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Non-linearities, catastrophic risk and thresholds in resource economics Eric Nævdal

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1 Non-linearities, catastrophic risk and thresholds in resource economics Eric Nævdal

2 Purpose of class Teach some advanced methods in optimal control theory Familiarize students with some applications of these methods to natural resource management. Enable students to solve simple problems numerically. A very applied course. No theorems; just methods!

3 Prerequisites A decent understanding of ordinary differential equations. A decent understanding of deterministic optimal control theory. E.g. Max c 0 ∫ T U(c,x)e -rt dt subject to x(0) = x0 and dx/dt=f(x,c). Here 0 < T ≤ ∞.

4 Cook book solution in easy steps 1)Define the Hamiltonian: H(c,x) = U(c,x) + λf(c,x) 2) Find optimality conditions –c = argmax H → c = c(x,λ) –dλ/dt = rλ – ∂H/∂x =rλ – ∂U/∂x – λ∂f/∂x –Insert c from into dx/dt and = dλ/dt 3) You now have two differential equations.

5 What to do with the differential equations Compute steady states x* and λ*. f(c(x,λ),x) = 0 and rλ – U’ x (c(x*,λ*),x*)– λf’ x (c(x*,λ*),x*) = 0 Draw a phase diagram. x λ dx/dt = 0 dλ/dt = 0 x* λ*λ* Paths converging to steady states candidates for optimal solution

6 Optimal Control cont’d The co-state λ(t) has an interesting economic interpretation. If somebody at time t gave you a present of 1 unit of x so that x(t) jumps to x(t) +1, then λ(t) is (roughly) the value of that present at time t. Entirely analogous to shadow prices in static theory. If T < ∞, then we have the transversality condition λ(t) = 0 if x(T) is free.

7 Transversality conditions for infinite Horizon problems The economic literature is full of flawed transversality conditions. The reason is that it is hard to find general conditions without using stuff like lim sup. For the purposes of this class we are satisfied if we can find at least one path where both the (current value) shadow price and the state variable converges to finite numbers. If T = ∞, then we really have a hard time with pinning down good transversality conditions. See Seierstad and Sydsæter (1987) for details.

8 Alternative approach that will (once) be used later Take the c(x,λ) function and differentiate it. Gives dc/dt = c’ x dx/dt + c’ λ dλ/dt = = c’ x dx/dt + c’ λ (rλ – ∂U/∂x – λ∂f/∂x) The solve c(x,λ) with respect to λ. Gives λ(c,x). Use dc/dt and dx/dt after inserting λ(c,x) as alternative differential equations

9 Example A firm has benefits from pollution given by: (u 0 – u) 2 The pollution accumulates in nature according to dx/dt = u – δx Damages from accumulated pollution given by –x 2. The problem: max u ∫ ∞ (-x 2 - (u 0 – u) 2 )e -rt dt subject to: dx/dt = u – δx, x(0) given Solution Define the Hamiltonian: H= -x 2 - (u 0 – u) 2 + λ(u – δx) 1.The value of u that maximizes the Hamiltonian is given by u = u 0 + ½λ for u 0 + ½λ > 0. Else u = 0. 2.dλ/dt = rλ + 2x + δλ 3.dx/dt = u 0 + ½λ – δx

10 Computing Steady States gives x* = u 0 (r+δ)/(1+δ(r+δ)) λ* = –2u 0 /(1+δ(r+δ)) From these expressions we see (for example) dx*/dr > 0 dx*/dδ < 0

11 Phase diagram

12 Phase diagram with paths

13 Phase diagram with paths and optimal paths for infinite horizon

14 The same solution seen as a function of time

15 Phase diagrams with finite vs inifinite time horizon

16 Optimal Solution for various time horizons

17 Crucial insight If T is chosen sufficiently large, there will be some value t* such that the optimal solution for the infinite horizon problem and the optimal solution for the finite horizon problem will be numerically indistinguishable over the interval [0,t*]. This allows us to solve infinite horizon problems on the computer

18 Getting a deeper understanding of Optimal Control Some mathematical background: Boundary value problems. A general class of differential equations. Consider the problem dL/dt =λL, L(0) = L 0. You should all know that the solution is: L 0 e λt. But what about this problem?: dL/dt =λL, L(T) = L T

19 Solving a boundary value problem dL/dt =λL implies that L(t) = Ce λt for some constant C. This constant is found by using the boundary value condition: Ce λT = L T Gives that C = L T e -λT. Therefore L(t) = L T e λ(t-T). Important. We can not independtly specify both L(0) and L(T). There is only one constant!

20 More boundary value problems dL/dt =λL, L(0) = 1. dN/dt = N – L, N(1) = 1. Solution: L(t) = Ce λt and N(t) = (λ – 1) -1 (e t – e λt )C +e t K. We have two constants C and K. Determined by: Ce λ0 = 1 and (λ – 1) -1 (e 1 – e λ1 )C +e 1 K = 1 C = 1 and K = (e(λ – 1)) -1 (1 + e – e λ – λ) Solution is L(t) = e λt and N(t) = (λ – 1) -1 ((e t – e λt ) +e t-1 (1 + e – e λ – λ)

21 Why boundary value problems? The solution to an optimal control problem may be written as a boundary value problem. Best seen in finite time problems: max 0 ∫ T U(c,x)e -rt dt subject to x(0) = x 0 and dx/dt=f(x,c). Here 0 < T< ∞. x(T) free. The maxmimum principle we know, but look at the transversality condition. λ(T)=0.

