ARROW-DEBREU SECURITIES

1

A simple 2 Period-2 State Example

Let us consider a simple economy that lasts 2 periods. The first period, denoted t = 1, is non-stochastic. The second period, denoted t = 2, is subject to an endowment shock which can either reach a low state, ω, with probability π, or high state, ω, with probability 1 − π. There exists a representative consumer that lives for 2 periods and has preferences over consumption that are represented by the following expected intertemporal utility function u(c1 ) + β [πu (c2 (ω)) + (1 − π)u (c2 (ω))]

(1)

where u(·) satisfies u0 (·) > 0 and u00 (·) < 0. Note the dependence of c2 (·) to the state of Nature. In particular, c2 (ω) (resp. c2 (ω)) denotes period-2 consumption in the low (resp. high) state of Nature. β ∈ (0, 1) is the psychological discount factor. In the first period, the household receives a constant non-stochastic endowment y1 . She has access to complete market and buys a portfolio of contingent claims to insure herself against the risk she faces in period t = 2. We denote by b(ω) (resp. b(ω)) the quantity of contingent claims that are bought at price q(ω) (resp. q(ω)) in period t = 1 and that will deliver 1 unit of good in period t = 2 in case state ω (resp. ω) occurs. The value of the portfolio she buys in t = 1 is therefore given by: q(ω)b(ω) + q(ω)b(ω). We also assume that she can buy a quantity B of a risk free bond that will yield a risk free return R in period t = 2. Finally she purchases consumption for the current period. Her period-1 budget constraint is therefore given by q(ω)b(ω) + q(ω)b(ω) + B + c1 = y1

(2)

In the second period, the household receives the stochastic endowment y2 (ω), and the proceeds of her financial investment I

The gains from the risky portfolio: b(ω) (resp. b(ω)) if state ω (resp. ω) realizes

Notes on Arrow-Debreu Securities

I

2

The capital income from the risk free bond: R B

These incomes are then used to purchase period-2 consumption. It should be clear that, from period-1 view point, the household faces 2 budget constraints, one for each state: c2 (ω) = b1 (ω) + RB + y2 (ω)

(3)

c2 (ω) = b1 (ω) + RB + y2 (ω)

(4)

The Lagrangian of the problem is given by L =u(c1 ) + β [πu (c2 (ω)) + (1 − π)u (c2 (ω))] + λ1 (y1 − q(ω)b(ω) − q(ω)b(ω) − B − c1 ) + λ2 (ω) (b1 (ω) + RB + y2 (ω) − c2 (ω)) + λ2 (ω) (b1 (ω) + RB + y2 (ω) − c2 (ω)) The set of first order conditions is given by u0 (c1 ) = λ1

(5)

βπu0 (c2 (ω)) = λ2 (ω)

(6)

β(1 − π)u0 (c2 (ω)) = λ2 (ω)

(7)

λ2 (ω) = λ1 q1 (ω)

(8)

λ2 (ω) = λ1 q1 (ω)

(9)

λ1 = R [λ2 (ω) + λ2 (ω)]

(10)

Using (5), (6) in (8), we obtain q1 (ω) = βπ

u0 (c2 (ω)) u0 (c1 )

(11)

Likewise, using (5), (7) in (9), we obtain q1 (ω) = β(1 − π)

u0 (c2 (ω)) u0 (c1 )

(12)

Finally, using (5)–(7) in (10), we get   u0 (c1 ) = βR πu0 (c2 (ω)) + (1 − π)u0 (c2 (ω)) = βRE(u0 (c2 ))

(13)

Hence, Equations (11) and (12) imply that the price of each contingent claim is given by the intertemporal rate of substitution between consumption in period 2 and 1 weighted by the probability of occurrence of the state. Note that Equation (13) implies  0  u (c2 (ω)) u0 (c2 (ω)) 1 =β π + (1 − π) = E[q] R u0 (c1 ) u0 (c1 )

Notes on Arrow-Debreu Securities

2

3

Infinite Horizon, N-states

We now consider the case of an infinite horizon model. To keep things tractable, we will assume that there exists N possible states of Nature. More precisely, in a given period t, state ωt ∈ Ω. From period 0 to period t the economy experienced History ω t = [ω0 , ω1 , . . . , ωt ]. This structure is illustrated in Figure 1, where only 2 states of Nature are considered (High, 1, and Low, 0). In this case, History ω t = (0, 1, 0, 1) corresponds to the green path on the tree. Figure 1: Nodes and History ω3 = 1

ω t = (0, 1, 1, 1)

