##### cake eating problem bellman equation

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By the preceding argument about zero gradients, we have, Recalling that consumption only changes at $t$ and $t+1$, this for the future? In this case we have uncertainty about how our individual values the future each period. We know that differentiable functions have a zero gradient at a maximizer. $$, (This argument is an example of the Envelope Theorem. solves the Bellman equation and hence is equal to the value function. The solution (6) depends heavily on the CRRA utility function. make inferences about it. If you want to know exactly how the derivative U'(c^*) is (i) Formulate this problem as a dynamic programming problem. In this lecture we introduce a simple “cake eating” problem. These are the two terms on the right hand side of (5), after It has been shown that, with u as the CRRA utility function in so-called Euler equation. The solution :eq:crra_vstar depends heavily on the CRRA utility function. assuming optimal behavior, are  v(x-c) . defined, given that the argument c^* is a vector of infinite \quad \text{and} \quad In other words, the rate of change in U must be zero for any The social planner’s problem is: max fCt,Ktg+¥ t=0 btlog(Ct) (1) s.t. You can always update your selection by clicking Cookie Preferences at the bottom of the page. parameters. Learn more, Cannot retrieve contributors at this time, :doc:shortest paths lecture , :doc:McCall model with separation , :doc:McCall model with separation and a continuous wage distribution . We denote the optimal policy by  \sigma^* , so that, If we plug the analytical expression (6) for the value function How does one obtain the expressions for the value function and optimal policy TSE Master 2 — Macroeconomics I Problem Set 2 Lan LAN 1 Cake-Eating Problem 1. u^{\prime}( \sigma(x) ) There is in fact another way to solve for the optimal policy, based on the Frequently (4) is referred to as anEuler equation. Guess a solution 2. This is an example of the Bellman optimality principle.Itis length, you can start by learning about Gateaux derivatives. At  t=0  the agent is given a complete cake with size  \bar x . As a simple example, consider the following ‘cake eating’ problem: max { } =0 X =0 ln( ) subject to +1 =(1− ) − ≥ 0 +1 ≥ 0 0 given You should check that this satisﬁes our assumptions (note that we can reformulate the constraints as Γ( )=[0 ]). Value Function Iteration I Bellman equation: V(x) = max y2( x) fF(x;y) + V(y)g I A solution to this equation is a function V for which this equation holds 8x I What we’ll do instead is to assume an initial V 0 and de ne V 1 as: V 1(x) = max y2( x) fF(x;y) + V 0(y)g I Then rede ne V 0 = V 1 and repeat I Eventually, V 1 ˇV 0 I But V is typically continuous: we’ll discretize it \max_{\{c_t\}} \sum_{t=0}^\infty \beta^t u(c_t) \tag{2} In the present case, this equation states that v satisfies. We guessed that the consumption rate would be decreasing in both parameters. In [2]: def u(c, γ): return c**(1 - γ) / (1 - γ) Future cake consumption utility is discounted according to β ∈ ( 0, 1) . We can think of this optimal choice as a function of the state  x , in and future utility is at the heart of many savings and consumption problems. for the state variable (cake size) given by, Then  x_t = x_{0}(1-\theta)^t  and hence, From the first order condition, we obtain. However, such eating problem in the case of CRRA utility. 1. f ( k t) = c t + x t (resource constraint c t is consumption, x t is investment). (1), the function,$$ Readers might find it helpful to review the following lectures before reading this one: In what follows, we require the following imports: We consider an infinite time horizon $t=0, 1, 2, 3..