22 An even simpler example Define the Hamiltonian: H= -ax - ½(u 0 – u) 2 + λ(u – δx) 1.The value of u that maximizes the Hamiltonian is given by u = u 0 + ½λ for u 0 + ½λ > 0. Else u = 0. 2.dλ/dt = rλ + a + δλ 3.dx/dt = u 0 + ½λ – δx

23 A Philosophical digression The difference between human ecology (AKA economics) and ecology. An ecosystem and its inhabitants are unemcumbered by precognition. Humans are not. In order to understand an ecosystem we need differential equations and initial values. In order to understand human behaviour we need transversality conditions. Humans operate by backwards induction

24 Numerical methods Standard Optimal Control Problems in the literature do the following: –Find explicit solutions. Works for very few problems. –Phase diagram. Only works for problems with one state variable. –Steady state analysis. May be hard to do for some problems. Some times steay states are uninteresting Alternative: Numerical analysis

25 Numerical Methods in finite time - Shooting The basic problem; The Maximum Principle gives us a set of differential equations. An optimal solution must start with the known and correct initial value of x(0). It must also start with an unknown correct value of λ(0) such that λ(T) = 0. Alternatively if x(T) is given, λ(0) must start from a value so that those constraint holds. The fundamental problem: Solve a rather complicated equation to find λ(0).

26 A solution in two steps First write computer code that solves the differential equations for arbitrary values of x(0) and λ(0). (As if we are solving an initial value problem.) Then write a routine that finds the value of λ(0) that sets λ(T) = 0, (or x(T) to the required value). Luckily, there are ways of doing this without writing much code. –Using solver functions in Excel –BVP4C routine in Matlab Both methods work well, but may have to be tweaked.

27 Step 1. The 4 th order Runge – Kutta Method General formulation. For an OC problem the vector y = [x, λ]. Let dy/dt = f(t, y). Let h be a small number. Then y(t) is usually well approximated by the following sequence: y(t+h) = y(t) +(h/6)×(k 1 + 2k 2 + 2k 3 + k 4 ) k 1 = f(t, y(t)), k 2 =f(t + h/2, y(t) +hk 1 /2) k 3 =f(t + h/2, y(t) +hk 2 /2), k4= f(t + h, y(t) + hk 3 )

28 Starting Example Let dy/dt =y (1- y), y(0) = ½. The solution to this differential equation : y(t) = Exp(t)/(1 + Exp(t)) No difference whatsoever!

29 Setting up the differential equations for a control problem We will return to our previous example. The differential equations are: dλ/dt = rλ + 2x + δλ dx/dt = u 0 + ½λ – δx 1.Start Excel. 2.Click on Sheet Tab to View Code 3.Open a module (Not class module! 4.Write code

30 May look like this:

31 Implemenent Runge Kutta in Spreadsheet. Time to load spread sheet Small Optimal Control Example

32 Step 2 – Finding λ(0) May in principle be done by programming some suitable search algorithm. We are going to let Excel take care of it. Two ways of doing this –Goal Seek Function. Slow, robust, only handles problems with one state variable –The Solver. Fast, will stop if the algorithm encounters errors, Handles a large number state variables. Let’s do it.

33 Alternative Use Matlab BVC4P function. Not really better or more robust. Good for when a large number of problems must be solved. Also, if the initial guess is far off, all solvers crash. BVC4P is good to generate a sequence of solutions that converges to the problem that one actually wants to solve.

34 Using Numerical Analysis Optimal Vaccination of non-persistent epidemics Very policy relevant Economists have made very limited contributions Shows the power of numerical analysis when out standard tool kit breaks down Solutions programmed in Matlab

35 Typical trajectory after outbreak – No vaccination McKendrick-Kermac model Suceptibles x, infected y and recovered/dead z dx/dt = –βxy, x(0) = N – ε dy/dt = βxy – γy y(0) = ε = ininital infected population dz/dt = γy, z(0) = 0 Essensial paramter γ/β.

36 Trajectory without vaccination Note the effect of reducing suceptibles before outbreak. - More suceptibles at the end of an epidemic episode. - Fewer infected

37 Model with vaccination Individuals may be vaccinated u. dx/dt = –βxy – u, x(0) = N – ε dy/dt = βxy – γy y(0) = ε = ininital infected dz/dt = γy + u, z(0) = 0 Objective function: ∫(-wy - ½cu 2 )e -rt dt K is the cost of disease. ½cu 2 is the cost of vaccination Must be solved numerically. Standard tools of optimal control useless.

38 Optimal vaccination - Low cost of disease (w)

39 Optimal vaccination - High cost of disease (w)

40 The value of reducing the stock of suceptibles The shadow price of x multiplied by -1 is the value of vaccinating one ”population unit.” The graph show the shadow price on x for different values of w What does it mean that some of the curves are not monotonic? Succeptibles at outbreak

41 Explaining “increasing returns” ”Brush fire” effekt. At high levels of x, the disease spreads so rapidly that the return on vaccination prior to outbreak is reduced. Flow with the punch (relatively speaking) becomes optimal strategy. High disease costs reduces brush fire effect

42 New Section Multiple Equilibria Many systems exhibit non-linear dynamics. May or may represent a technical challenge Here we look at convexo-concave differential equations. Important to note that systems that naturally exhibit multiple equilibria may not do so when controlled optimally

43 Example – Eutrophication Let x be the nutrient (phosphoros and nitrogen) loading in a lake. Let u be the deposition of nutrients. The ecologists claim that the dynamics of the lake may be reasonably modeled by: Analysis taken from W.A. Brock and D. Starrett and K-G Mäler, A. Xepapadeas, Zeeuw

44 Dynamics with low loading (u)

45 The effect of increased (constant) loading

46 The flip is irreversible even if u is set to zero!

47 Management For economic analysis we need some evaluation of consequences. Instantaneous benefits from nutrient use given by ½ln(u) Instantaneous damages from eutrophication given by –cx 2.

