Intro to Quantitative Investment (Lecture 4 of 6)
- 1. Optimal Trading Strategies Stochastic Control Haksun Li haksun.li@numericalmethod.com www.numericalmethod.com
- 2. Speaker Profile Dr. Haksun Li CEO, Numerical Method Inc. (Ex-)Adjunct Professors, Industry Fellow, Advisor, Consultant with the National University of Singapore, Nanyang Technological University, Fudan University, the Hong Kong University of Science and Technology. Quantitative Trader/Analyst, BNPP, UBS PhD, Computer Sci, University of Michigan Ann Arbor M.S., Financial Mathematics, University of Chicago B.S., Mathematics, University of Chicago 2
- 3. References Optimal Pairs Trading: A Stochastic Control Approach. Mudchanatongsuk, S., Primbs, J.A., Wong, W. Dept. of Manage. Sci. & Eng., Stanford Univ., Stanford, CA. 2008. Optimal Trend Following Trading Rules. Dai, M., Zhang, Q., Zhu Q. J. 2011 3
- 5. Stochastic Control We model the difference between the log-returns of two assets as an Ornstein-Uhlenbeck process. We compute the optimal position to take as a function of the deviation from the equilibrium. This is done by solving the corresponding the Hamilton-Jacobi-Bellman equation. 5
- 6. Formulation Assume a risk free asset 𝑀𝑡, which satisfies 𝑑𝑀𝑡 = 𝑟𝑀𝑡 𝑑𝑑 Assume two assets,𝐴 𝑡 and 𝐵𝑡. Assume 𝐵𝑡follows a geometric Brownian motion. 𝑑𝐵𝑡 = 𝜇𝐵𝑡 𝑑𝑑 + 𝜎𝐵𝑡 𝑑𝑧𝑡 𝑥 𝑡is the spread between the two assets. 𝑥𝑡 = log 𝐴 𝑡 − log 𝐵𝑡 6
