Download Discounting, values, decisions

Discounting, values, decisions
a short review
by John Quah and Bruno Strulovici
November 11, 2012
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Introduction
Develop a methodology to compare utility processes with
respect to discounting functions.
[+] Results: more patience ⇒
i. stop later
ii. higher payoff
iii. larger continuation domain
[−] Results: contrary to the the [+] results.
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We continue our investigation of comparative
statics
• There are objective functions f, g : X → R, where
X ∈ R.
• We want the optimizers to be well-behaved:
arg max g(x) ≥ arg max f (x).
x∈X
x∈X
• Answers: SCP and IDO.
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Terminology
• Let T = [ 0, t̄ ].
• Let the probability space with a filtration be
(Ω, P, {Ft }t∈T ).
• α(t) denotes a discount rate, e.g. α(t) = e−rt .
• τ denotes a stopping time w.r.t. the filtration.
• u(t) denotes a payoff stream.
• G(t) is the termination payoff if the process is
stopped at time t.
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A lattice order () on the set of discounting
functions
Considering the problem: maxτ ∈T U (τ ; α)
———
The general formulation:
βα⇔
β(s)
is increasing in s
α(s)
A simple example:
0
Let α(s) = e−rs and β(s) = e−r s , then
β α ⇔ r ≥ r0 .
That is, the agent with the β discounting function is
more patient.
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Setup: the optimal stopping problem
A general formulation:
V (α) = sup U (τ ; α)
τ ∈T
Z τ
α(s)u(s)ds + α(τ )G(τ )
= sup E
τ ∈T
0
A simple example:
Z
V (r) = sup
τ ∈T
τ
e−rt u(t)dt ,
0
where u(t) is a deterministic function of t.
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Main idea: comparing optimal stopping rules
and value functions
Theorem (1)
Let Gt be a non-negative process. If β α, then
τ (β) > τ (α) and
V (β)
V (α)
≥
.
β(0)
α(0)
Proposition (1)
Let α and β be continuous. If β 6 α, the there exists a
deterministic cash-flow stream (πt )t∈T such that, with
zero termination value,
τ (α) > τ (β) and
V (β)
V (α)
≥
.
α(0)
β(0)
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Example: The IOD perspective – 1
Let the payoff stream to be deterministic. The discounted
payoff payoff function is
Z τ
U (τ ; r) =
e−rt u(t)dt.
0
Claim: If r00 > r0 > 0, then
arg max U (τ ; r0 ) ≥ arg max U (τ ; r00 ).
τ >0
τ >0
Statement of Theorem (1):
Lower discount rate
⇒ later stopping (and higher utility)
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Example: The IOD perspective – 2
Recall two facts: First, let J be an interval in X, where
X ⊂ R, then
g I f ⇒ arg max g(x) ≥ arg max f (x).
x∈J
x∈J
Second, a sufficient condition for g I f is that ∃ a
positive function d(·) such that
g 0 (x) ≥ d(x)f 0 (x).
Proof:
0
Ur (t; r)|r=r0 = e−r t u(t) = e(r
≥ e(r
00 −r 0 )t
00 −r 0 )t
00
e−r t u(t)
Ur (t; r)|r=r00
⇒ U (t; r0 ) I U (t; r00 )
9/1
Theorem (1) and Proposition (1)
Theorem (1)
Let G be a non-negative process. If β α, then
τ (β) > τ (α) and
V (α)
V (β)
≥
.
β(0)
α(0)
Proposition (1)
Let α and β be continuous. If β 6 α, the there exists a
deterministic payoff stream (πt )t∈T and zero termination
value such that,
τ (α) > τ (β) and
V (β)
V (α)
≥
.
α(0)
β(0)
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Proof of the theorem – 1
Lemma (1)
Let τ is optimal w.r.t. the discounting function α(t).
Then,
"Z
#
τ (ω)
P{ω : t ≤ τ (ω) and E
α(s)u(s)ds|Ft < 0} = 0.
t
Lemma (2)
def Let v(s) = E u(s)χ{s≤τ } . Then for all t ∈ [0, t̄),
Z
t̄
α(s)v(s)ds ≥ 0.
t
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Proof of the theorem – 2
Lemma (3)
R x00
Let γ be increasing and x h(s) > 0 for all x ∈ [x0 , x00 ].
Then,
Z x00
Z x00
0
γ(s)h(s)ds ≥ γ(x )
h(s)ds.
x0
x0
Proof: (The part about the value function)
Z
V (β) ≥
t̄
Z
β(s)v(s)ds =
0
≥
β(0)
α(0)
0
Z
t̄
β(s)
α(s)v(s)ds
α(s)
t̄
α(s)v(s)ds =
0
β(0)
V (α) ≥ 0.
α(0)
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Proof of the theorem – 3
Proof: (The part about the optimal stopping time)
Suppose not; then there is a set of non-zero
measure such that
B = {ω : τ (β) < τ (α)}.
