Optimal Sojourn Time Control within an Interval 1
|
|
- Tracey Greene
- 5 years ago
- Views:
Transcription
1 Optimal Sojourn Time Control within an Interval Jianghai Hu and Shankar Sastry Department of Electrical Engineering and Computer Sciences University of California at Berkeley Berkeley, CA Abstract In this paper we study the problem of find optimal feedback control that can keep an agent within an interval for at least a certain amount of epected time with the least energy. The dynamics of the agent are subject to the perturbation of random noises, and thus are given by a stochastic differential equation. By formulating the problem as an optimal control problem, we propose a solution by using the Maimum Principle and finding a first integral to the resulting differential equations. Some numerical simulations are presented to illustrate the results. Introduction In many practical applications where safety is the principal concern, for eample, in traffic systems such as air traffic management system [] and automated highway system [5], the evolution of the state of the agents to be controlled can often be modeled as a proper dynamical system subject to the perturbation of random noises, or more precisely, by a stochastic differential equation, with the control to the system of the form of state feedback control. The system is defined to be safe as long as its state evolves inside a certain subset of the state space called the safe region, and whenever this is violated, some emergency procedures have to be evoked in order to reset the state within the safe region. Due to the usually much higher cost of the emergency procedures, it is preferable to design a feedback control law of a reasonable cost that can keep the state of the system within the safe region for as long as possible. Or equivalently, one wishes to find a feedback control with the least cost that can keep the system safe long enough. In this paper we study a simple instance of such a problem. As a concrete eample, consider the following scenario. Suppose that there are three consecutive cars driving in the same direction on a road, numbered,, and 3 from front to end. The body length of each car is This material is based upon work supported by the National Science Foundation under Grant No. EIA d D d Figure : Three cars on a highway. d. Denote by D the distance between car and car 3 and by the distance between car and car 3, both ecluding body length. Hence the distance between car and car is D d (see Fig. ). Suppose that in a period of time, D remains constant, i.e., car and car 3 have identical velocities, while is given by the equation: d(t) = f[(t)] + w(t). dt Here f is the feedback control law for car that depends on the state, and w(t) is the white noise modeling random perturbations such as air resistance and road frictions. Car is safe if its distance from either car or car 3 is at least ɛ >, i.e., if [ɛ, D d ɛ]. Suppose that at time t = car is at its ideal position () = (D d)/. Due to the presence of noises, (t) will eventually venture outside of the safe region. Denote by T the first time (t) escapes from the safe region, which is a positive random variable. A natural question arises: among all the feedback control laws f satisfying the constraint E[T ] > t for some threshold t, find the one with the minimal cost D d ɛ ɛ f () d. The solution to a version of this problem will be discussed in this paper. This paper is organized as following. First of all, the problem under study is defined in its most general form in Section. In Section 3, some properties of its optimal solutions are established, which inspire us to propose a simplified version of the problem. By formulating the simplified problem as an optimal control problem in Section, we use the Maimal Principle to find the equations for the optimal solutions. Due to presence of the two-point boundary conditions, these equations are hard to solve directly. However, by finding
2 a first integral, we can significantly simplify their solution. These results are illustrated through numerical simulations. Finally, conclusions and possible etensions are discussed in Section 5. Problem Formulation g() µ= µ= µ= Given an interval [ a, a] on the real line, consider the solution X t to the following stochastic differential equation (SDE): dx t = f(x t )dt + db t, where B t is the standard Brownian motion, and f is a piecewise Lipschitz continuous function. Denote by T a = inf{t : X t = a} (respectively, T a = inf{t : X t = a}) the first time X t hits a (respectively, a). Then T a T a min{t a, T a } is the first time X t escapes from the interval [ a, a]. For each [ a, a], define g() = E [T a T a ], where E means that the epectation is taken under the initial condition that X =. Thus g() is the epected escaping time from [ a, a] given that X t starts from Figure : Plots of g() for µ =,,. J *(t ) J(f) It is a standard result in stochastic calculus (see, e.g., []) that g satisfies the following ordinary differential equation (ODE): g () + f()g () + =, [ a, a], () with the boundary condition g( a) = g(a) =. Note that g belongs to C, the family of functions on [ a, a] with piecewise Lipschitz continuous second order derivative. Remark For certain special cases, equation () can be solved eplicitly [3]. For eample, if f u on [ a, a], then X t = B t + ut is the BM with drift u, and g() = (a )eua + ( + a)e ua ae u u(e ua e ua. ) In particular, if f on [ a, a], then X t = B t is the standard BM. Let u in the above equation, we have g() = a. Figure plots g() for different values of u. We now formulate our problem. Problem Among all the control functions f for which the corresponding g satisfies g ma ma{g() : [ a, a]} t o a t g() Figure 3: Feasible region of control function f in (g(), J(f)) coordinates. for some fied threshold t >, find the one that minimizes the energy J(f) = a a f () d. In other words, we try to find the minimal energy control f subject to the constraint that the epected escaping time from at least one point in the safe region [ a, a] is no less than t. By Remark, if f has the minimum energy J(f) =, i.e., if f, then g() = a, hence g ma = a. As a result, unless t > a, the solution to Problem is always f. Therefore, we shall in the following assume that t a. For each threshold t, denote by J (t ) the energy of