ω2 = 1

ω t = (0, 1, 1, 0) ω3 = 0 ω1 = 1

ω3 = 1

ω t = (0, 1, 0, 1) ω2 = 0

ω t = (0, 1, 0, 0) ω3 = 0

ω0 = 0 ω3 = 1

ω t = (0, 0, 1, 1)

ω2 = 1

ω t = (0, 0, 1, 0) ω3 = 0

ω1 = 0 ω3 = 1

ω t = (0, 0, 0, 1) ω2 = 0

ω t = (0, 0, 0, 0) ω3 = 0

The expected intertemporal utility function is then given by "∞ # ∞ X X X t t U(c) = E0 β u(ct (ω )) = β t u(ct (ω t ))π(ω t |ω0 ) t=0

(14)

t=0 ω t

where π(ω t |ω0 ) denotes the probability that the economy reaches period t with History ω t conditional on starting in state ω0 . In order to ease exposition, we will make the Markov assumption, which implies that π(ω t |ω0 ) can be rewritten as π(ω t |ω0 ) = π(ωt |ωt−1 )π(ωt−1 |ωt−2 ) × . . . × π(ω1 |ω0 ) This will prove useful in the sequel. Just like in the 2-state case, the period-t budget constraint writes X ωt+1 ∈Ω

q(ωt+1 |ω t )b(ωt+1 |ω t ) + Bt (ω t ) + ct (ω t ) = b(ω t ) + Rt−1 (ω t−1 )Bt−1 (ω t−1 ) + y(ω t )

(15)

Notes on Arrow-Debreu Securities

4

Just like the 2 period-2 state case, we form the Lagrangian " ∞ X X t L = β u(ct (ω t ))π(ω t |ω0 ) t=0 ω t

!# X

+ λt (ω t ) b(ω t ) + Rt−1 (ω t−1 )Bt−1 (ω t−1 ) + y(ω t ) −

q(ωt+1 |ω t )b(ωt+1 |ω t ) − Bt (ω t ) − ct (ω t )

ωt+1 ∈Ω

which leads to the following set of first order conditions λt (ω t ) = u0 (ct (ω t ))π(ω t |ω0 )

(16)

q(ωt+1 |ω t )λt (ω t ) = βλt+1 (ω t+1 ) X λt (ω t ) = βRt (ω t ) λt+1 (ω t+1 )

(17) (18)

ω t+1

Combining Equations (16) and (17), we get q(ωt+1 |ω t ) = β

u0 (ct+1 (ω t+1 )) π(ω t+1 |ω0 ) u0 (ct (ω t )) π(ω t |ω0 )

Recall that we made the Markov assumption assumption such that π(ωt+1 |ωt ) π(ωt |ωt−1 ) π(ωt−1 |ωt−2 ) × . . . × π(ω1 |ω0 ) π(ω t+1 |ω0 ) = t π(ω |ω0 ) π(ωt |ωt−1 ) π(ωt−1 |ωt−2 ) × . . . × π(ω1 |ω0 ) which simplifies to π(ω t+1 |ω0 ) = π(ωt+1 |ωt ) π(ω t |ω0 ) Hence we have q(ωt+1 |ω t ) = β

(19)

u0 (ct+1 (ω t+1 )) π(ωt+1 |ωt ) u0 (ct (ω t ))

(20)

which is essentially the same as in (11) or (12). Likewise, plugging Equation (16) in (18), we get X  u0 (ct (ω t ))π(ω t |ω0 ) = βRt (ω t ) u0 (ct+1 (ω t+1 ))π(ω t+1 |ω0 ) ω t+1

which rewrites 0

t

t

u (ct (ω )) = βRt (ω )

X ω t+1

0

u (ct+1 (ω

t+1

π(ω t+1 |ω0 ) )) π(ω t |ω0 )



Using the Markov property and the result from Equation (19), this simplifies to X    u0 (ct (ω t )) = βRt (ω t ) u0 (ct+1 (ω t+1 ))π(ωt+1 |ωt ) = βRt (ω t )Et u0 (ct+1 (ω t+1 )) ω t+1

which is the equivalent to Equation (13). Rewriting the last equation, we get  X  u0 (ct+1 (ω t+1 )) 1 = β π(ω |ω ) t+1 t Rt (ω t ) u0 (ct (ω t )) t+1 ω

Using Equation (20), we get X    1 = qt+1 (ω t+1 ))π(ωt+1 |ωt ) = Et qt+1 (ω t+1 ) t Rt (ω ) t+1 ω

(21)