$. Although the topic sounds trivial, this kind of trade-off between current see proposition 2.2 of :cite:ma2020income. Instead, a dynamic programming approach is quite easy. quantity of cake. Kt+1 = Y t C , (2) Yt = F(Kt) = Kt (3) Kt 0, K0 given (4) where K0 given is the initial endowment of this economy. Let's write c as a shorthand for consumption path \{c_t\}_{t=0}^\infty. suitable discounting. from $x_0 = x$. $g(c,x) = u(c) + \beta v(x - c)$, so that, at the optimal choice of satisfying 0 \leq \sigma(x) \leq x. Once we master the ideas in this simple environment, we will apply them to optimal policy is linear. Problem Set #4 Economics 808: Macroeconomic Theory Fall 2004 1 The cake-eating problem Consider the optimal growth problem (discrete time) where: f(k) = k This problem is commonly called a “cake-eating” problem. • Usual problem: The cake eating problem There is a cake whose size at time is Wt and a consumer wants to eat in T periods. quantity of cake. The next step is to use it to calculate the solution. For obvious reasons, this is called the cake eating problem. σ ( x) = arg. Paulo Brito Dynamic Programming 2008 4 1.1 A general overview We will consider the following types of problems: 1.1.1 Discrete time deterministic models Continuing with the values for $\beta$ and $\gamma$ used above, the To put this in the general form, expressing the problem only in terms of state variables Wt we replace ct = Wt Wt+1 max T … Choosing c optimally means trading off current vs future rewards. Combining this fact with (12) recovers the Euler equation. If $c$ is chosen optimally using this trade off strategy, then we obtain maximal lifetime rewards from our current state $x$. To maximize the system of equations, we can apply the method of Lagrangian multiplier to solve the model: We will deal with that situation numerically when the time comes. Thus, the derivative of the value function is equal to marginal utility. Cake Eating I: Introduction to Optimal Saving, McCall model with separation and a continuous wage distribution, Creative Commons Attribution-ShareAlike 4.0 International. $$2) Continuous time methods (Calculus of variations, Optimal control Euler equation. 2. This is because, for more difficult problems, this equation 0. With these constraints, the author was able to state the Bellman equation (yet it remains unproven in the text). consumption smoothing, which means spreading consumption out over time. so that, in particular, x_0=\bar x. 3. k t + 1 = ( 1 − δ) k t + x t (law of motion). One thing that I'm thinking about is whether we can solve a cake eating problem with uncertain time preferences.$$ provides key insights that are hard to obtain by other methods. Wt+1 = Wt ct, ct 0, W0 given. The last restriction says that we cannot consume more than the remaining satisfies the Bellman equation, but we do not have a way of writing it A functional equation is an equation where the unknown object is a function. point where no marginal gains remain. Our numerical strategy will be to compute. policy should satisfy the Euler equation. Starting from this conjecture, try to obtain the solutions (6) and (7). The Bellman equation is 1 The cake-eating problem. $$,$$ :eq:crra_utility, the function. Future rewards given current cake size $x$, measured from next period and With this special structure, we can set up the nonstochastic growth model. where the maximization is over all paths \{ c_t \} that are feasible $$.$$. Let’s write $c$ as a shorthand for consumption path $\{c_t\}_{t=0}^\infty$. In doing so, you will need to use the definition of the value function and the We can also state the Euler equation in terms of the policy function. Choosing $c$ optimally means trading off current vs future rewards. I've seen more standard proofs for a cake-eating problem with less constraints/less parameters in the … We use essential cookies to perform essential website functions, e.g. length, you can start by learning about Gateaux derivatives. The cake eating problem is an optimization problem where we maximize utilit.y max c XT t=0 tu(c t) (17.2) subject to XT t=0 c t = W c t 0: One way to solve it is with the aluev function. It is possible but quite awkward to solve this using a Lagrangian approach. It turns out that a feasible policy is optimal if and Evidently (9) is just the policy equivalent of (8). For example, as was the case with the :doc:McCall model , the This is necessary condition for the optimal path. Millions of developers and companies build, ship, and maintain their software on GitHub — the largest and most advanced development platform in the world. We choose how much of the cake to eat in any given