3 the solution to Problem, i.e., J (t ) = inf{j(f) : f such that g ma t }. Then it is clear that J (t ) is an increasing function of t. J (t ) has the following interpretation. Represent each control function f as a point on a plane such that its coordinate is given by g ma and its y coordinate is given by J(f). In this coordinate system, the feasible region of all possible f is plotted as the shaded region in Fig. 3. The region is bounded from the right by the graph of the function J (t ) for t a, and from the left by another curve of no practical interest. The verte of the region corresponds to the control function f. As a by-product of the above analysis, we have an equivalent (dual) formulation of the problem as follows. Problem Among all the control functions f whose energy is at most J >, find the one for which the corresponding g ma is maimized. The value of g ma for the solution to Problem as a function of the threshold J is eactly the inverse of the function J. We shall focus on Problem only in the rest of the paper. To prove this proposition, we need two technical lemmas on the monotonicity of solutions to the ODE (). Lemma For a fied control function f and R, denote by g(; g, u ) the solution to the ODE () with initial condition g( ) = g and g ( ) = u. Then for any fied and g, g(; g, u ) is an increasing function of u if >, and a decreasing function of u if <. Proof: Suppose that for some u < v, >, we have g () g(; g, u ) > g () g(; g, v ). Then since g () < g () for > in a neighborhood of, there eists an y (, ] where g g achieve its maimum over [, ]. At y we have g (y) < g (y) and g (y) = g (y). Now g () and g () g (y) + g (y) are two solutions to the ODE () with identical values of g(y) and g (y), but nonetheless differ at and, a contradiction to the the uniqueness of solutions to the ODE (). Lemma Suppose that g and g are the solutions to the ODE () with initial condition g( ) = g, g ( ) =, that correspond to the control functions f and f, respectively, and that f f on [, ). Then g g on [, ). 3 Some Properties of Optimal Solutions We first show some useful properties of the optimal solutions to Problem. Proposition For any control function f, the solution g to equation () is strictly concave. Therefore, g, and there is a unique [ a, a] such that g( ) = g ma. Proof: We adopt a probabilistic proof using the definition of g. For any a y < < z a, g() = E [T a T a ] = P (T y < T z )g(y)+ P (T z < T y )g(z) + E [T y T z ], where T y and T z denote the first time X t hits y and z respectively. Note that P (T y < T z )+P (T z < T y ) = and E [T y T z ] >. Therefore, g is strictly concave. Proposition Suppose that f is an optimal solution to Problem, and the corresponding g assumes its maimum at. Then f on [ a, ] and f on [, a]. In other words, the optimal control function f will direct toward from both side. The proof of Lemma follows the same line as that of Lemma, hence is omitted. We now prove Proposition. Suppose otherwise that the optimal solution f is positive on some nontrivial subinterval I = [, ] of [, a]. Define a new control function ˆf such that ˆf = f on I and ˆf = f at everywhere else. Note that ˆf and f have the same energy. We claim that by adopting ˆf as the control function, the epected escaping time, ĝ, is higher at, a contradiction to our assumption that f is optimal. To prove this claim, choose = and g = g( ) in Lemma to conclude that the solution to equation () with initial condition g( ) and g ( ) = corresponding to the new control function ˆf is no larger than g everywhere on [, a]. Piece such a solution on [, a] with g on [ a, ] together to obtain a solution to () on [ a, a] with initial condition g( a) = and g(a) <. Now Lemma shows that a solution to equation () with initial condition g( a) = and g(a) = is smaller than the pieced together solution at any ( a, a]. In other words, the epected escaping time corresponding to the control function ˆf starting from is larger than g( ), a contradiction. The other half of the proposition can be proved similarly. We conjecture that for optimal solution f to Problem, the corresponding g assumes its maimum at =. However, a rigorous proof is yet to be found. If this is
4 indeed the case, then it is clear that there is a version of f that is odd on [ a, a] which generates an escaping time of the same value at. In fact, starting any optimal solution f, if a f () d a f () d, then by choosing ˆf() = f() on [ a, ] and ˆf() = f( ) on [, a], we have a control function ˆf for which the escaping time ĝ satisfies ĝ() = g() on [ a, ] and ĝ() = g( ) on [, a]. Inspired by the above discussions, we propose a restricted version of Problem as follows. Problem 3 Among all the control functions f that satisfy g ma t and that are odd on [ a, a], find the one with minimal energy. Because of the restriction, f is odd and g is even on [ a, a]. Hence we need only to design the functions f and g on [ a, ], and then etend them to the whole interval [ a, a] by their respective symmetries. The equation that g satisfies is now given by g () + f()g () + =, [ a, ], () with initial condition g( a) =, g() = t, g () =. Here we use the obvious fact that the optimal solution f satisfies g() = t. Then the optimal control f satisfies []: and λ = H y =, (5) λ = H y = fλ λ, (6) with the boundary conditions H f = f λ y =, (7) y ( a) =, y () = t, y () =, λ ( a) =. By (5), λ is constant on [ a, ]. From (7) we have f = λ y. (8) Substituting this into () and (6), we obtain y = λ y, (9) λ = λ y λ, () whose boundary conditions are y () = and λ ( a) =. We now try to solve the above dynamic equations. Using equations (9) and (), we have Solution Using Optimal Control Theory Define y () = g(), y () = g (). Then equation () implies and (λ y ) = λ y + λ y = (λ y λ )y λ (λ y + ) = (λ y + λ ), y = y, (3) y = fy, () for [ a, ]. The boundary conditions are y ( a) =, y () = t, and y () =. The above equations specify the dynamics of a control system whose states are (y, y ) and whose input is f. The cost of the system is a f () d = J(f). So Problem is equivalent to Problem Solve the following optimal control problem: Minimize subject to Define the Hamiltonian f () d a { y = y, y = fy, [ a, ], y ( a) =, y () = t, y () =. H(y, y, λ, λ, u) = λ y + λ ( fy ) + f = λ y λ (fy + ) + f. (λ λ y ) = λ λ y = (λ y λ ) + λ (λ y + ) Therefore, = λ y (λ y + λ ). (λ λ y ) = (λ y )(λ y ) = (λ y ). Integrating, we have a first integral λ λ y = λ y + C, [ a, ], () for some constant C. The initial conditions y () = and λ ( a) = imply that C = λ () = λ y ( a). Now λ and y can be solved from () as λ = ± + y (λ y + C) y, y = λ ± λ λ (λ C) λ,
5 Φ[y ( )] 8 6 f opt.5.5 g opt y ( ).5.5 Figure : Plot of Φ[y ( a)] for a =. which, after substitution into (9) and (), yield y = ± + y (λ y + C), λ = ± λ λ (λ C), By Proposition, the sign in the first equation can be easily determined as: y = + y (λ y + C) = + λ y [y y ( a)], () with boundary conditions y () =. Write equation () as dy / + λ y [y y ( a)] = d, and integrating from = a to =, we have Ψ(λ, y ( a)) y( a) dy + λ y [y y ( a)] = a, (3) which defines λ as an implicit function of y ( a) [a, ). In fact, for each fied y ( a) a, since y[y y ( a)] < for < y < y ( a), Ψ(λ, y ( a)) is strictly increasing with λ. Note that Ψ(, y ( a)) = y ( a) a, and lim λ Ψ(λ, y ( a)) = a. So there eists a unique λ, denoted as ϕ[y ( a)], such that Ψ(ϕ[y ( a)], y ( a)) = a. From the above discussion, if y ( a) a, then the solution y to equation () satisfies the boundary condition y () = if and only if we choose λ = ϕ[y ( a)]. To determine y (), note that y satisfies y = y, y ( a) =, y () = t. Therefore, if we define Φ[y ( a)] = a y () d, f opt.5.5 g opt f opt g opt Figure 5: Plots of optimal f and g when a =, with t = (first two rows),t = 3 (middle two rows), and t = 6 (last two rows). where y is the solution to equation () with initial condition y ( a) a and with λ = ϕ[y ( a)], then Φ[y ( a)] = y () y ( a) = t. () From eperiments, Φ[y ( a)] is strictly increasing for y ( a) a. See Figure for a plot of Φ[y ( a)] when a =. Therefore we can find a y ( a) such that Φ[y ( a)] = t if t a. Using this y ( a) and λ = ϕ[y ( a)], the solution to Problem can be found by integrating the following differential equations: y = y, y ( a) =, y = + λ y [y y ( a)]. Note that other boundary conditions y () = t and y () = are automatically satisfied by our choice of y () and λ. Hence the difficulty of solving the two-
6 point boundary problem specified by (9) and () is avoided. Figure 5 plots the optimal f and g computed as above when a = for t =, 3, 6 respectively. The first two rows are the optimal f and g for t =, the middle two rows for t = 3, and the last two rows for t = 6. As epected, as t increases, the optimal feedback control law f becomes more aggressive. An interesting fact is that the optimal f is nearly zero around the center of the interval, while much of the effort is spent near / and 3/ positions of the interval. 5 Conclusion In this paper the problem of optimal sojourn time control for an agent moving in an interval is studied. Using tools from optimal control theory, and by finding a first integral to the resulting equations, we propose a solution to the problem under the symmetric assumption. As future directions, we are currently trying to prove the conjecture that solutions to the symmetric version of the problem are also solutions to the general problem. It is also interesting to see if our approach can be etended to the higher dimensional case, for eample, when the safe region is a disk in R n. References [] A. E. Bryson, Jr. and Y. C. Ho. Applied Optimal Control. Hemisphere Publishing Corp. Washington, D. C., 975. Optimization, Estimation, and Control, Revised printing. [] R. Durrett. Stochastic Calculus: A Practical Introduction. CRC Press, 996. [3] L. C. G. Rogers and D. Williams. Diffusions, Markov Processes, and Martingales, volume I, II. John Wiley & Sons, 99. [] C. Tomlin. Hybrid control of air traffic management system. Ph.D. thesis, EECS Department, UC Berkeley, 997. [5] P. Varaiya. Smart cars on smart roads: problems of control. IEEE Transaction on Automatic Control, AC-38(), 993.
1 Brownian Local Time
1 Brownian Local Time We first begin by defining the space and variables for Brownian local time. Let W t be a standard 1-D Wiener process. We know that for the set, {t : W t = } P (µ{t : W t = } = ) =
More informationFUNDAMENTALS OF REAL ANALYSIS by. IV.1. Differentiation of Monotonic Functions
FUNDAMNTALS OF RAL ANALYSIS by Doğan Çömez IV. DIFFRNTIATION AND SIGND MASURS IV.1. Differentiation of Monotonic Functions Question: Can we calculate f easily? More eplicitly, can we hope for a result
More informationThe equivalence of two tax processes
The equivalence of two ta processes Dalal Al Ghanim Ronnie Loeffen Ale Watson 6th November 218 arxiv:1811.1664v1 [math.pr] 5 Nov 218 We introduce two models of taation, the latent and natural ta processes,
More information1.2 Functions What is a Function? 1.2. FUNCTIONS 11
1.2. FUNCTIONS 11 1.2 Functions 1.2.1 What is a Function? In this section, we only consider functions of one variable. Loosely speaking, a function is a special relation which exists between two variables.