48 The optimal management problem

49 Optimality Conditions

50 Transforming into equations in (u,x) space

51 Draw differential equations We proceed to draw a phase diagram in (x,u) space.

52 Phase diagram – Low loading optimal

53 Phase diagram – High loading optimal

54 Phase diagram –Three steady states

55 Words of warning The differential equation for dλ/dt was wrong on the slide (but correct in the calculations). The right one is: When doing numerical analysis, you should still use your brain! I didn’t and it took me ages to figure out why numeric methods didn’t work on the eutrophication example. Can you see why it doesn’t work?

56 Modify the Eutrophication model

57 High cost of eutrophication – Only oligotrophic Equilibrium

58 Intermediate cost – The Skiba story

59 High cost of eutrophication A B

60 Oh dear! For low values of x we have two solution candidates! Which one to choose. I havn’t proved this formally (although it may be worth a try) that as long as we choose a finite horizon, then the we stay close to A manifold in order to satisfy transversality conditions. I therefore believe this is the relevant manifold

61 Genetic Management Many people are concerned with biodiversity. I am, for various reasons, not. But I am concerned about outcomes. Harvesting living resources leads to evolutionary genetic selection. How to regulate this? Also, an example of non-standard analysis.


63 Objective Construct a bioeconomic model where we analyze the effect of selective harvesting on genetic frequency for one specific gene in terms of the socially optimal long-term management of the resource. This object is determined solely through the profits generated by harvesting

64 Genetic Dynamics Standard model of Mendelian genetics Two alleles, A and a of the same gene. The homozygotes AA and the heterozygote Aa are of phenotype G (for good) The homozygote aa are of phenotype B (for bad) The frequency of a is q. (q = 1 is bad. q = 0 is good.) Only individuals of type G are of commercial interest, and harvesting is totally selective

65 The model Determination of phenotype frequency is determined by gene frequency. Population dynamics. In this model B will lose in the absence of harvesting.

66 Comparing Phenotype Productivity Let h G =  x G E and h B = 0. One can calculate the difference in productivity between B and G. The difference is:

67 Genetic Dynamics Changes in x = x B + x G is determined by Because of evolutionary pressure imposed by selection, the Hardy-Weinberg law may not apply and q will be a non- constant function of time. q is determined by Where s is measure of the extent to which natural selection is acting to reduce the relative contribution of a given genotype to the next generation. Note that it is also the productivity difference!

68 Optimal Management Focus on how selection affects the dynamics of q. Assume a sole owner that wants to maximize the discounted profits from harvesting the resource. Harvesting is assumed costless and valued at an exogenous price Gives objective function

69 Optimization Problem s.t

70 Solution We had to invent some stability analysis. From the expression for, s = 0 in steady state. From the expression for : Inserting these expressions into yields a key expression  (q), which happens to be a 4th order polynomial.

71 Results No explicit solution, but the structure of the problem enables us to determine steady state values of q. Also possible to find the stability properties of the steady states. Boils down to the initial magnitude of q and the relationship between:

72 Results, key expressions the intrinsic productivity of phenotype B, measures how fast phenotype B regenerates gives the selection in absence of harvesting or the rate at which nature selects for phenotype G. Can also be interpreted as the internal rate of return on preserving phenotype G the discount rate measuring the opportunity cost of capital

73 Case 1 q* is a stable steady state.

74 Case 2 Optimal to let the gene A become extinct

75 Case 3a Optimal to let the gene a become extinct

76 Case 3b q1 is a Skibapoint, i.e., unstable steady state. q2* is a stable steady state

77 Case 4 Optimal to conserve A at low levels of q because the low productivity of phenotype B, but at some point the low internal rate of return makes the extinction of phenotype G optimal

78 The value function We have our original OC problem: max c 0 ∫ T U(c,x)e -rt dt subject to x(0) = x0 and dx/dt=f(x,c). The value that we get depends on x0. Thus we have J(0,x0,T) = max c 0 ∫ T U(c,x)e -rt dt. The function J(0,x0,T) is called the value function. It is what we get when our system is run as optimized.

79 The embedded problem Now consider the problem: max c t ∫ T U(c,y)e -rs ds subject to y(t) = x and dy/dt=f(y,c). What have we done? Renamed the state variable from x to y. A new arbitrary starting point t rather than zero. (t> 0) Have a new initial condition y(t) = x. We now have a new value function J(t,x) =J(0,x) e -rt.

80 What is J(t,x)? It is the same problem as the original, except that we have an arbitrary starting point. This function has some economically important properties. Shadow price: Interpretation: If somebody gives you a present of one unit of x at time t, what is the value of that present.

81 Another property The value of living. This has an important interpretation. At every point in time you eat the Hamiltonian. This has given rise to the interpretation of the Hamiltonian as the Net National Product. See recent book by Weitzman. HΔt = U(c,x) Δt + λf(c,x) Δt The value of Consumption today The marginal value of “capital” The amount of investment

82 Some useful extensions of deterministic control theory (1) Variable final time. Max c,T 0 ∫ T U(c,x)e -rt dt subject to x(0) = x0 and dx/dt=f(x,c). T is a choice variable. Extra condition required to fix T. It turns out that Necessary condition is then H(T,c(T), x(T)) = 0

83 Very simple example Max u,T ∫ T (–9 – ¼u 2 )dt subject to dx/dt=u and x(0) = 0 and x(T) = 16. Hamiltonian H=(–9 – ¼u 2 )+λu. Maximising H w.r.t. implies u = 2λ. dλ/dt = -dH/dx = 0 → λ = constant K. dx/dt=u = 2λ = 2K → x(t) = 2Kt + C. We now have two constants, K and C, that we need to determine. We cant use λ(T) = 0 as we have constrained x(T).

