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)