Consider the stopping rule τ ∗ = max{τ (α), τ (β)}, then
U (τ ∗ ; β) = E[χ{Ω\B}
Z
τ (β)
Z
τ (β)
ds + χ{B}
0
ds
0
Z
+ χ{B}
|
τ (α)
τ (β)
β(s)u(s)ds]
{z
}
>0
which is greater than V (τ (β); β).
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The problem of optimal stopping and control
Agents simultaneously decide when to stop (τ ) and how
to control (λ) her payoff streams.
V (α) = sup U (τ, λ; α)
τ, λ
Z
= sup E
τ, λ
τ
α(s)u(x(λ), λ(s), s)ds + α(τ )G(τ ) ,
0
where {x(λ(t)}t∈T , {λ(t)}, and τ are adapted to the
filtration {Ft }.
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Existence of a solution
The existence of solution to the problem
V (α) = supτ ∈T U (τ ; α) is relatively easy: Continuity
condition on G(t) and {Ft }t is sufficient.
The existence of optimal control and optimal stopping to
the problem V (α) = supτ, λ U (τ, λ; α) is less obvious. The
stopping part can be parsed out by the DDP principle,
but finding the optimal control is not easy. One approach
is to use Hamilton-Jacobi-Bellman PDE. For the finite
horizon problem: A function satisfies
−Vt − sup µVx + 1/2σ 2 Vxx + u(t) = 0.
λ
is only a candidate to be the valuation function.
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Continuation Domain – 1 [+]
The continuation domain is the set of states that the
decision maker should wait, that is,
C(α, t) = {x : V (α, t, x) > G(t, x)},
where
V (α, t, x) = sup E
τ ≥t
1
α(t)
Z
τ
α(s)u(s)ds + α(τ )G(τ )
t
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Continuation domain – 2 [+]
Theorem (3)
If β α, then
C(α, t) ⊂ C(β, t).
Proof:
x ∈ C(α, t) ⇒ V (β, t, x) ≥ V (α, t, x) > G(t, x)
⇒ x ∈ C(β, t)
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Timing of decisions– 1 [−]
With both stopping and control decisions to be made, a
more patient player do not always stop later.
•
•
•
•
•
controls: λ ∈ {1, 2}
For x ∈ [1, 10), u(x, λ) = M
For x ∈ [10, ∞), u(x, λ) = −M
For x ∈ [0, 1), u(x, 1) = 1 and u(x, 2) = −0.01.
dx
dt = λ(t) > 0; hence x(t) is strictly increasing.
Then,
1. An impatient player (α = > 0) chooses control 1
and stops at t = 10.
2. A moderate patient player chooses control 2 to get to
the M payoff sooner, and also stops at t = 10.
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Timing of decisions– 2.a [+]
The problem:
Z
max U (τ, λ; r) = max E
τ, λ
τ, λ
τ
e−rs u(xt , λt , t)dt
0
Assume that
• x(t) is an Ito process with drift, i.e.
dx(t) = µdt + σdB(t).
• m∗ (t; r) = arg maxm {u(x, λ, t) : µ(x, λ, t) = m}
• u is increasing in x for all λ and t.
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Timing of decisions – 2.b [+]
Theorem (4)
V (r), m∗ (t; r), x∗ (t; r), and τ (r) are all decreasing in r.
In words, the value function, the optimal drift, the state
path, and the optimal stopping time are all increasing
with patience. This is analogous to Theorem (1).
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Time consistency – 1
An agent has a time in-homogenous discounting function.
At time s, the agent has a discounting function α(s, ·).
Let’s define a slightly different partial order, and let
β α if at each time s, the ratio
β(s, t)
α(s, t)
is increasing in t, for all t > s.
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Time consistency – 2 : naive agents [+]
Theorem (6)
If β α, then
τ (β) > τ (α) and
V (α(s, ·)
V (β(s, ·)
≥
.
β(s, s)
α(s, s)
Almost the same as Theorem (1). In fact, it is implied by
Theorem (1).
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Time consistency – 3 : sophisticated agents [−]
A agent with more patience stops earlier.
Let the cash flow be:
1, −M, M/n, M/n, · · · , M/n, 0, 0, 0, · · ·
|
{z
}
n+1
1. Agent A has α(s, t) = (1/2)t . She stops at period
t = 1.
2. Agent B’s α(0, t) = (1/2)t and α(s, t) = 1, for all
s ≥ 1. She stops at period t = 0.
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A few ideas
• Apply similar comparative statics results for a
different class of optimization problems
• Extend the analysis here to a repeated game setting,
e.g. different players have different but comparable
discounting functions.
• For dx(t) = µdt + σdB(t), put a ordering structure
() on σ.
• A large literature in finance that use a stochastic
control approach to solve the problems of portfolio
optimization and asset pricing. Most solutions are
numerical, and comparative results would be very
useful.
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