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

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

of 31

  • date post

    18-Jul-2020
  • Category

    Documents

  • view

    4
  • download

    0

Embed Size (px)

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