84 Therefore… x(t) = 2Kt + C. → x(0) = 0 → C = 0. x(T) =16 = 2KT → K = 8/T. We then have that λ =K = 8/T and u = 2×K = 16/T Now find optimal T. The Hamiltonian is now H*= -9 - ¼(16/T) 2 + 2×(8/T) 2 = 0 Solving this for T gives T=8/3

85 Some useful extensions of deterministic control theory (2) Scrap value problems - a complicated problem finite horizon problem. E.g. Max c,T 0 ∫ T U(c,x)e -rt dt +S(x(T)) e -rT. subject to x(0) = x0 and dx/dt=f(x,c). Conditions: The same conditions on c and λ. Different transversality conditions: λ(T) = S’(x(T)) e -rT. U(c,x)e -rt + λf(x,c) = r S(x(T)) e -rT.

86 A closer look at these conditions Why are these conditions not surprising? λ(T) = S’(x(T)). Says that the shadow price of x should be continuous. U(c,x) + λf(x,c) = r S(x(T)). Says that the change in utility from continuing one more unit of time should be the same as the loss from not cashing in on the scrap value. Remember that U(c,x)e -rT + λe -rT f(x,c) = J’(T,x).

87 Load Excel file

88 Deterministic Threshold Problems Irreversible case. A threshold is here defined to be some curve in state space such that crossing this curve leads to a discrete jump in state-variables. Preliminary observation; if the location of the threshold is known one can choose to cross the threshold or not cross the threshold. An irreversible threshold effect can be modeled as a scrap value problem with endogenous time horizon

89 Mathematical description From Seierstad and Sydsæter Theorem 3 and Note 2, pp 182. Let x be of dimension n Let the threshold be given by R(x) = 0 Let τ be the first point in time when the threshold is reached. We assume that is R(x(τ)) = 0. Let the threshold effect be x(τ + )-x(τ) = g(x(τ)) –x(τ + ) =x(τ - ) = lim t→τ x(t) from below –x(τ + ) = lim t→τ x(t) from above

90 Formulating the Optimization Problem(s) - 1 Going over the threshold Max c,τ 0 ∫ τ U(c,x)e -rt dt + S(x(τ)+ g(x(τ))) e -rτ subject to x(0) = x0, dx/dt=f(x,c) and R(x(τ)) = 0. Here we should think of S S(x(τ)+ g(x(τ))) as a value function. In fact S(x) is given by Max c,τ τ ∫ ∞ U(c,y)e -rt dt subject to y(0) = x and dy/dt=f(y,c).

91 Solve the problem recursively From Max c τ ∫ ∞ U(c,y)e -rt dt subject to y(0) = x and dy/dt=f(y,c) we get the shadow price λ(x|τ).

92 Optimality conditions (Present value) u maximizes the Hamiltonian dλ/dt = –∂H/∂x λ(τ) = λ(x(τ)|τ)(1+g’(x)) + γ∂R/∂x Note that all these may be vectors If τ lies in (0, ∞), then H + ∂(S(x(τ)+ g(x(τ))) e -rτ )/∂t = 0

93 Optimality conditions (Present value) u maximizes the Hamiltonian dλ/dt = –∂H/∂x λ(τ) = λ(x(τ)|τ)(1+g’(x)) + γ∂R/∂x Note that all these may be vectors If τ lies in (0, ∞), then H + ∂(S(x(τ)+ g(x(τ))) e -rτ )/∂t = H – r(S(x(τ)+ g(x(τ))) e -rτ ) = 0.

94 Formulating the Optimization Problem(s) - 2 Going over the threshold Max c,τ 0 ∫ τ U(c,x)e -rt dt + S(x(τ)))e -rτ subject to x(0) = x0, dx/dt=f(x,c) and R(x(τ)) = 0. Again we should think of S(x(τ)) as a value function. In fact S(x) is given by Max c,τ τ ∫ ∞ U(c,y)e -rt dt subject to y(0) = x, dy/dt=f(y,c) and R(x(t)) = 0.

95 Solve the problem recursively From Max c τ ∫ ∞ U(c,y)e -rt dt subject to y(0) = x, dy/dt=f(y,c), and R(x(t)) = 0. We get the shadow price λ*(x|τ). Note: Here I have excluded the possibility that R(x(t)) ≠ 0 for some t > τ. It is however perfectly possible that we may “turn away” from the threshold at some point in time. In particular if we are studying a finite horizon problem

96 Optimality conditions (Present value) u maximizes the Hamiltonian dλ/dt = –∂H/∂x λ(τ) = λ*(x(τ)|τ) + γ∂R/∂x Note that all these may be vectors If τ lies in (0, ∞), then H + ∂(S(x(τ))) e -rτ )/∂t = 0

97 A simple threshold model (Nævdal 2003) Let dx/dt = u – δx. Let the threshold be x’. Let B(u) be maximised at B(u*) for u* > 0. Let the let the utility function be A + B(u) if the threshold is not crossed and let the benefit function be B(u) if it is crossed.