More informationI forgot to mention last time: in the Ito formula for two standard processes, putting
I forgot to mention last time: in the Ito formula for two standard processes, putting dx t = a t dt + b t db t dy t = α t dt + β t db t, and taking f(x, y = xy, one has f x = y, f y = x, and f xx = f yy
More informationHarmonic Functions and Brownian motion
Harmonic Functions and Brownian motion Steven P. Lalley April 25, 211 1 Dynkin s Formula Denote by W t = (W 1 t, W 2 t,..., W d t ) a standard d dimensional Wiener process on (Ω, F, P ), and let F = (F
More informationUniversidad Carlos III de Madrid
Universidad Carlos III de Madrid Eercise 1 2 3 4 5 6 Total Points Department of Economics Mathematics I Final Eam January 22nd 2018 LAST NAME: Eam time: 2 hours. FIRST NAME: ID: DEGREE: GROUP: 1 (1) Consider
More informationBrownian Motion. An Undergraduate Introduction to Financial Mathematics. J. Robert Buchanan. J. Robert Buchanan Brownian Motion
Brownian Motion An Undergraduate Introduction to Financial Mathematics J. Robert Buchanan 2010 Background We have already seen that the limiting behavior of a discrete random walk yields a derivation of
More informationLecture 3: September 10
CS294 Markov Chain Monte Carlo: Foundations & Applications Fall 2009 Lecture 3: September 10 Lecturer: Prof. Alistair Sinclair Scribes: Andrew H. Chan, Piyush Srivastava Disclaimer: These notes have not
More informationReflected Brownian Motion
Chapter 6 Reflected Brownian Motion Often we encounter Diffusions in regions with boundary. If the process can reach the boundary from the interior in finite time with positive probability we need to decide
More informationDuality and dynamics in Hamilton-Jacobi theory for fully convex problems of control
Duality and dynamics in Hamilton-Jacobi theory for fully convex problems of control RTyrrell Rockafellar and Peter R Wolenski Abstract This paper describes some recent results in Hamilton- Jacobi theory
More informationApproximating diffusions by piecewise constant parameters
Approximating diffusions by piecewise constant parameters Lothar Breuer Institute of Mathematics Statistics, University of Kent, Canterbury CT2 7NF, UK Abstract We approximate the resolvent of a one-dimensional
More informationEconomics 205 Exercises
Economics 05 Eercises Prof. Watson, Fall 006 (Includes eaminations through Fall 003) Part 1: Basic Analysis 1. Using ε and δ, write in formal terms the meaning of lim a f() = c, where f : R R.. Write the
More informationIntroduction to Random Diffusions
Introduction to Random Diffusions The main reason to study random diffusions is that this class of processes combines two key features of modern probability theory. On the one hand they are semi-martingales
More informationUS01CMTH02 UNIT-3. exists, then it is called the partial derivative of f with respect to y at (a, b) and is denoted by f. f(a, b + b) f(a, b) lim
Study material of BSc(Semester - I US01CMTH02 (Partial Derivatives Prepared by Nilesh Y Patel Head,Mathematics Department,VPand RPTPScience College 1 Partial derivatives US01CMTH02 UNIT-3 The concept of
More informationA Stochastic Paradox for Reflected Brownian Motion?
Proceedings of the 9th International Symposium on Mathematical Theory of Networks and Systems MTNS 2 9 July, 2 udapest, Hungary A Stochastic Parado for Reflected rownian Motion? Erik I. Verriest Abstract
More informationHelpful Concepts for MTH 261 Final. What are the general strategies for determining the domain of a function?
Helpful Concepts for MTH 261 Final What are the general strategies for determining the domain of a function? How do we use the graph of a function to determine its range? How many graphs of basic functions
More informationThe Hilbert transform
The Hilbert transform Definition and properties ecall the distribution pv(, defined by pv(/(ϕ := lim ɛ ɛ ϕ( d. The Hilbert transform is defined via the convolution with pv(/, namely (Hf( := π lim f( t
More informationf(s)ds, i.e. to write down the general solution in
1. First order ODEs. Simplest examples. Existence and uniqueness theorem Example 1.1. Consider the equation (1.1) x (t) = 1 This is an equation because there is an unknown: the function x(t). This is a
More informationFunctions with orthogonal Hessian
Functions with orthogonal Hessian B. Dacorogna P. Marcellini E. Paolini Abstract A Dirichlet problem for orthogonal Hessians in two dimensions is eplicitly solved, by characterizing all piecewise C 2 functions
More information14.6 Spring Force Energy Diagram
14.6 Spring Force Energy Diagram The spring force on an object is a restoring force F s = F s î = k î where we choose a coordinate system with the equilibrium position at i = 0 and is the amount the spring
More informationTHE INVERSE FUNCTION THEOREM
THE INVERSE FUNCTION THEOREM W. PATRICK HOOPER The implicit function theorem is the following result: Theorem 1. Let f be a C 1 function from a neighborhood of a point a R n into R n. Suppose A = Df(a)
More informationSolution to Exercise 3
Spring 207 MATH502 Real Analysis II Solution to Eercise 3 () For a, b > 0, set a sin( b ), 0 < f() = 0, = 0. nπ+π/2 Show that f is in BV [0, ] iff a > b. ( ) b Solution. Put n = for n 0. We claim that
More informationSTATIC LECTURE 4: CONSTRAINED OPTIMIZATION II - KUHN TUCKER THEORY
STATIC LECTURE 4: CONSTRAINED OPTIMIZATION II - KUHN TUCKER THEORY UNIVERSITY OF MARYLAND: ECON 600 1. Some Eamples 1 A general problem that arises countless times in economics takes the form: (Verbally):
More informationContinuous Functions on Metric Spaces
Continuous Functions on Metric Spaces Math 201A, Fall 2016 1 Continuous functions Definition 1. Let (X, d X ) and (Y, d Y ) be metric spaces. A function f : X Y is continuous at a X if for every ɛ > 0
More informationAn introduction to Mathematical Theory of Control
An introduction to Mathematical Theory of Control Vasile Staicu University of Aveiro UNICA, May 2018 Vasile Staicu (University of Aveiro) An introduction to Mathematical Theory of Control UNICA, May 2018
More informationDifferential Calculus
Differential Calculus. Compute the derivatives of the following functions a() = 4 3 7 + 4 + 5 b() = 3 + + c() = 3 + d() = sin cos e() = sin f() = log g() = tan h() = 3 6e 5 4 i() = + tan 3 j() = e k()
More informationPiecewise Smooth Solutions to the Burgers-Hilbert Equation
Piecewise Smooth Solutions to the Burgers-Hilbert Equation Alberto Bressan and Tianyou Zhang Department of Mathematics, Penn State University, University Park, Pa 68, USA e-mails: bressan@mathpsuedu, zhang
More information1. Sets A set is any collection of elements. Examples: - the set of even numbers between zero and the set of colors on the national flag.