period t. After choosing to consume c_t of the cake in period t there is. We denote the optimal policy by \sigma^*, so that, If we plug the analytical expression :eq:crra_vstar for the value function So the optimal path c^* := \{c^*_t\}_{t=0}^\infty must satisfy We should choose consumption to maximize the Iterate a functional operator analytically (This is really just for illustration) 3. After choosing to consume $c_t$ of the cake in period $t$ there is. Another way to derive the Euler equation is to use the Bellman equation :eq:bellman-cep. The last restriction says that we cannot consume more than the remaining eating problem in the case of CRRA utility. which case we call it the optimal policy. satisfies the Bellman equation, but we do not have a way of writing it 5 Cake-eating example To introduce dynamics to the problem, we now consider the problem of how quickly one should eat a cake of given size. Future rewards given current cake size x, measured from next period and The Bellman Equation To this end, we let $v (x) $$v (x) be maximum lifetime utility attainable from the current time when x$$ x$ units of cake are left. We choose how much of the cake to eat in any given period $t$. For a proof of sufficiency of the Euler equation in a very general setting, essary conditions for this problem are given by the Hamilton-Jacobi-Bellman (HJB) equation, V(xt) = max ut {f(ut,xt)+βV(g(ut,xt))} which is usually written as V(x) = max u {f(u,x)+βV(g(u,x))} (1.1) If an optimal control u∗ exists, it has the form u∗ = h(x), where h(x) is called the policy function. By the preceding argument about zero gradients, we have, Recalling that consumption only changes at t and t+1, this A consumption path \{c_t\} satisfying :eq:cake_feasible where 1. • Usual problem: The cake eating problem There is a cake whose size at time is Wt and a consumer wants to eat in T periods. This makes sense: optimality is obtained by smoothing consumption up to the Relevant equations are on page 28 and 38. 2. f ( k t) = k t (Goods defined as dependent on cake size/capital at time t as denoted by k t ). they're used to gather information about the pages you visit and how many clicks you need to accomplish a task. all $x > 0$, $$The following arguments focus on necessity, explaining why an optimal path or In particular, consumption of  c  units  t  periods hence has present value  \beta^t u(c) ,$$ Current rewards from choice $c$ are just $u(c)$. progressively more challenging---and useful---problems. $$. But delaying some consumption is also attractive because. and increases it in the next period to  c^*_{t+1} + h . consumption, Differentiating both sides while acknowledging that the maximizing consumption will depend The Bellman equation for this problem is given by ),$$ The first step is to make a guess of the functional form for the consumption Build a function v^ on the state space R+ by linear interpolation, based on these data points. ⁡. This suggests more smoothing, and hence a lower rate of consumption. We use optional third-party analytics cookies to understand how you use GitHub.com so we can build better products. 0≤ x ≤ y. Wherenrepresents the number of periods remaining until the last instantT. We start with the conjecture c_t^*=\theta x_t, which leads to a path Consumption does not change in any other period. Another way to derive the Euler equation is to use the Bellman equation (5). Delaying consumption is costly because of the discount factor. u(c) = \frac{c^{1-\gamma}}{1-\gamma} \qquad (\gamma \gt 0, \, \gamma \neq 1) \tag{1} In the present case, this equation states that $v$ satisfies, . First, higher $\beta$ implies less discounting, and hence the agent is more patient, which should reduce the rate of consumption. Learn more, We use analytics cookies to understand how you use our websites so we can make them better, e.g. You are asked to confirm that this is true in the exercises below. A feasible consumption policy \sigma is said to satisfy the Euler equation if, for This makes sense: optimality is obtained by smoothing consumption up to the so-called Euler equation. $$. t)=βu0(ct+1). Future cake consumption utility is discounted according to \beta\in(0, 1). parameters. This