98 Avoid Crossing the threshold? Let T e be the time at which the threshold is crossed. If we do not cross the threshold, then after getting to x(T e ) = x’ we must freeze u. Thereafter there is no need to reduce u so u = δx’ for all t > T e. Then S e (x(T e ))e -rT e = e -rT e T e ∫ ∞ (A+B(u*))e -rt dt

99 Crossing the threshold? Let T a be the time at which the threshold is crossed. If we cross the threshold, then we accept the damage. Thereafter there is no need to reduce u so u = u* for all t > T a. Then S a (x(T a ))e -rT a = e -rT a T a ∫ ∞ B(u*)e -rt dt

100 Optimality conditions prior to hitting the threshold Hamiltonian is the same for both problems H = (A + B(u))e -rt + p(u –δu) Gives that B’(u i )e -rt + p i = 0 dp i /dt = δp i → p i (t) = C i e -δt x(T i ) = x’ i = a,e The shadow prices after hitting or crossing the threshold are both zero. u is a decreasing function of time. Pollute early and then reduce

101 Note Both are problems with variable final time and the requirement that xi(T) = x’. We need condition to find Ti.

102 Going over the threshold Condition for optimal T a. We can prove that u a is discontinuous at T a.

103 Optimal paths of u a and x a

104 Staying on the edge Condition for optimal T e. This condition can be used to prove that u e (t) is continuous at T e.

105 Optimal paths of u e and x e

106 Important to note There are kinks (ue) or jumps (ua) in the optimal paths of the control. The previous literature on deterministic thresholds had overlooked: 1.That a threshold could either be crossed or observed 2.That optimal controls may be discontinuous at the time the threshold is crossed. (In some models this can happen when the threshold is reached even if it is not crossed.)

107 Comparing Scenarios I said no proofs, but this one is rather instructive and a good example of how simple proofs can give interesting results

108 Incorporating uncertainty into optimal control Most processes are subject to uncertainty One class is Brownian motion. I will not talk about that at all. Catastrophic events. –Tsunamis –Floods –Car breakdown

109 Categories of catastrophic uncertainty Endogenous vs. exogenous risk –Some processes we can (partially) control. This is endogenous risk. –Some processes are truly beyond our abilities. Exogenous risk. State-space distributed risk vs time-distributed risk. –State space distributed risk requires movement through state space. (thresholds) –Time distributed risk depends on location in state- space. For a fixed location in state-space, uncertainty is invariant with respect to time. This will perhaps become clearer. (I hardly understand it myself)

110 Preliminaries – Poisson processes in continuous time We are here concerned with events that occur at random points in time. In order to deal with these problems they must have some kind of distribution that we actually know. We shall refer to the point in time when the event occurs as τ (tau) and it is distributed over a subset of the positive real numbers [0, β) where 0 < β ≤ ∞. The pdf is g(t) and the cdf is G(t) = 0 ∫ t g(s)ds.

111 Poisson processes and conditional updating. Our process starts t = 0. We make it to t* >0. What is the distribution of τ conditional on us having made it to t*? Answer: We need optimality conditions that reflect updating of the distribution as long as τ does not happen.

112 The hazard rate - A very useful concept We ask the question; What if we have made it to time t? What is the probability that τ occurs in the time interval (t, t + Δt)? If dt is small, then that probability is roughly Pr(τ (t, t + Δt))=g(t|τ>t)×dt. g(t|τ>1) is of course:

113 Hazard rate μ(t)– Formal definition

114 Example - the exponential distribution Let μ 0 be a positive constant number. g(t) = μ 0 e -μ 0 t. → G(t) = 1 - e -μ 0 t. Then μ(t) = μ 0. We have a constant hazard rate. This is only true for the exponential distribution. Interesting fact: E(τ - t) = t ∫ ∞ sµe -µs ds = 1/µ. The expected conditional waiting time is constant We can in principle calculate hazard functions for any g(t). They are frequently messy.

115 We have the hazard rate – What is the pdf? A very useful property. Let μ(t) be any function that is: –continuous and positive for all t –has the property that 0 ∫ β μ(t)dt=∞. Then we can construct a cdf by solving the following differential equation. μ(t) = y’(t)/(1 – y(t)), y(β) = 1

116 Finding a pdf μ(t) = y’(t)/(1 – y(t)), y(β) = 1 implies that y(t) = 1 – C×Exp(-∫ t μ(s)ds) Here C is a constant determined by the boundary condition. For many choices of hazard rate this constant will be equal to 1. Counter example: μ(t) = Exp(t) Having found y(t), the cdf, we can find the pdf y’(t) = C×μ(t)×Exp(-∫ t μ(s)ds)

117 Example – linear hazard rate Assume that μ(t) = μ 0 t defined for t € [0, ∞). Then Exp(-∫ t μ(s)ds) = ½μ 0 t 2. Therefore y(t) = 1 – C× Exp(-½μ 0 t 2 ). y(∞) = 1 implies that C = 1, so the cdf is 1 – Exp(-½μ 0 t 2 ) and the pdf is y’(t) = μ 0 t×Exp(-½μ 0 t 2 ).

118 Hazard rate continued The previous results also hold if the hazard rate can be written μ(t) = φ(x(t)) (or φ(t,x(t)) for that matter) where x(t) is any function of t. This holds as long as: –φ(x(t)) only takes positive values –The integral ∫φ(x(t))dt does not converge Thus we can work with very general density pdfs over time over time: φ(x(t))Exp(- ∫ t φ(x(s))ds)

119 Exogenous vs endogenous uncertainty Slightly confusing literature. Here the difference is as follows. If φ(x(t))is the hazard rate and x(t) is determined by a controllable differential equation, then the stochastic process is endogenous. If not, the process is exogenous his is not clear cut. Hurricanes may be (in part) endogenous to US policymakers but exogenous New Orleans

120 Controlling exogenous catastrophic uncertainty Basic problem: Nature or (somebody we can’t affect) triggers a catastrophic event. We can not control the probability of the event occurring, but we can control: –preparedness (what to do before the event) –consequences (how to act after the event) No strict boundary between preparedness and consequence management.

