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### Transcript of Mathematical Economics Discrete time: calculus of ... Discrete time: intuition Assume we have a cake

• Mathematical Economics Discrete time: calculus of variations problem

Paulo Brito

1pbrito@iseg.ulisboa.pt University of Lisbon

November 23, 2018

• Discrete time: intuition Assume we have a cake of initial size W0 = ϕ and we want to eat it completely by time t = T, then WT = 0 The size of the cake evolution is:

Wt+1 = Wt − Ct ⇒ Wt = W0 − t−1∑ s=0

Cs

By imposing the two constraints W0 = ϕ and WT = 0 we have

T−1∑ s=0

Ct = ϕ

There is an infinite  number of paths {C0, . . .CT−1} verifying this condition Example: if we chose Ct = C̄ constant for all t we get C̄ = ϕT . Is this optimal ? Depends on the value functional

• Discrete time dynamic optimization: Intuition

A dynamic model is defined over sequences x (or (x, u)) in which there is some form of intertemporal interaction: actions today have an impact in the future; Types of time interactions: - intratemporal: iinteraction within one period - intertemporal: interaction across periods Admissible sequences: the set X contains a large number (possibly infinite) of admissible paths verifying a given intratemporal relation  together with some information regarding initial and/or terminal data ; Optimal sequences: an intertemporal optimality criterium  allows for choosing the best admissible sequence.

• The discrete time calculus of variations problem (CV)

Calculus of variations problem: find x = {xt}Tt=0 ∈ X , i.e., a path belonging to the set of admissible paths X , that maximises an intertemporal objective function (or value functional)

J(x) = T−1∑ t=0

F(xt+1, xt, t), x ∈ X

where F(t, xt, xt+1) specifies an intratemporal relation for the value of the action upon the state variable, within period t (i.e., between times t and t + 1).

• time

period

state

flow

0

x0

1

x1

2

x2

t

xt

t+1

xt+1

T−2

xT−2

T−1

xT−1

T

xT

0

F0 1

F1 t

Ft T−2

FT−2 T−1

FT−1

The value for changing the state variable in period t  is

Ft = F(xt+1, xt, t).

The value of the a given sequence of actions across T − 1 periods  is

J(x) = F(0, x0, x1) + . . .+ F(t, xt, xt+1) + . . .+ F(T − 1, xT−1, xT)

The optimal sequence  x∗ has value

J∗ = J(x∗) = max x

{J(x) : x ∈ X}

• The discrete time optimal control problem (OC)

Optimal control problem: find the pair (x, u) = ({xt}Tt=0, {ut}T−1t=0 ) ∈ D = X × U where D is defined by the sequence of intratemporal relations for periods 0 to T − 1

xt+1 = G(xt, ut, t), t = 0, 1, . . . , t, . . .T − 1

plus other conditions, that maximises an intertemporal objective functional

J(x, u) = T−1∑ t=0

F(t, xt, ut)

Observation: J(x) has terminal time T − 1 because if we know the optimal x∗T−1 and u∗T−1, we know the optimal x∗T, because x∗T = G(x∗T−1, u∗T−1,T − 1).

• time

period

state

control value flow

0

x0

1

x1

2

x2

t

xt

t+1

xt+1

T−2

xT−2

T−1

xT−1

T

xT

0

u0 1

u1 t

ut T−2

uT−2 T−1

uT−1

0

F0

1

F1

t

Ft

T−2

FT−2

T−1

FT−1

The value associated to choosing the control ut, given the state xt,  throughout period t  is

Ft = F(xt, ut, t).   Choosing the control ut, generates a the value for the state variable at the end of period t, xt+1 = g(xt, ut, t) The value of a given sequence of controls across T − 1 periods  is

J(u, x) = F(0, x0, u0) + . . .+ F(t, xt, ut) + . . .+ F(T − 1, xT−1, uT−1)   The optimal sequence  u∗ has value

J∗ = J(u∗) = max u

{J(x, u) : (x, u) ∈ D}

• Calculus of variations problems

We consider the following problems: Simplest problem: X = {T, x0, and xT given} Free terminal state problem: X = {x0, and T given}, xT free Constrained terminal state problem: X = {x0, and T, and h(xT) ≥ 0 given} Discounted infinite horizon problems: T = ∞ and limt→∞ xt free or constrained

• Calculus of variations: simplest problem The problem : Find x∗ = {x∗t }Tt=0  that maximizes

J(x) = T−1∑ t=0

F(xt+1, xt, t), s. t. x0 = ϕ0, xT = ϕT (1)

where T, ϕ0 and ϕT are known. First order necessary conditions

Proposition if x∗ ≡ {x∗0, x∗1, . . . , x∗T} is a solution of problem (1) , it verifies the Euler-Lagrange equation and the admissibility conditions

∂F ∂xt (x

∗ t , x∗t−1, t − 1) + ∂F∂xt (x

∗ t+1, x∗t , t) = 0, t = 1, 2, . . . ,T − 1

x∗0 = ϕ0, t = 0 x∗T = ϕT, t = T

Proof .