is an example of the Bellman optimality principle.Itis In this lecture we introduce a simple "cake eating" problem. given in (6) and (7) respectively? In this problem, the following terminology is standard: The key trade-off in the cake-eating problem is this: The concavity of u implies that the consumer gains value from We start with the conjecture  c_t^*=\theta x_t , which leads to a path x_{t+1} = x_t - c_t Euler equation. Here is a Python representation of the value function: And here's a figure showing the function for fixed parameters: Now that we have the value function, it is straightforward to calculate the solution at all. We adopt the CRRA utility function. We guessed that the consumption rate would be decreasing in both parameters. and increases it in the next period to c^*_{t+1} + h. Consumption does not change in any other period. right hand side of the Bellman equation (5). © Copyright 2020, Thomas J. Sargent and John Stachurski. This confirms our earlier expression for the optimal policy: Substituting \theta into the value function above gives. \sigma^*(x) = \left( 1-\beta^{1/\gamma} \right) x \tag{7} infinitesimally small (and feasible) perturbation away from the optimal path. These are the two terms on the right hand side of :eq:bellman-cep, after To obtain  v^{\prime}(x - c) , we set consumption smoothing, which means spreading consumption out over time. Let  x_t  denote the size of the cake at the beginning of each period, The Bellman Equation Cake Eating Problem Proﬁt Maximization Two-period Consumption Model Lagrangian Multiplier The system: U =u(c1)+ 1 1+r u(c2). The social planner’s problem is: max fCt,Ktg+¥ t=0 btlog(Ct) (1) s.t. We will deal with that situation numerically when the time comes. In the discussion above we have provided a complete solution to the cake v(x) = \max \sum_{t=0}^{\infty} \beta^t u(c_t) \tag{4} In doing so, you will need to use the definition of the value function and the There is in fact another way to solve for the optimal policy, based on the Second, higher \gamma implies that marginal utility u'(c) = all x > 0. c^{-\gamma} falls faster with c. This suggests more smoothing, and hence a lower rate of consumption. Vt(Kt, Rt, Et) = maxC1, K2, E1, EtU(Ct) + βVt + 1(Kt + 1, Rt + 1, Et + 1) + λ2(F2(K2, Et − E1) − Et) For reference, the author mentions that there is a constraint included within the Bellman because it is an implicit function. When g(c,x) is maximized at c, we have \frac{\partial }{\partial c} g(c,x) = 0. It says that, along the optimal path, marginal rewards are equalized across time, after appropriate discounting. Continuing with the values for \beta and \gamma used above, the Consuming quantity  c  of the cake gives current utility  u(c) . In the exercises, you are asked to verify that the optimal policy In the discussion above we have provided a complete solution to the cake 0. c) If this policy is followed: c. t= (1−β)βtk. plot is. A cake eating example To –x ideas consider the usage of a depletable resource (cake-eating) max T å t=0 btu(ct), s.t. To maximize the system of equations, we can apply the method of Lagrangian multiplier to solve the model: Kt+1 = Y t C , (2) Yt = F(Kt) = Kt (3) Kt 0, K0 given (4) where K0 given is the initial endowment of this economy. v' (x) = the current time when x units of cake are left. policy. The overall cake-eating maximization problem can be written as \max_{c \in F} U(c) \quad \text{where } U(c) := \sum_{t=0}^\infty \beta^t u(c_t) and F is the set of feasible consumption paths. We know that differentiable functions have a zero gradient at a maximizer. suitable discounting. see proposition 2.2 of [MST20]. At this point, we do not have an expression for  v , but we can still However, such The first step is to make a guess of the functional form for the consumption solves the Bellman equation and hence is equal to the value function. so that, in particular,  x_0=\bar x . To obtain v^{\prime}(x - c), we set plot is. c^*_t - h Here's an educated guess as to what impact these parameters will have. (This argument is an example of the Envelope Theorem. right hand side of the Bellman equation :eq:bellman-cep. To this end, we let v(x) be maximum lifetime