San Francisco State University Math Review Notes Michael Bar Sets A set is any collection of elements Eamples: a A {,,4,6,8,} - the set of even numbers between zero and b B { red, white, bule} - the set
More informationSolutions to Math 41 First Exam October 12, 2010
Solutions to Math 41 First Eam October 12, 2010 1. 13 points) Find each of the following its, with justification. If the it does not eist, eplain why. If there is an infinite it, then eplain whether it
More informationStochastic Dominance-based Rough Set Model for Ordinal Classification
Stochastic Dominance-based Rough Set Model for Ordinal Classification Wojciech Kotłowski wkotlowski@cs.put.poznan.pl Institute of Computing Science, Poznań University of Technology, Poznań, 60-965, Poland
More informationRoberto s Notes on Differential Calculus Chapter 1: Limits and continuity Section 7. Discontinuities. is the tool to use,
Roberto s Notes on Differential Calculus Chapter 1: Limits and continuity Section 7 Discontinuities What you need to know already: The concept and definition of continuity. What you can learn here: The
More information2015 Math Camp Calculus Exam Solution
015 Math Camp Calculus Exam Solution Problem 1: x = x x +5 4+5 = 9 = 3 1. lim We also accepted ±3, even though it is not according to the prevailing convention 1. x x 4 x+4 =. lim 4 4+4 = 4 0 = 4 0 = We
More informationDirectional derivatives and gradient vectors (Sect. 14.5). Directional derivative of functions of two variables.
Directional derivatives and gradient vectors (Sect. 14.5). Directional derivative of functions of two variables. Partial derivatives and directional derivatives. Directional derivative of functions of
More informationStochastic Optimization with Inequality Constraints Using Simultaneous Perturbations and Penalty Functions
International Journal of Control Vol. 00, No. 00, January 2007, 1 10 Stochastic Optimization with Inequality Constraints Using Simultaneous Perturbations and Penalty Functions I-JENG WANG and JAMES C.
More informationRates of Convergence to Self-Similar Solutions of Burgers Equation
Rates of Convergence to Self-Similar Solutions of Burgers Equation by Joel Miller Andrew Bernoff, Advisor Advisor: Committee Member: May 2 Department of Mathematics Abstract Rates of Convergence to Self-Similar
More informationMath 75B Practice Problems for Midterm II Solutions Ch. 16, 17, 12 (E), , 2.8 (S)
Math 75B Practice Problems for Midterm II Solutions Ch. 6, 7, 2 (E),.-.5, 2.8 (S) DISCLAIMER. This collection of practice problems is not guaranteed to be identical, in length or content, to the actual
More informationA MODEL FOR THE LONG-TERM OPTIMAL CAPACITY LEVEL OF AN INVESTMENT PROJECT
A MODEL FOR HE LONG-ERM OPIMAL CAPACIY LEVEL OF AN INVESMEN PROJEC ARNE LØKKA AND MIHAIL ZERVOS Abstract. We consider an investment project that produces a single commodity. he project s operation yields
More informationLAW OF LARGE NUMBERS FOR THE SIRS EPIDEMIC
LAW OF LARGE NUMBERS FOR THE SIRS EPIDEMIC R. G. DOLGOARSHINNYKH Abstract. We establish law of large numbers for SIRS stochastic epidemic processes: as the population size increases the paths of SIRS epidemic
More informationFirst Variation of a Functional
First Variation of a Functional The derivative of a function being zero is a necessary condition for the etremum of that function in ordinary calculus. Let us now consider the equivalent of a derivative
More informationOPTIMAL STOPPING OF A BROWNIAN BRIDGE
OPTIMAL STOPPING OF A BROWNIAN BRIDGE ERIK EKSTRÖM AND HENRIK WANNTORP Abstract. We study several optimal stopping problems in which the gains process is a Brownian bridge or a functional of a Brownian
More informationMATH 220 solution to homework 1
MATH solution to homework Problem. Define z(s = u( + s, y + s, then z (s = u u ( + s, y + s + y ( + s, y + s = e y, z( y = u( y, = f( y, u(, y = z( = z( y + y If we prescribe the data u(, = f(, then z
More informationNash equilibria of threshold type for two-player nonzero-sum games of stopping
Nash equilibria of threshold type for two-player nonzero-sum games of stopping De Angelis, T; Ferrari, G; MORIARTY, JM Institute of Mathematical Statistics This is a pre-copyedited, author-produced version
More informationProving the Regularity of the Minimal Probability of Ruin via a Game of Stopping and Control
Proving the Regularity of the Minimal Probability of Ruin via a Game of Stopping and Control Erhan Bayraktar University of Michigan joint work with Virginia R. Young, University of Michigan K αρλoβασi,
More informationReview of Multi-Calculus (Study Guide for Spivak s CHAPTER ONE TO THREE)
Review of Multi-Calculus (Study Guide for Spivak s CHPTER ONE TO THREE) This material is for June 9 to 16 (Monday to Monday) Chapter I: Functions on R n Dot product and norm for vectors in R n : Let X
More informationModule 2: Reflecting on One s Problems
MATH55 Module : Reflecting on One s Problems Main Math concepts: Translations, Reflections, Graphs of Equations, Symmetry Auxiliary ideas: Working with quadratics, Mobius maps, Calculus, Inverses I. Transformations
More informationOptimal stopping of a mean reverting diffusion: minimizing the relative distance to the maximum
Optimal stopping of a mean reverting diffusion: minimiing the relative distance to the maimum Romuald ELIE CEREMADE, CNRS, UMR 7534, Université Paris-Dauphine and CREST elie@ceremade.dauphine.fr Gilles-Edouard
More informationOptimal exit strategies for investment projects. 7th AMaMeF and Swissquote Conference
Optimal exit strategies for investment projects Simone Scotti Université Paris Diderot Laboratoire de Probabilité et Modèles Aléatories Joint work with : Etienne Chevalier, Université d Evry Vathana Ly
More information2. Metric Spaces. 2.1 Definitions etc.