121 The mathematical formulation In infinite time: Max E{∫ ∞ (U(x,u)e -rt )dt + h(x(τ - )) subject to dx/dt=f(x,u) for all t ≠ τ x(τ + )-x(τ - ) = g(x(τ - )) τ distributed μe -μt over [0, ∞) g(x(τ - )) is the physical description of the event h(x(τ - )) is the instantaneous value of the shock. x(τ - ) = lim t→τ x(t) from below x(τ + ) = lim t→τ x(t) from above

122 Extreme example A tsunami in a community –We have utility before the disaster. Depends on consumption and stock of capital. –The instantaneous cost of the disaster, depends on stochasticity and the stock of preparedness capital. –Utility after the disaster, depends on consumption and the stock of capital that survived the disaster

123 Solving the problem a recursive algorithm First find the optimal policy after the disaster. Find optimal policy before the disaster. Sounds pretty simple…

124 Post-disaster control There is no stochasticity. Simply solve the following problem: J(t,x|τ = t)e rt = max u e rt t ∫ ∞ U(u,y)e -rs ds subject to y(0) = x and dy/dt=f(y,u). This looks like our old friend the embedded problem. Here t and x are arbitrary. The indicate the possible state(s) of the world after a disaster. The notation J(t,x|τ = t) indicates that J is evaluated conditional on τ happening at time t.

125 Pre-disaster control From the post disaster control we will need J’ x (t,x|τ = t)e rt = λ(x|τ). This is the shadow price of x at the instant after the catastrophic event occurs. In this problem, that is really all we need from the post-disaster program.

126 Pre-disaster program- Necessary conditions Hazard rate depends on t. Define the Hamiltonian: H = U(u,x) + λf(u,x) + µ(t)(J(x+g(x)|τ) –J(x) + h(x)) Utility today Marginal value of X × increase in x Hazard rate Change in utility if Catastrophic event occurs

127 Optimality Conditions Apply the maximum principle to the Hamiltonian. u =argmax H dλ/dt = rλ - ∂H/∂x dx/dt=f(x,u) Transversality condition as in deterministic models.

128 The differential equation for λ dλ/dt = rλ - ∂H/∂x = rλ - ∂U/∂x - λ∂f/∂x - µ((∂J(x+g(x)|τ)/∂x –∂J(x)/∂x)+h’(x)) But ∂J(x+g(x)|τ)/∂x = λ(x|τ)(1 + g’(x)) and ∂J(x)/∂x = λ, so: dλ/dt = rλ - ∂U/∂x - λ∂f/∂x + µ(λ - λ(x|τ)(I n + g’(x)) – h’(x)) Note: I n is the identiy matrix where n is the number of elements in x(t)

129 Example Let U(x,u) = -ax - ½(u 0 – u) 2 and f(x,u) + u – δx Let τ be exponentially distributed with hazard rate µ. Let x(τ + )-x(τ - ) = βx(τ - ). h(x) = 0. Could be a model of pollution where there is a possibility that a shock increases the stock of pollutants.

130 Optimality conditions after event H= -ax - ½(u 0 – u) 2 + λ(u – δx) 1.The value of u that maximizes the Hamiltonian is given by u = u 0 + λ for u 0 + λ > 0. Else u = 0. 2.dλ/dt = rλ + a + δλ 3.dx/dt = u 0 + λ – δx dλ/dt = rλ + a + δλ → λ(t|τ) = -a/(r + δ) for all t. We make note of λ(t|τ) and proceed to the pre-event control

131 Pre event conditions 1.The value of u that maximizes the Hamiltonian is given by u = u 0 + ½λ for u 0 + ½λ > 0. Else u = 0. Same as before! 2.dx/dt = u 0 + ½λ – δx. Same as before! 3.dλ/dt = rλ + a + δλ +µ(λ – (–a(1+β)/(r + δ))) Here is the difference in the conditions. Obviously this implies that u and x will be affected by the risk

132 We can solve for λ Solution is: Here K is a constant that must be determine by transversality conditions. We just not that K ≠ 0 implies no convergence to steady state so we just set K = 0. Note that if β = 0, then pre-event solution is same as post-event

133 The we have that…. The optimal value of u(t) is found by inserting λ(t) into u 0 + ½λ. The optimal value of x(t) is fund by integrating dx/dt = u 0 + ½λ – δx I will not do this as the resulting expression is a bugger. Numeric solutions are straightforward to find

134 Entering Equations in Excel

135 Result

136 The effect of uncertainty – Let us increase µ

137 Weird hazard rates The setup is good enough to include time dependent hazard rates. That is, the hazard rate depends on time. Let us consider a hazard rate a×(Sin(t) + 1). This implies a cdf given by 1 – exp(-1- t+Cos(t)). Cdf looks like this:

138 Pre-disaster program- Necessary conditions Hazard rate depends on t. Define the Hamiltonian: H = U(u,x) + λf(u,x) + µ(t)(J(x+g(x)|τ) –J(x) + h(x)) Utility today Marginal value of X × increase in x Hazard rate Change in utility if Catastrophic event occurs

139 Optimality Conditions Apply the maximum principle to the Hamiltonian. u =argmax H dλ/dt = rλ - ∂H/∂x dx/dt=f(x,u) Transversality condition as in deterministic models. NO CHANGE!