• Application: cake eating problem The problem

max C

{ T∑ t=0

βt ln(Ct) : Wt+1 = Wt − Ct, W0 = ϕ, WT = 0 }

Assumptions: 0 < β < 1, T ≥ 1, ϕ > 0 Intuition:

at time t = 0, we have a cake of initial size ϕ, and we want to consume it completely until time T (known) by choosing a sequence of bites such that: (1) we value independently each bite (the value functional is a sum); (2) the instantaneous pleasure of each bite is increasing with the size of the bite, but at a decreasing rate (the utility function u(Ct) is concave); (3) we are impatient: when we plan to eat the cake we value more the immediate bites rather than future bites ( there is a discount factor βt which decreases with time). How should we eat the cake ?

• Application: cake eating as a CV problem

We can transform the previous problem into a calculus of variations problem

max W

{ T∑ t=0

βt ln(Wt − Wt+1) : W0 = ϕ, WT = 0 }

• Application: cake eating problem - solution The optimal cake size is W∗ = {ϕ, . . . ,W∗t , . . . ,W∗T−1, 0} where

W∗t = ( βt − βT

1 − βT

) ϕ, t = 0, 1, . . .T

It slightly bends down from a linear path because of time discounting, as 0 < β < 1 Proof .

• Calculus of variations: free terminal state problem Find x∗ = {x∗t , }Tt=0  that maximizes

J(x) = T−1∑ t=0

F(xt+1, xt, t) s.t. x0 = ϕ0 (2)

where ϕ0 and T is given and xT is free. First order necessary conditions:

Proposition if x∗ is a solution of problem (2) , it verifies the Euler-Lagrange equation the admissibility condition and the transversality condition

∂F ∂xt

(x∗t , x∗t−1, t − 1) + ∂F∂xt (x ∗ t+1, x∗t , t) = 0, t = 1, 2, . . . ,T − 1

x∗0 = ϕ0, t = 0 ∂F ∂xT (x

∗ T, x∗T−1,T − 1) = 0, t = T

Proof .

• Cake eating problem: free terminal state

Problem

max {C}

T∑ t=0

βt ln(Ct), subject to Wt+1 = Wt − Ct, W0 = ϕ, WTfree

The problem is ill-posed: there is no solution with economic meaning Proof . Reason: the transversality condition only holds if C0 = ∞ this is intuitive: if we had no restriction on the terminal size of the case (or if could borrow freely) we would overeat. We have to redefine the problem to obtain a reasonable solution.

• Calculus of variations: constrained terminal state Find x∗ = {x∗t }Tt=0  that maximizes

J(x) = T−1∑ t=0

F(xt+1, xt, t) s.t. x0 = ϕ0, xT ≥ ϕT (3)

where T, ϕ0 and ϕT are given. First order necessary conditions:

Proposition if x∗ is a solution of problem (3), it verifies

∂F ∂xt

(x∗t , x∗t−1, t − 1) + ∂F ∂xt

(x∗t+1, x∗t , t) = 0, t = 1, 2, . . . ,T − 1

x∗0 = ϕ0, t = 0 ∂F ∂xT

(x∗T, x∗T−1,T − 1) · (ϕT − x∗T) = 0, t = T

Proof .

• Cake eating problem: constrained terminal state

Problem

max {C}

T∑ t=0

βt ln(Ct), subject to Wt+1 = Wt − Ct, W0 = ϕ, WT ≥ 0

F.o.c  W∗t+2 = (1 + β)W∗t+1 − βW∗t , t = 0, 1, . . . ,T − 2 W∗0 = ϕ, t = 0

βT−1

W∗T − W∗T−1 W∗T = 0, t = T

has the same formal solution as the simplest problem: W∗T = 0 is determined endogenously not by assumption. Proof .

• Calculus of variations: discounted infinite horizon 1

The problem: find the infinite sequence x∗ = {x∗t }∞t=0

max x

∞∑ t=0

βtf(xt+1, xt), s.t. x0 = ϕ0 (4)

where, 0 < β < 1, and ϕ0 are given and the terminal state is free (i.e., limt→∞ xt is free). The first order conditions are:

∂f ∂xt

(x∗t , x∗t−1) + β ∂f ∂xt

(xt+1, xt) = 0, t = 0, 1, . . .

x∗0 = x0, t = 0 limt→∞ β

t−1 ∂f(x ∗ t ,x

∗ t−1)

∂xt = 0, t → ∞

• Calculus of variations: discounted infinite horizon 2

The problem: find the infinite sequence x∗ = {x∗t }∞t=0

max x

∞∑ t=0

βtf(xt+1, xt), s.t. x0 = ϕ0, limt→∞ xt ≥ 0 (5)

given β and ϕ0 F.o.c. 

∂f ∂xt