2. Metric Spaces 2.1 Definitions etc. The procedure in Section for regarding R as a topological space may be generalized to many other sets in which there is some kind of distance (formally, sets with
More information{σ x >t}p x. (σ x >t)=e at.
3.11. EXERCISES 121 3.11 Exercises Exercise 3.1 Consider the Ornstein Uhlenbeck process in example 3.1.7(B). Show that the defined process is a Markov process which converges in distribution to an N(0,σ
More informationMath 104: Homework 7 solutions
Math 04: Homework 7 solutions. (a) The derivative of f () = is f () = 2 which is unbounded as 0. Since f () is continuous on [0, ], it is uniformly continous on this interval by Theorem 9.2. Hence for
More informationCLEP Calculus. Time 60 Minutes 45 Questions. For each question below, choose the best answer from the choices given. 2. If f(x) = 3x, then f (x) =
CLEP Calculus Time 60 Minutes 5 Questions For each question below, choose the best answer from the choices given. 7. lim 5 + 5 is (A) 7 0 (C) 7 0 (D) 7 (E) Noneistent. If f(), then f () (A) (C) (D) (E)
More informationBounded uniformly continuous functions
Bounded uniformly continuous functions Objectives. To study the basic properties of the C -algebra of the bounded uniformly continuous functions on some metric space. Requirements. Basic concepts of analysis:
More informationDensities for the Navier Stokes equations with noise
Densities for the Navier Stokes equations with noise Marco Romito Università di Pisa Universitat de Barcelona March 25, 2015 Summary 1 Introduction & motivations 2 Malliavin calculus 3 Besov bounds 4 Other
More informationXt i Xs i N(0, σ 2 (t s)) and they are independent. This implies that the density function of X t X s is a product of normal density functions:
174 BROWNIAN MOTION 8.4. Brownian motion in R d and the heat equation. The heat equation is a partial differential equation. We are going to convert it into a probabilistic equation by reversing time.
More informationCHAPTER 1-2: SHADOW PRICES
Essential Microeconomics -- CHAPTER -: SHADOW PRICES An intuitive approach: profit maimizing firm with a fied supply of an input Shadow prices 5 Concave maimization problem 7 Constraint qualifications
More informationREAL AND COMPLEX ANALYSIS
REAL AND COMPLE ANALYSIS Third Edition Walter Rudin Professor of Mathematics University of Wisconsin, Madison Version 1.1 No rights reserved. Any part of this work can be reproduced or transmitted in any
More informationQuadratic Forms. Ferdinand Vanmaele. July 5, 2017
Quadratic Forms Ferdinand Vanmaele July 5, 2017 1 Introduction Question. Consider Pythagorean triples (, y, z), that is integers for which 2 + y 2 = z 2. More generally, we wish to find all integers n
More informationIntroduction to Reachability Somil Bansal Hybrid Systems Lab, UC Berkeley
Introduction to Reachability Somil Bansal Hybrid Systems Lab, UC Berkeley Outline Introduction to optimal control Reachability as an optimal control problem Various shades of reachability Goal of This
More information2. The Concept of Convergence: Ultrafilters and Nets
2. The Concept of Convergence: Ultrafilters and Nets NOTE: AS OF 2008, SOME OF THIS STUFF IS A BIT OUT- DATED AND HAS A FEW TYPOS. I WILL REVISE THIS MATE- RIAL SOMETIME. In this lecture we discuss two
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/15.070J Fall 2013 Lecture 7 9/25/2013
MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.65/15.070J Fall 013 Lecture 7 9/5/013 The Reflection Principle. The Distribution of the Maximum. Brownian motion with drift Content. 1. Quick intro to stopping times.
More informationMATH 51H Section 4. October 16, Recall what it means for a function between metric spaces to be continuous:
MATH 51H Section 4 October 16, 2015 1 Continuity Recall what it means for a function between metric spaces to be continuous: Definition. Let (X, d X ), (Y, d Y ) be metric spaces. A function f : X Y is
More informationThe Kuhn-Tucker and Envelope Theorems
The Kuhn-Tucker and Envelope Theorems Peter Ireland EC720.01 - Math for Economists Boston College, Department of Economics Fall 2010 The Kuhn-Tucker and envelope theorems can be used to characterize the
More informationAPPLICATIONS OF DIFFERENTIABILITY IN R n.