140 The differential equation for λ dλ/dt = rλ - ∂H/∂x = rλ - ∂U/∂x - λ∂f/∂x - µ(t)((∂J(x+g(x)|τ)/∂x –∂J(x)/∂x)+h’(x)) But ∂J(x+g(x)|τ)/∂x = λ(x|τ)(1 + g’(x)) and ∂J(x)/∂x = λ, so: dλ/dt = rλ - ∂U/∂x - λ∂f/∂x + µ(t)(λ - λ(x|τ)(I n + g’(x)) – h’(x)) Note: I n is the identiy matrix where n is the number of elements in x(t)

141 Modifying the code (Hazard rate is actually Sin(2*t – 1)+1) Old code Tiny change

142 Results

143 Endogenous risk – Time distributed Recall that we know face a controllable hazard rate. The hazard rate depends on a state variable

144 Endogenous time distributed risk Here the hazard rate will depend on the state variable. µ = µ(x). Example µ(x) = µ 0 x Remember that x is a function of t because it is determined by an differential equation This is the pdf

145 Endogenous risk continued Say that x(t) is driven by dx/dt = - ax + b, x(0) = X. Then x(t) = e -at (X – b/a) + b/a. The hazard rate is then transformed from a function of x to a function of time. µ = µ0×(e -at (X – b/a) + b/a) Illustration for different starting points of x(0)

146 Controlling Endogenous Uncertainty The general problem

147 Recursivity - Working yourself backwards As before. First solve the post disaster problem. Get the shadow prices

148 Post-disaster control (Copy of previous slide) There is no stochasticity. Simply solve the following problem: J(t,x|τ = t)e rt = max u e rt t ∫ ∞ U(u,y)e -rs ds subject to y(0) = x and dy/dt=f(y,u). This looks like our old friend the embedded problem. Here t and x are arbitrary. The indicate the possible state(s) of the world after a disaster. The notation J(t,x|τ = t) indicates that J is evaluated conditional on τ happening at time t.

149 Pre Disaster Solution Define the risk Augmented Hamiltonian Then apply the maximum principle Note: λ is here the hazard rate and µ is the shadow price

150 Applying the Maximum Principle Find the control that maximizes the Hamiltonian and use d Compared to the case with exogenous risk we have an additional term. If risk is exogenous, we have that λ’(x) = 0

151 A closer look at dµ/dt As in the exogenous risk case, we have that We have J(x+g(x)|τ) from the post-disaster problem. But what about J(x)?

152 I turns out that: J(x) = z is given by: So we can rewrite dµ/dt as: Additional transversality condition z(T) = 0 (Why?)

153 A pedagogic problem After inserting for the optimal u, we have three differential equations in x, µ and z. Tough to do. In fact, I know of no application where one can find a closed form solution or even a reasonably manageable steady state. Numerical methods are all that is left. Oh, and general statements.

154 The Problem we are going to solve

155 The post event solution Really simple. No more damage from x so post event shadow price is zero. This implies that u = u0 for all t > τ. J(x|t) = -K/r

156 The pre-event problem Hamiltonian given by H= α – ½(u 0 -u) 2 + λ(u – δx) + µ×x×(– K/r - z) Apply maximum principle to get: u=argmax H = u 0 + λ dλ/dt = (r+δ)λ + µxλ – µ(–K/r – z) dz = rz + ½(u 0 -u) 2 – µ(-K/r – z) Transversality conditions λ(T) = z(T) = 0

157 The code

158 The results

159 Play around to deal with instability Load Excel file

160 Endogenous, state-space distributed risk – AKA stochastic threshold We move a state variable through time. If we move in the “right” direction we may trigger a threshold effect. Possible threshold locations t x(t) Risky time segment

161 Preliminary math. We all know the chain rule. We should also know that

162 Let f(x) be a distribution If the event x = x’ is distributed f(x) over [a, b] and… x(t) is such that x’(t)  0 for all t and Range[x(t)] = [a, b] Then the cdf for the event x = x’ and the event x(τ) = x are interchangeable. The pdf for the event is distributed over time is f(x(t))x’(t). Regardless of whether Range[x(t)] = [a, b], the hazard rate is

163 The hazard rate continued Note how the hazard rate depends on the rate of increase in x. dx/dt = 0 implies that the hazard rate is zero.

164 Example. The hazard rate in state- space is linear The hazard rate is linear in state space. This is the pdf: Assume that x’(t) = -ax(t) + b

165 However, the hazard rate in time behaves differently x(t) Hazard rate as a funtion of time t

166 Properties of the hazard rate If we move away from the threshold, the hazard rate is zero If we let x take values that have happened before, then the hazard rate is also zero.

167 Optimal control Use exactly the same conditions as with endogenous time distributed problems, but with modified hazard rate. Technical note: Previously the hazard rate depended only on state variables. Now it will in general also depend on control variables. (Why?) This does not matter as long as we can restrict ourselves to look at continuous controls,

168 Example Same as before. However, this time we will consider a threshold problem.

169 The Problem we are going to solve

170 The post event solution Same as before. No more damage from x so post event shadow price is zero. This implies that u = u0 for all t > τ. J(x|t) = -K/r

171 The pre-event problem Hamiltonian given by H= α – ½(u 0 -u) 2 + λ(u – δx) + µ× (u-δx)(– K/r - z) Apply maximum principle to get: u=argmax H = u 0 + λ - µ(K/r + z) dλ/dt = (r+δ)λ + µ(u – δx)λ +µδ(– K/r - z) dz = rz + ½(u 0 -u) 2 – µ(u – δx)(-K/r – z) Note the difference Hazard rate now directly affects control! Derivative of hazard rate Optimal u from maximization

172 Analytics One can actually compute steady states here. In steady state x’(t) is zero so lots of stuff disappears. To whit: u= u 0 + λ - µ(K/r + z) dλ/dt = (r+δ)λ - µδ( K/r + z)=0 dz/dt = rz + ½(u 0 -u) 2 = 0 dx/dt = u – δx = 0

173 Steady state solution Note the following: K = 0 implies x and u equal to unregulated levels Otherwise K > 0 implies small x and u. K large enough implies x and u less than zero in steady state.