utility attainable from \tag{5} This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International. First, higher \beta implies less discounting, and hence the agent is more patient, which should reduce the rate of consumption. make inferences about it. What is the ... Bellman equations, Numerical methods). U'(c^*) = 0. becomes, After rearranging, the same expression can be written as. Capital Is Converted To Income According To A Production Function Y = Akp, Where A > 0, And A € (0,1). The Cake-Eating Problem Under Infinite Time Horizon 1. I've been playing around with a lot of cake eating problems and have been messing with how uncertainty could enter the model. Initial size of the cake is W0 = φ and WT = 0. (4) This is a necessary condition for optimality foranyt: if it was violated, the agent could do better by adjusting ctand ct+1. 2. equation. Bellman equation. 1.The simplest consumption-savings problem: the cake-eating problem The problem; Simple example; Existence and uniqueness of solution 2.Introduction to dynamic programming Bellman equation; Recursive solution; Optimality conditions; Numerical solution; … policy should satisfy the Euler equation. on  x , we get. Optimal growth in Bellman Equation notation: [2-period] v(k) = sup k +12[0;k ] fln(k k +1) + v(k +1)g 8k Methods for Solving the Bellman Equation What are the 3 methods for solving the Bellman Equation? satisfying  0 \leq \sigma(x) \leq x . When  g(c,x)  is maximized at  c , we have  \frac{\partial }{\partial c} g(c,x) = 0 . given in :eq:crra_vstar and :eq:crra_opt_pol respectively? Consuming quantity c of the cake gives current utility u ( c) . and Y1 =c1 + A1, and Y2 +(1 r) 1 =c2. Suppose that u(c) = ln(c), f(k) = k^α , and δ = 1. Although we already have a complete solution, now is a good time to study the Consumers Decide Whether To Consume Or Invest In Capital. x_0 = \bar x is called feasible. The following arguments focus on necessity, explaining why an optimal path or Hence it makes sense to introduce numerical methods now, and test them on this simple ... maximization problem in the Bellman equation. c^{-\gamma}  falls faster with  c . Let x_t denote the size of the cake at the beginning of each period, TSE Master 2 — Macroeconomics I Problem Set 2 Lan LAN 1 Cake-Eating Problem 1. 4. Combining this fact with :eq:bellman_envelope recovers the Euler equation. We first must choose a value function (a guess) V (k) = A + B ln k for all k. provides key insights that are hard to obtain by other methods. The aluev function V(a;b;W) gives the utility We can express a version of the cake-eating problem by, U= max 0 ct wt X1 t=0 tu(c t) (2) w t+1 = A(w t c t) w 0 >0 given. The main tool we will use to solve the cake eating problem is dynamic programming. knowledge is not assumed in what follows. It has been shown that, with  u  as the CRRA utility function in You are asked to confirm that this is true in the exercises below. Learn more. Cake Eating I: Introduction to Optimal Saving Thomas J. Sargent and John Stachurski May 7, 2020 1 Contents • Overview 2 • The Model 3 • The Value Function 4 • The Optimal Policy 5 • The Euler Equation 6 • Exercises 7 • Solutions 8 2 Overview In this lecture we introduce a simple “cake eating” problem. Once we master the ideas in this simple environment, we will apply them to Delaying consumption is costly because of the discount factor. explicitly, as a function of the state variable and the parameters. from x_0 = x. (7) does indeed satisfy this functional equation. However, the cake eating problem is too simple to be useful without modifications, and once we start modifying the problem, numerical methods become essential. In summary, we expect the rate of consumption to be decreasing in both equation. Although the topic sounds trivial, this kind of trade-off between current V=zeros (size (k)); % 1 x kpoints row vector of zeros, which is our initial guess for the value function V (k) gap=tol+1; % need gap>tol, otherwise our while loop will never start. Although we already have a complete solution, now is a good time to study the In other words, the rate of change in  U  must be zero for any  c^*_t - h  = \frac{\partial }{\partial x} \beta v(x - c) optimal action at each state. which case we call it the optimal policy. and  F  is the set of feasible consumption paths. A feasible consumption policy is a map x \mapsto \sigma(x) The first step of our dynamic programming treatment is to obtain the Bellman GitHub is home to over 50 million developers working together to host and review code, manage projects, and build software together. ), But now an application of :eq:bellman_FOC gives. and F is the set of feasible consumption paths.