1. Consider the standard cake eating problem with a modification of the transition equation for the cake

W′ = RW − c  (1)

if R > 1. the cake yields a positive return, whereas if Rϵ(0, 1) the cake depreciates.

(a) If preferences are of the form u(c) = \frac{c^{1−σ}}{1 − σ}, calculate the optimal sequence of consumption. What happens when σ = 1? and when σ = 0?

To solve the constrained optimization problem, need to find the sequence of {c_{t}}_{1}^{T} {W_{t}}_{2}^{T +1} that satisifies

V_{t}(W ′) = max ∑_{t=1}^{T} β^{(t−1)} u(c_{t})  (2)

subject to the transition equation (1), which holds for t = 1, 2, 3, …, T. Also, subject to nonegative constrants on comsuming the cake given by c_{t} ≥ 0 and W_{t+1} ≥ 0

Adding period 0 consumption, the solution to (2) can be rewitten as

V_{t+1}(W) = max_{{c_{t}}_{t}^{T}} {u(c_{t}) + βV_{t}(W ′)}  (3)

By using the Legrange multipliers, combining the first order conditions we obtain the Euler equation that is necessary condition of optimality for any t.

u′(c_{t}) = R β u′(c_{t+1})  (4)

By letting u(c) = \frac{c^{1−σ}}{1 − σ} , ⇒ u′(c) = c^{−σ}, subsitute in (4) obtain the following:

(c_{t})^{−σ} = R β (c_{t+1})^{−σ}  (5)
c_{t+1} = (R β)^{\frac{1}{σ}} c_{t}  (6)

Let β̃ = (R β)^{\frac{1}{σ}}, and using properites of geometric sequence then (6) can be expressed as consumpution of period T + 1 in terms of period 0.

c_{t+1} = (β̃)^{T} c_{0}  (7)

Using the sequence of transition equations from (1), multiply each equation by R^{T −i} where i = 0, 1, …, T.

at t = 0  R^{T} W^{1} = R_{T +1} W_{0} − R_{T} c_{0}
at t = 1  R^{T −1} W_{2} = R^{T} W_{1} − R^{T −1} c_{1}
at t = 2  R^{T −2} W_{3} = R^{T −1} W_{0} − R^{T −2} c_{2}
⋮  ⋮
at t = T  R^{0} W_{T +1} = R^{0} W_{T} − R^{0} c_{T}

When sum all the transaction equations obtain the following:

R^{0} W_{T +1} = R^{T +1} W_{0} − ∑_{t=0}^{T} R^{T −t} c_{t}  (8)

Let consumption at period W_{T +1} = 0 and intial cake size is known, W_{0} = \overline{W}_{0}. By subsituting (7) into (8) for c_{i}, obtain the following expression for c_{0}.

R^{T +1} \overline{W}_{0} = ∑_{t=0}^{T} R^{T −t} (β̃)^{t} c_{0}

R^{T +1} \overline{W}_{0} = R^{T} c_{0} ∑_{t=0}^{T} R^{−t} (β̃)^{t}

R \overline{W}_{0} = c_{0} ∑_{t=0}^{T} (\frac{β̃}{R})^{t}

c_{0} = \frac{R \overline{W}_{0}}{∑_{t=0}^{T} (\frac{β̃}{R})^{t}}  (9)

From (7), subsititing (9), obtain a solution for the consumption at any period t = 1, 2, …, T on the finite horizon.

c_{t} = \frac{β̃^{t} R \overline{W}_{0}}{∑_{t=0}^{T} (\frac{β̃}{R})^{t}}  (10)

Recall that β̃ = (R β)^{\frac{1}{σ}}.

c_{t} = \frac{(R β)^{\frac{t}{σ}} R \overline{W}_{0}}{∑_{t=0}^{T} (\frac{(R β)^{\frac{1}{σ}}}{R} )^{t}}  (11)

When σ = 1 obtain:

c_{t} = \frac{(R β)^{t} R \overline{W}_{0}}{∑_{t=0}^{T} (β)^{t}}  (12)

When σ = 0 obtain:

u(c) = c ⇒ u′(c) = 1

From (4), obtain the following:

1 = R β  (13)

(b) Show that we can reduce the cake eating problem to a system of nonlinear equations of the form F (W) = 0, the optimal sequence of consumption, {c_{t} ^{∗}}_{1}^{T}.

Note: If find the sequence {W_{t} ^{∗}}_{1}^{T +1}, can find sequence of consumption by using (1).

Using (3), subject to the constaints (1), Note: c_{t} = RW_{T} − W_{t+1}, consumption at period W_{T +2} = 0 and intial cake size is known, W_{0} = \overline{W}_{0}, subsituting in (4), obtain the following system of nonlinear equations with t equations and t unknowns:

F (W) = [ u′(R W_{0} − W_{1}) = R β u′(R W_{1} − W_{2}) ; u′(R W_{1} − W_{2}) = R β u′(R W_{2} − W_{3}) ; u′(R W_{2} − W_{3}) = R β u′(R W_{3} − W_{4}) ; ⋮ ; u′(R W_{T} − W_{T +1}) = R β u′(R W_{T +1} − W_{T +2}) ]

(c)

2. Consider the previous version of the cake eating problem with a law of motion of the form,

W ′ = y + δ − c , where δ ϵ (0, R)  (14)

(a) Formulate the Bellman equation of the infinite horizon problem.