174 A caveat This solution assumed that x(0) ≤ x ss. Otherwise we get negative hazard rate. What happens if we have x(0) > x ss Answer: Freeze x at x(0) indefinitely.

175 Application – Disintegration of the Western Antarctic Ice Sheet Oppenheimer (1998) estimates that a WAIS disintegration could increase sea levels by as much as 4-6 meters. Oppenheimer evaluates that there is a threshold temperature increase above pre-industrial levels where this event could occur. This threshold lies in the range 2.5° to 8° degrees Celsius. What does this imply for human welfare?

176 Consequences of WAIS disintegration Considerable impact on real state markets. For some areas such as Bangladesh, the picture is even grimmer

177 There are several GHGs that contribute to global warming. Methane (1/5) and Carbon (3/5) being the most important. Because of different retention periods in the atmosphere they must be treated separately. Let c be carbon and m be methane. Then there is a threshold defined by: ac(t) + bm(t) = A + . The functions c and m are determined by and A +  is the temperature threshold,  is a random variable. Emissions are given by ui Some relevant geoscience

178 Possible paths of the system The Main Technical Problem. How to translate the event that the threshold is crossed from a probability distribution in State-space to a probability distribution over time? Answer: Change of variable techniques from integration theory.

179 We can now state society’s optimization problem as follows: The E conomy The instantanous cost of emission reduction The cost of crossing Wais disintegration:  = G if threshold is crossed,  = 0 otherwise Steady states are calculated in the paper

180 Optimal Paths

181 Optimal Stopping in Exogenous Risk Problems

182 Some numerical comparative dynamics

183 The Economics of the Thermohaline Circulation – A Problem with two Thresholds

184 Some Unpleasant Facts and Possibilities Historical Geophysical data suggests that the Thermohaline Circulation has been disrupted in the past. Scientific results suggests that this may be triggered by Global Warming. The consequences, although uncertain, will be a bugger. They include but are not limited to: –Regional disruptions in weather patterns –Permanent regional climate change –May trigger other global catastrophic events.

185 Some Science Global Warming will affect ocean temperatures and salinity which are the main drivers of the thermohaline circulation. THC disruption depends on temperature levels and the rate of change in temperature. If temperature levels or rates of temperature change exceed certain thresholds a shutdown may occur. The location of these thresholds are unknown.

186 A Simple Model of Global Warming c – Atmospheric carbon concentration F – Increase in global temperature average u – CO 2 emissions If F or increase above certain thresholds, the THC collapses.

187 Illustration of the Risk Structure

188 The Economic Problem(s) What is the optimal CO2 emission pattern in the presence of this risk? How to optimize such a system? Piecewise Deterministic Optimal Control! Some technical issues must be resolved before optimization can be done. –The events distributed in state space must be transformed so that they are distributed over time. –A distribution for the minimum of these events must be derived.

189 The Optimization Problem Climate cost. 0 if THC is OK, G if THC collapses Cost of reducing emissions below unregulated emission level u 0.

190 The Optimization Problem contnd. Subject to the differential equations and two concurrent stochastic processes. As long as both F and dF/dt are increasing the hazard rate is the sum of two hazard rates. When only F is increasing, only this hazard rate is active

191 The Hamiltonian for the present problem First terms are the standard Hamiltonian This term is the expected loss/gain from disaster in the interval (t, t + dt)

192 The Optimal Path

193 Optimal Time path for Temperature

194 Optimal stopping We have a risky process and we wonder when to stop. In deterministic problems, we use H = 0. Here too. In Finite time: Max E{∫ T (U(x,u)e -rt )dt + h(x(τ - ))e -rτ + G(x(T))e -rT subject to dx/dt=f(x,u) for all t ≠ τ x(τ + )-x(τ - ) = g(x(τ-)) τ distributed μe -μt over [0, ∞)

195 A quick word about the value function If we have a scrap value, then we have that J(T,x(T)) = Scrapvalue. Intuitively obvious, but needs to be pointed out

196 Optimal Stopping (Eksogenous Risk) Apply the maximum principle to the Hamiltonian: H = U(u,x) + λf(u,x) + µ(t)(h(x)e -rt +J(x+g(x)|τ) – J(x)) u =argmax H dλ/dt = rλ - ∂H/∂x λ(T) = ∂G(x(T))e -rT /∂x dx/dt=f(x,u)

197 Optimality Conditions The Stopping Condition U(u,x) + λf(u,x) +∂ G(x(τ - ))e -rτ /∂t – µ(t)(h(x)e -rt +J(x+g(x)|τ) –G(x(T))e -rT )

198 Nice ting to know about hazard rates If we have two stochastic events τ 1 and τ 2, with hazard rates λ 1 and λ 2, then…. The event τ = (τ 1, τ 2 ) has a hazard rate λ = λ 1 + λ 2. Practical as it enables us to treat two stochastic processes as one!

199 Optimal harvesting of illegal drugs A drug producer grows marihuana. At optimally chosen time T he sells the stuff at a price q, unless… –He gets raided by the cops, in which case he receives a punishment –c per weight unit of weed –The local college boys finds the farm and nicks a fraction 1 – b of the drugs. The farmer then sells the remaining.

200 Mathematical formulation Equations of motion Maximization problem max E((λ 1 (-cx 1 e -rτ )+λ 2 bqx 1 e -rτ )/λ) +qx 1 (T)e -rT

201 The Jump x 2 is a just a way of saying that if τ occurs, the dope stops growing

202 The Maximum Principle We only need p(T) = qe -rT.

203 The Stopping condition H Which can be solved and yield:

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