APPLICATIONS OF DIFFERENTIABILITY IN R n. MATANIA BEN-ARTZI April 2015 Functions here are defined on a subset T R n and take values in R m, where m can be smaller, equal or greater than n. The (open) ball
More informationCHAPTER 3: OPTIMIZATION
John Riley 8 February 7 CHAPTER 3: OPTIMIZATION 3. TWO VARIABLES 8 Second Order Conditions Implicit Function Theorem 3. UNCONSTRAINED OPTIMIZATION 4 Necessary and Sufficient Conditions 3.3 CONSTRAINED
More informationPrinciple of Mathematical Induction
Advanced Calculus I. Math 451, Fall 2016, Prof. Vershynin Principle of Mathematical Induction 1. Prove that 1 + 2 + + n = 1 n(n + 1) for all n N. 2 2. Prove that 1 2 + 2 2 + + n 2 = 1 n(n + 1)(2n + 1)
More informationApproximate inference, Sampling & Variational inference Fall Cours 9 November 25
Approimate inference, Sampling & Variational inference Fall 2015 Cours 9 November 25 Enseignant: Guillaume Obozinski Scribe: Basile Clément, Nathan de Lara 9.1 Approimate inference with MCMC 9.1.1 Gibbs
More informationThe Skorokhod reflection problem for functions with discontinuities (contractive case)
The Skorokhod reflection problem for functions with discontinuities (contractive case) TAKIS KONSTANTOPOULOS Univ. of Texas at Austin Revised March 1999 Abstract Basic properties of the Skorokhod reflection
More informationReview Problems for Test 3
Review Problems for Test 3 Math 6-3/6 7 These problems are provided to help you study. The presence of a problem on this handout does not imply that there will be a similar problem on the test. And the
More informationCOFINITE INDUCED SUBGRAPHS OF IMPARTIAL COMBINATORIAL GAMES: AN ANALYSIS OF CIS-NIM
#G INTEGERS 13 (013) COFINITE INDUCED SUBGRAPHS OF IMPARTIAL COMBINATORIAL GAMES: AN ANALYSIS OF CIS-NIM Scott M. Garrabrant 1 Pitzer College, Claremont, California coscott@math.ucla.edu Eric J. Friedman
More informationAbe Mirza Graphing f ( x )
Abe Mirza Graphing f ( ) Steps to graph f ( ) 1. Set f ( ) = 0 and solve for critical values.. Substitute the critical values into f ( ) to find critical points.. Set f ( ) = 0 and solve for critical values.
More informationarxiv: v1 [math.pr] 14 Aug 2017
Uniqueness of Gibbs Measures for Continuous Hardcore Models arxiv:1708.04263v1 [math.pr] 14 Aug 2017 David Gamarnik Kavita Ramanan Abstract We formulate a continuous version of the well known discrete
More informationI have not checked this review sheet for errors, hence there maybe errors in this document. thank you.
I have not checked this review sheet for errors, hence there maybe errors in this document. thank you. Class test II Review sections 3.7(differentials)-5.5(logarithmic differentiation) ecluding section
More informationStochastic Calculus. Kevin Sinclair. August 2, 2016
Stochastic Calculus Kevin Sinclair August, 16 1 Background Suppose we have a Brownian motion W. This is a process, and the value of W at a particular time T (which we write W T ) is a normally distributed
More informationMATH 131A: REAL ANALYSIS (BIG IDEAS)
MATH 131A: REAL ANALYSIS (BIG IDEAS) Theorem 1 (The Triangle Inequality). For all x, y R we have x + y x + y. Proposition 2 (The Archimedean property). For each x R there exists an n N such that n > x.
More informationBASICS OF CONVEX ANALYSIS
BASICS OF CONVEX ANALYSIS MARKUS GRASMAIR 1. Main Definitions We start with providing the central definitions of convex functions and convex sets. Definition 1. A function f : R n R + } is called convex,
More information3.3.1 Linear functions yet again and dot product In 2D, a homogenous linear scalar function takes the general form:
3.3 Gradient Vector and Jacobian Matri 3 3.3 Gradient Vector and Jacobian Matri Overview: Differentiable functions have a local linear approimation. Near a given point, local changes are determined by
More informationu( x)= Pr X() t hits C before 0 X( 0)= x ( ) 2 AMS 216 Stochastic Differential Equations Lecture #2
AMS 6 Stochastic Differential Equations Lecture # Gambler s Ruin (continued) Question #: How long can you play? Question #: What is the chance that you break the bank? Note that unlike in the case of deterministic
More informationMAC 2311 Final Exam Review Fall Private-Appointment, one-on-one tutoring at Broward Hall
Fall 2016 This review, produced by the CLAS Teaching Center, contains a collection of questions which are representative of the type you may encounter on the eam. Other resources made available by the
More informationSOLUTIONS TO ADDITIONAL EXERCISES FOR II.1 AND II.2
SOLUTIONS TO ADDITIONAL EXERCISES FOR II.1 AND II.2 Here are the solutions to the additional exercises in betsepexercises.pdf. B1. Let y and z be distinct points of L; we claim that x, y and z are not
More information2 (Bonus). Let A X consist of points (x, y) such that either x or y is a rational number. Is A measurable? What is its Lebesgue measure?
MA 645-4A (Real Analysis), Dr. Chernov Homework assignment 1 (Due 9/5). Prove that every countable set A is measurable and µ(a) = 0. 2 (Bonus). Let A consist of points (x, y) such that either x or y is
More informationLecture 1: Brief Review on Stochastic Processes
Lecture 1: Brief Review on Stochastic Processes A stochastic process is a collection of random variables {X t (s) : t T, s S}, where T is some index set and S is the common sample space of the random variables.