- 7. Ornstein-Uhlenbeck Process We assume the spread, the basket that we want to trade, follows a mean-reverting process. 𝑑𝑥𝑡 = 𝑘 𝜃 − 𝑥𝑡 𝑑𝑑 + 𝜂𝜂𝜔 𝑡 𝜃 is the long term equilibrium to which the spread reverts. 𝑘 is the rate of reversion. It must be positive to ensure stability around the equilibrium value. 7
- 8. Instantaneous Correlation Let 𝜌 denote the instantaneous correlation coefficient between 𝑧 and 𝜔. 𝐸 𝑑𝜔 𝑡 𝑑𝑧𝑡 = 𝜌𝜌𝜌 8
- 9. Univariate Ito’s Lemma 9 Assume 𝑑𝑋𝑡 = 𝜇 𝑡 𝑑𝑑 + 𝜎𝑡 𝑑𝐵𝑡 𝑓 𝑡, 𝑋𝑡 is twice differentiable of two real variables We have 𝑑𝑑 𝑡, 𝑋𝑡 = 𝜕𝜕 𝜕𝜕 + 𝜇 𝑡 𝜕𝜕 𝜕𝜕 + 𝜎𝑡 2 2 𝜕2 𝑓 𝜕𝑥2 𝑑𝑑 + 𝜎𝑡 𝜕𝜕 𝜕𝜕 𝑑𝐵𝑡
- 10. Log example For G.B.M., 𝑑𝑋𝑡 = 𝜇𝑋𝑡 𝑑𝑑 + 𝜎𝑋𝑡 𝑑𝑧𝑡, 𝑑 log 𝑋𝑡 =? 𝑓 𝑥 = log 𝑥 𝜕𝑓 𝜕𝑡 = 0 𝜕𝑓 𝜕𝑥 = 1 𝑥 𝜕2 𝑓 𝜕𝑥2 = − 1 𝑥2 𝑑 log 𝑋𝑡 = 𝜇𝑋𝑡 1 𝑋𝑡 + 𝜎𝑋𝑡 2 2 − 1 𝑋𝑡 2 𝑑𝑑 + 𝜎𝑋𝑡 1 𝑋𝑡 𝑑𝐵𝑡 = 𝜇 − 𝜎2 2 𝑑𝑑 + 𝜎𝑑𝐵𝑡 10
- 11. Multivariate Ito’s Lemma 11 Assume 𝑋𝑡 = 𝑋1𝑡, 𝑋2𝑡, ⋯ , 𝑋 𝑛𝑛 is a vector Ito process 𝑓 𝑥1𝑡, 𝑥2𝑡, ⋯ , 𝑥 𝑛𝑛 is twice differentiable We have 𝑑𝑑 𝑋1𝑡, 𝑋2𝑡, ⋯ , 𝑋 𝑛𝑛 = ∑ 𝜕 𝜕𝑥 𝑖 𝑛 𝑖=1 𝑓 𝑋1𝑡, 𝑋2𝑡, ⋯ , 𝑋 𝑛𝑛 𝑑𝑋𝑖 𝑡 + 1 2 ∑ ∑ 𝜕2 𝜕𝑥 𝑖 𝜕𝑥 𝑗 𝑛 𝑗=1 𝑛 𝑖=1 𝑓 𝑋1𝑡, 𝑋2𝑡, ⋯ , 𝑋 𝑛𝑛 𝑑 𝑋𝑖, 𝑋𝑗 𝑡