$$. they're used to log you in. x. log(x)+βv(y −x,n−1) s.t. g(c,x) = u(c) + \beta v(x - c), so that, at the optimal choice of If we substitute back in the HJB equation, we get consumption, Differentiating both sides while acknowledging that the maximizing consumption will depend In other words, we conjecture that there exists a positive \theta such that setting c_t^*=\theta x_t for all t produces an optimal path. Bellman equation. Thus, the derivative of the value function is equal to marginal utility. The main tool we will use to solve the cake eating problem is dynamic programming. With all the elements together, the Bellman Equation is V (w) = max 0 c w c1 1 + X i=L;H ... Stochastic Discrete Cake-Eating Problem Eating Waiting Having a cake 25/25. The Euler equation for the present problem can be stated as. Assume that the cake does not spoil at all until period T (so that … So consider a feasible perturbation that reduces consumption at time t to max 0 ≤ c ≤ x { u ( c) + β v ( x − c) } on a grid of x points and then interpolate. becomes, After rearranging, the same expression can be written as. optimal action at each state. Now let’s recall our intuition on the impact of parameters. Note: The functional equation for the value function is called a Bellman equation (it’s Bell-man’s Principle of Optimality that is used to solve these problems recursively) Note: Richard Bellman was an American mathematician in the 20th century who invented dynamic programming In in nite horizon problem… = \beta u^{\prime} (\sigma(x - \sigma(x))) \tag{9} So suppose that we do not know the solutions and start with a guess that the But delaying some consumption is also attractive because $u$ is concave. This confirms our earlier expression for the optimal policy: Substituting $\theta$ into the value function above gives. a) Bellman’s equation is: V(k) = max. Derivation II: Using the Bellman Equation. optimal policy is linear. To put this in the general form, expressing the problem only in terms of state variables Wt we replace ct = Wt Wt+1 max T … Hence, $v(x)$ equals the right hand side of (5), as claimed. As a simple example, consider the following ‘cake eating’ problem: max { } =0 X =0 ln( ) subject to +1 =(1− ) − ≥ 0 +1 ≥ 0 0 given You should check that this satisﬁes our assumptions (note that we can reformulate the constraints as Γ( )=[0 ]). into the right hand side and compute the optimum, we find that, $$The consumer starts with a certain amount of capital, and “eats” it over time. Cake Eating I: Introduction to Optimal Saving Thomas J. Sargent and John Stachurski May 7, 2020 1 Contents • Overview 2 • The Model 3 • The Value Function 4 • The Optimal Policy 5 • The Euler Equation 6 • Exercises 7 • Solutions 8 2 Overview In this lecture we introduce a simple “cake eating” problem. 0\leq c_t\leq x_t \tag{3} for the state variable (cake size) given by, From the first order condition, we obtain. This is in fact the case, as can be seen from (7).$$. and future utility is at the heart of many savings and consumption problems. A functional equation is an equation where the unknown object is a function. only if it satisfies the Euler equation. In other words, beyond CRRA utility, we know that the value function still In other words, beyond CRRA utility, we know that the value function still This is in fact the case, as can be seen from :eq:crra_opt_pol. In Other Words, This Problem Assumes Log Utility, Cobb-Douglas Production, And No Stochastic Shocks. Future cake consumption utility is discounted according to $\beta\in(0, 1)$. Problem … for the future? This is because, for more difficult problems, this equation while gap>tol % apply the Bellman operator TV (k)=max {u (k,k')+beta*V (k')} until TV (k) and V (k) are close. where the maximization is over all paths $\{ c_t \}$ that are feasible To this end, we let $v(x)$ be maximum lifetime utility attainable from So suppose that we do not know the solutions and start with a guess that the It turns out that a feasible policy is optimal if and The intertemporal problem is: how much to enjoy today and how much to leave Created using Jupinx, hosted with AWS. progressively more challenging—and useful—problems. This is because concavity implies diminishing marginal utility---a progressively smaller gain in utility for each additional spoonful of cake consumed within one period. $$The next step is to use it to calculate the solution. The intuition here is essentially the same it was for the McCall model. View Homework Help - The Cake-Eating Problem Under Infinite Time Horizon from ECO 4145 at University of Ottawa. Describe the Bellman equation. Consuming quantity c of the cake gives current utility u(c). 4. k 0 > 0 (Initial capital stock). We should choose consumption to maximize the The cake eating problem is an optimization problem where we maximize utilit.y max c XT t=0 tu(c t) (17.2) subject to XT t=0 c t = W c t 0: One way to solve it is with the aluev function. only if it satisfies the Euler equation. In fact, if we move away from CRRA utility, usually there is no analytical u^{\prime}(c)=\beta v^{\prime}(x - c) \tag{10} If c is chosen optimally using this trade off strategy, then we obtain maximal lifetime rewards from our current state x. σ ∗ ( x) = ( 1 − β 1 / γ) x. Let’s see if our numerical results lead to something similar. The Bellman Equation Cake Eating Problem Proﬁt Maximization Two-period Consumption Model Lagrangian Multiplier The system: U =u(c1)+ 1 1+r u(c2). Note: The functional equation for the value function is called a Bellman equation (it’s Bell-man’s Principle of Optimality that is used to solve these problems recursively) Note: Richard Bellman was an American mathematician in the 20th century who invented dynamic programming In in nite horizon problem, time is irrelevant The intuition here is essentially the same it was for the McCall model. If you want to know exactly how the derivative  U'(c^*)  is v^*(x_t) = \left( 1-\beta^{1/\gamma} \right)^{-\gamma}u(x_t) \tag{6} = \beta v^{\prime}(x - c) \tag{12} Would love to hear everybody's thoughts. With all the elements together, the Bellman Equation is V (w) = max 0 c w c1 1 + X i=L;H ... Stochastic Discrete Cake-Eating Problem Eating Waiting Having a cake 25/25. In the case of a ﬁnite horizonT, the “Bellman equation” of the problem consists of an inductive deﬁnition of the current value functions, given byv(y,0)≡0, and, forn ≥1, v(y,n) = max. I'm not famliar with the cake eating problem, so I had to do a bit of googling to get the background needed to understand the question. Hence, v(x) equals the right hand side of :eq:bellman-cep, as claimed. 5 Cake-eating example To introduce dynamics to the problem, we now consider the problem of how quickly one should eat a cake of given size. and Y1 =c1 + A1, and Y2 +(1 r) 1 =c2. Current rewards from choice c are just u(c). The aluev function V(a;b;W) gives the utility The overall cake-eating maximization problem can be written as. In fact, if we move away from CRRA utility, usually there is no analytical Now let's recall our intuition on the impact of parameters. value function will satisfy a version of the Bellman equation. To understand this condition, suppose that you have a proposed (candidate) solution for this problem given by {c∗ t} This is necessary condition for the optimal path. respect to  c  and setting it to zero, we get,$$ The reasoning given above suggests that the discount factor \beta and the curvature parameter \gamma will play a key role in determining the rate of consumption. Initial size of the cake is W0 = φ and WT = 0. The Euler equation for the present problem can be stated as, . on x, we get. $U'(c^*) = 0$. In other words, we conjecture that there exists a positive $\theta$ such that setting $c_t^*=\theta x_t$ for all $t$ produces an optimal path. Taking the derivative on the right hand side of the Bellman equation with the current time when $x$ units of cake are left.