}  (13)

Using (13), able to compute the average growth rate of consumption, before having observed λ_{0}.

E[λ] = P (λ_{t} = \overline{λ_{1}} )\overline{λ_{1}} + P (λ_{t} = \overline{λ_{2}} )\overline{λ_{2}}
= (\frac{3}{7}) (.98) + (\frac{4}{7}) (1.03)
= 1.0086  (14)

In order to compute the time-0 prices for the for each of the three bonds promising to pay one unit of time-j = 0, 1, 2. Assume that given intial state, λ_{0} = .98 = \overline{λ_{1}}. Thus, has corresponding values, p_{11} = .8, and 1 − p_{11} = .2 on first branch of tree.

[figure: j = 0; j = 1; j = 2; p_{11} = .8; 1 − p_{11} = .2; .8; .2; p_{22} = .85; 1 − p_{22} = .15; state 1; λ_{0} = .98; y_{0} = 1; state 1; \overline{λ_{1}} = .98; price = 0.7997; state 2; \overline{λ_{1}} = 1.03; price = .1810; state 1; \overline{λ_{2}} = .98; price = 0.6394; state 2; \overline{λ_{2}} = 1.03; price = 0.1447; state 1; \overline{λ_{1}} = .98; price = 0.027; state 2; \overline{λ_{2}} = 1.03; price = 0.1392]

From (8), using the following given information, u(c) = \frac{c^{1−γ}}{1 − γ}, where γ = 2, β = .96, and y_{0} = 1.

Also known from (6), y_{t} = ∏_{i=1}^{t} λ_{i} y_{0} → y_{t} = ∏_{i=1}^{t} λ_{i}

From the given conditions, will obtain

p _{t}^{0}(s^{t}) = β^{t} π(s^{t}|s_{0}) \frac{u′(y_{t})}{u′(y_{0})}
p _{t}^{0}(s^{t}) = (.96)^{t} π(s^{t}|s_{0}) (\frac{y_{0}}{y_{t}} )^{2}  (15)

I will use (15), to compute the price at each time t = 0, 1, 2 t=1

• p _{1}^{0}(\overline{s^{1}}|s_{0}) = (.96) (.8)(.2) (\frac{1}{.98})^{2} = .7997

• p _{1}^{0}(\underline{s^{1}}|s_{0}) = (.96) (.2) (\frac{1}{1.03})^{2} = .1810

t=2

• p _{2}^{0}(\overline{s^{2}}|s_{0}) = (.96)^{2} (.8)(.8) (\frac{1}{(.98)(.98)})^{2} = .6394

• p _{2}^{0}(\overline{\underline{s^{2}}}|s_{0}) = (.96)^{2} (.2)(.15) (\frac{1}{(1.03)(.98)})^{2} = .027

• p _{2}^{0}(\underline{s^{2}}|s_{0}) = (.96)^{2} (.8)(.2) (\frac{1}{(1.03)(.98)})^{2} = .1447

• p _{2}^{0}(\underline{\underline{s^{2}}}|s_{0}) = (.96)^{2} (.85)(.15) (\frac{1}{(1.03)(1.03)})^{2} = .1392

From Cochrane, (pg.52 1ed.), using to what he refers to as the happy-meal theorem (3.1):

p(x) = ∑_{s} pc(s)x(s)  (16)

where x(s) denotes an asset’s payoff in state of nature s, pc is the price of a contingent claim and (s) is used to denote which state s the claim pays off. In problems (d)-(f), the bond promises to pay one unit, x(s) = 1. From (16), able to compute the price of the bond promising to pay one unit of time-j comsumption for j = 0, 1, 2.

(d) Compute the time-0 prices of three risk-free discount bonds, in particular, those promising to pay one unit of time-j consumption for j = 0, 1, 2, respectively. t= 0

• p(x) = 1 t= 1

• p(x) = .7997 + .1810 = .9807 t= 2

• p(x) = .6394 + .027 + .1447 + .1392 = .9508

(e) Compute the time-0 prices of three bonds, in particular, those promising to pay one unit of time-j contingent consumption on λ_{j} = \overline{λ_{1}} for j = 0, 1, 2, respectively. t= 0

• p(x) = 1 from λ_{0} = \overline{λ_{1}} t= 1

• p(x) = .7997 t= 2

• p(x) = .6394 + .027 = .666 (f) Compute the time-0 prices of three bonds, in particular, those promising to pay one unit of time-j contingent

consumption on λ_{j} = \overline{λ_{2}} for j = 0, 1, 2, respectively. t= 0

• p(x) = 0 from λ_{0} = \overline{λ_{1}} t= 1

• p(x) = .1810 t= 2

• p(x) = .1447 + .1392 = .2839

(g) Compare the prices that you computed in parts (d)-(f).

From above, can see that at each time t, the sum of each price in answers (e) and (f), is equal to the price of the risk free discount bond, (d).

At each time, t = 0, 1, 2 (e)+(f)=(d).