More informationUC Merced: MATH 21 Final Exam 16 May 2006
UC Merced: MATH 2 Final Eam 6 May 2006 On the front of your bluebook print () your name, (2) your student ID number, (3) your instructor s name (Bianchi) and () a grading table. Show all work in your bluebook
More informationNotes for Introduction to Partial Di erential Equations, Fall 2018
Notes for Introduction to Partial Di erential Equations, Fall 2018 Esteban G. Tabak 1 Tra c Flow We eplore a simple mathematical model for tra c flow. This will be our introduction to systems of conservation
More informationLearning Targets: Standard Form: Quadratic Function. Parabola. Vertex Max/Min. x-coordinate of vertex Axis of symmetry. y-intercept.
Name: Hour: Algebra A Lesson:.1 Graphing Quadratic Functions Learning Targets: Term Picture/Formula In your own words: Quadratic Function Standard Form: Parabola Verte Ma/Min -coordinate of verte Ais of
More informationComments and Corrections
1386 IEEE TRANSACTIONS ON AUTOMATIC CONTROL, VOL 59, NO 5, MAY 2014 Comments and Corrections Corrections to Stochastic Barbalat s Lemma and Its Applications Xin Yu and Zhaojing Wu Abstract The proof of
More informationAPPLICATIONS OF INTEGRATION
6 APPLICATIONS OF INTEGRATION APPLICATIONS OF INTEGRATION 6.5 Average Value of a Function In this section, we will learn about: Applying integration to find out the average value of a function. AVERAGE
More informationMA2223 Tutorial solutions Part 1. Metric spaces
MA2223 Tutorial solutions Part 1. Metric spaces T1 1. Show that the function d(,y) = y defines a metric on R. The given function is symmetric and non-negative with d(,y) = 0 if and only if = y. It remains
More informationHW #1 SOLUTIONS. g(x) sin 1 x
HW #1 SOLUTIONS 1. Let f : [a, b] R and let v : I R, where I is an interval containing f([a, b]). Suppose that f is continuous at [a, b]. Suppose that lim s f() v(s) = v(f()). Using the ɛ δ definition
More informationMath 2250 Final Exam Practice Problem Solutions. f(x) = ln x x. 1 x. lim. lim. x x = lim. = lim 2
Math 5 Final Eam Practice Problem Solutions. What are the domain and range of the function f() = ln? Answer: is only defined for, and ln is only defined for >. Hence, the domain of the function is >. Notice
More informationCalculus One variable
Calculus One variable (f ± g) ( 0 ) = f ( 0 ) ± g ( 0 ) (λf) ( 0 ) = λ f ( 0 ) ( (fg) ) ( 0 ) = f ( 0 )g( 0 ) + f( 0 )g ( 0 ) f g (0 ) = f ( 0 )g( 0 ) f( 0 )g ( 0 ) f( 0 ) 2 (f g) ( 0 ) = f (g( 0 )) g
More informationLIST COLORING OF COMPLETE MULTIPARTITE GRAPHS
LIST COLORING OF COMPLETE MULTIPARTITE GRAPHS Tomáš Vetrík School of Mathematical Sciences University of KwaZulu-Natal Durban, South Africa e-mail: tomas.vetrik@gmail.com Abstract The choice number of
More information1/12/05: sec 3.1 and my article: How good is the Lebesgue measure?, Math. Intelligencer 11(2) (1989),
Real Analysis 2, Math 651, Spring 2005 April 26, 2005 1 Real Analysis 2, Math 651, Spring 2005 Krzysztof Chris Ciesielski 1/12/05: sec 3.1 and my article: How good is the Lebesgue measure?, Math. Intelligencer
More informationSection 4.4 The Fundamental Theorem of Calculus
Section 4.4 The Fundamental Theorem of Calculus The First Fundamental Theorem of Calculus If f is continuous on the interval [a,b] and F is any function that satisfies F '() = f() throughout this interval
More informationSimulating Zeno Hybrid Systems Beyond Their Zeno Points
Simulating Zeno Hybrid Systems Beyond Their Zeno Points Haiyang Zheng Electrical Engineering and Computer Sciences University of California at Berkeley Technical Report No. UCB/EECS-006-4 http://www.eecs.berkeley.edu/pubs/techrpts/006/eecs-006-4.html
More informationUniformly Uniformly-ergodic Markov chains and BSDEs
Uniformly Uniformly-ergodic Markov chains and BSDEs Samuel N. Cohen Mathematical Institute, University of Oxford (Based on joint work with Ying Hu, Robert Elliott, Lukas Szpruch) Centre Henri Lebesgue,
More information3 (Due ). Let A X consist of points (x, y) such that either x or y is a rational number. Is A measurable? What is its Lebesgue measure?
MA 645-4A (Real Analysis), Dr. Chernov Homework assignment 1 (Due ). Show that the open disk x 2 + y 2 < 1 is a countable union of planar elementary sets. Show that the closed disk x 2 + y 2 1 is a countable
More informationMATH 202B - Problem Set 5
MATH 202B - Problem Set 5 Walid Krichene (23265217) March 6, 2013 (5.1) Show that there exists a continuous function F : [0, 1] R which is monotonic on no interval of positive length. proof We know there
More informationLecture 11 Hyperbolicity.
Lecture 11 Hyperbolicity. 1 C 0 linearization near a hyperbolic point 2 invariant manifolds Hyperbolic linear maps. Let E be a Banach space. A linear map A : E E is called hyperbolic if we can find closed
More informationMathematical Methods for Neurosciences. ENS - Master MVA Paris 6 - Master Maths-Bio ( )
Mathematical Methods for Neurosciences. ENS - Master MVA Paris 6 - Master Maths-Bio (2014-2015) Etienne Tanré - Olivier Faugeras INRIA - Team Tosca November 26th, 2014 E. Tanré (INRIA - Team Tosca) Mathematical
More information