As in the case of the previous problem, calculate the optimal sequences, {c_{t}}_{1}^{∞} {W_{t}}_{2}^{∞}, that solves the following equation, subject to the constraint in (14).

V (W) = max ∑_{t=0}^{T} β^{t} u(c_{t})  (15)

Equation (15) can be rewritten as, subject to (14), where 0 ≤ W′ ≤ W:

V (W) = max{ u(y + δ W − W′) + β V (W′) }  (16)

(b) If preferences are of the form u(c) = ln(c) and y = 0, calculate the optimal value function and the implied decision rules. Does the size of δ matter for the optimal values?

u(c) = ln(c) ⇒ u′(c) = \frac{1}{c} and y = 0

W ′ = δ W − c ⇒ c = δ W − W′  (17)

Using the following form below for the value function, derive the first order conditions to find the optimal policy functions.

V (W) = α_{0} + α_{1} ln(W)  with ⇒ V ′(W) = \frac{α_{1}}{W}  (18)

First Order Conditions (16)

[W′] :  u′(c) = β V ′(W′)
\frac{1}{δ W − W′} = β (\frac{α_{1}}{W′} )  subsituting (17) and (18).
W ′ = β α_{1} δ W − α_{1} β W′

Solve for W′ to obtain the optimal policy function for savings:

W ′ = (\frac{β α_{1}δ}{1 + β α_{1}} ) W  (19)

Subsituting (19) in (17) to obtain the optimal policy function for consumption:

c = δ W − (\frac{β α_{1}δ}{1 + β α_{1}} ) W = (\frac{δ + α_{1} δ β − α_{1} δ β}{1 + α_{1} β} ) W

c = (\frac{δ}{1 + β α_{1}} ) W  (20)

Need to solve for the coefficents of the value function in (18). Setting (18) equal to (16), and solve for coefficients of (18) {α_{0}, α_{1}}. This will be done by subsituting (20) and (19) into (16). Solving for {α_{0}, α_{1}} and subsituting into (20) and (19) will obtain the optimal value function.

α_{0} + α_{1} ln(W) = u(c) + β^{t} V (W ′)  from (18) and (16)
α_{0} + α_{1} ln(W) = ln(c) + β (α_{0} + α_{1} ln(W ′))  from given u(c) = ln(c) and (18)

α_{0} + α_{1} ln(W) = ln (\frac{δ W}{1 + β α_{1}} ) + β α_{0} + β α_{1} ln (\frac{β α_{1}δ W}{1 + β α_{1}} )  from (20) and (19)

α_{0} + α_{1} ln(W) = \underbrace{ln (\frac{δ}{1 + β α_{1}} ) + β α_{0} + β α_{1} ln (\frac{β α_{1}δ}{1 + β α_{1}} )}_{constant = α_{0}} + \underbrace{(1 + β α_{1})}_{α_{1}} ln(W)

Therefore, able to setup the following equations and solve for {α_{0}, α_{1}}.

α_{1} = 1 + β α_{1}

α_{1} = \frac{1}{1 − β}  (21)

α_{0} = ln (\frac{δ}{1 + β α_{1}} ) + β α_{0} + β α_{1} ln (\frac{β α_{1}δ}{1 + β α_{1}} )

α_{0} = ln (\frac{δ}{1 + \frac{β}{1 − β}} ) + β α_{0} + \frac{β}{1 − β} ln (\frac{\frac{β δ}{1 − β}}{1 + \frac{β}{1 − β}} )  Subsituting (21)

α_{0}(1 − β) = ln (δ (1 − β)) + \frac{β}{1 − β} ln(β δ)

α_{0} = \frac{1}{1 − β} ln (δ (1 − β)) + \frac{β}{(1 − β)^{2}} ln(β δ)  (22)

Using (22) and (21), able to obtain the optimal policy function and the optimal value function

c = (\frac{δ}{1 + α_{1} β} ) W  from (20)

c = (\frac{δ}{1 + \frac{β}{1 − β}} ) W  subsituting (21)

c = δ(1 − β)W  (23)

W ′ = δ W − C  from (17)

W ′ = δ β W  (24)

Therefore, (24) and (23) is the optimal policy function. Using the solutions for {α_{0}, α_{1}} from (21) and (22) able to obtain the following optimal value function

V (w) = \frac{1}{1 − β} ln (δ (1 − β)) + \frac{β}{(1 − β)^{2}} ln(β δ) + \frac{1}{1 − β} ln(W)  (25)

Does

(c)

3. Consider the instanteous utility function that depends on previous consumption u(c_{t−1}, c_{t}). Previous consumption affects present consumption, so consumers have to take into account that today’s consumption choice will also affect future utility besides the size of the cake.

(a.) Solve the T -period cake eating problem using this preferences.

The value function, using the conditions where the previous consumption affects present consumption, Calculate the optimal sequence, {c_{t}}_{0}^{∞}, that solves the following equation, subject to the constraint listed.

max ∑_{t=0}^{T} β^{t} u(c_{t−1}, c_{t})  (26)

subject to ∑_{t=0}^{T} c_{t} + W_{T +1} = W_{0}  (27)