- 12. Multivariate Example log 𝐴 𝑡 = 𝑥 𝑡 + log 𝐵𝑡 𝐴 𝑡 = exp 𝑥 𝑡 + log 𝐵𝑡 𝜕𝐴 𝑡 𝜕𝑥 𝑡 = exp 𝑥 𝑡 + log 𝐵𝑡 = 𝐴 𝑡 𝜕𝐴 𝑡 𝜕𝐵𝑡 = exp 𝑥 𝑡 + log 𝐵𝑡 1 𝐵𝑡 = 𝐴 𝑡 𝐵𝑡 𝜕2 𝐴 𝑡 𝜕𝑥 𝑡 2 = 𝜕𝐴 𝑡 𝜕𝑥 𝑡 = 𝐴 𝑡 𝜕2 𝐴 𝑡 𝜕𝐵𝑡 2 = 𝜕 𝜕𝐵𝑡 𝐴 𝑡 𝐵𝑡 = 0 𝜕2 𝐴 𝑡 𝜕𝐵𝑡 𝜕𝑥 𝑡 = 𝜕 𝜕𝐵𝑡 𝜕𝐴 𝑡 𝜕𝑥 𝑡 = 𝜕𝐴 𝑡 𝜕𝐵𝑡 = 𝐴 𝑡 𝐵𝑡 12
- 13. What is the Dynamic of Asset At? 𝜕𝐴 𝑡 = 𝜕𝐴 𝑡 𝜕𝜕 𝑑𝑥 𝑡 + 𝜕𝐴 𝑡 𝜕𝐵𝑡 𝑑𝐵𝑡 + 1 2 𝜕2 𝐴 𝑡 𝜕𝑥2 𝑑𝑥 𝑡 2 + 𝜕2 𝐴 𝑡 𝜕𝐵𝑡 𝜕𝑥 𝑑𝑥 𝑡 𝑑𝐵𝑡 = 𝐴 𝑡 𝑑𝑥 𝑡 + 𝐴 𝑡 𝐵𝑡 𝑑𝐵𝑡 + 1 2 𝐴 𝑡 𝑑𝑥 𝑡 2 + 𝐴 𝑡 𝐵𝑡 𝑑𝑥 𝑡 𝑑𝐵𝑡 = 𝐴 𝑡 𝑘 𝜃 − 𝑥 𝑡 𝑑𝑑 + 𝜂𝜂𝜔 𝑡 + 𝐴 𝑡 𝐵𝑡 𝜇𝐵𝑡 𝑑𝑑 + 𝜎𝐵𝑡 𝑑𝑧𝑡 + 1 2 𝐴 𝑡 𝜂2 𝑑𝑑 + 𝐴 𝑡 𝐵𝑡 𝜌𝜂𝜂𝐵𝑡 𝑑𝑑 13
- 14. Dynamic of Asset At 𝜕𝐴 𝑡 = 𝐴 𝑡 𝑘 𝜃 − 𝑥 𝑡 𝑑𝑑 + 𝜂𝜂𝜔 𝑡 + 𝐴 𝑡 𝜇𝑑𝑑 + 𝜎𝑑𝑧𝑡 + 1 2 𝐴 𝑡 𝜂2 𝑑𝑑 + 𝐴 𝑡 𝜌𝜂𝜂𝑑𝑑 = 𝐴 𝑡 𝑘 𝜃 − 𝑥 𝑡 + 𝜇 + 1 2 𝜂2 + 𝜌𝜌𝜌 𝑑𝑑 + 𝐴 𝑡 𝜂𝜂𝜔 𝑡 + 𝐴 𝑡 𝜎𝑑𝑧𝑡 = 𝐴 𝑡 𝜇 + 𝑘 𝜃 − 𝑥 𝑡 + 1 2 𝜂2 + 𝜌𝜌𝜌 𝑑𝑑 + 𝜎𝑑𝑧𝑡 + 𝜂𝜂𝜔 𝑡 14
- 15. Notations 𝑉𝑡: the value of a self-financing pairs trading portfolio ℎ 𝑡:the portfolio weight for stock A ℎ 𝑡 � = −ℎ 𝑡:the portfolio weight for stock B 15
- 16. Self-Financing Portfolio Dynamic 𝑑𝑉𝑡 𝑉𝑡 = ℎ 𝑡 𝑑𝐴 𝑡 𝐴 𝑡 + ℎ 𝑡 � 𝑑𝐵𝑡 𝐵𝑡 + 𝑑𝑀𝑡 𝑀𝑡 = ℎ 𝑡 𝜇 + 𝑘 𝜃 − 𝑥 𝑡 + 1 2 𝜂2 + 𝜌𝜌𝜌 𝑑𝑑 + 𝜎𝑑𝑧𝑡 + 𝜂𝜂𝜔 𝑡 − ℎ 𝑡 𝜇𝑑𝑑 + 𝜎𝑑𝑧𝑡 + 𝑟𝑟𝑟 = ℎ 𝑡 𝑘 𝜃 − 𝑥 𝑡 + 1 2 𝜂2 + 𝜌𝜌𝜌 𝑑𝑑 + 𝜂𝜂𝜔 𝑡 + 𝑟𝑟𝑟 = ℎ 𝑡 𝑘 𝜃 − 𝑥 𝑡 + 1 2 𝜂2 + 𝜌𝜌𝜌 + 𝑟 𝑑𝑑 + ℎ 𝑡 𝜂𝜂𝜔 𝑡 16
- 17. Power Utility Investor preference: 𝑈 𝑥 = 𝑥 𝛾 𝑥 ≥ 0 0 < 𝛾 < 1 17
- 18. Problem Formulation max ℎ 𝑡 𝐸 𝑉𝑇 𝛾 , s.t., 𝑉 0 = 𝑣0, x 0 = 𝑥0 𝑑𝑥𝑡 = 𝑘 𝜃 − 𝑥𝑡 𝑑𝑑 + 𝜂𝜂𝜔 𝑡 𝑑𝑉𝑡 = ℎ 𝑡 𝑑𝑥 𝑡 = ℎ 𝑡 𝑘 𝜃 − 𝑥 𝑡 𝑑𝑑 + ℎ 𝑡 𝜂𝜂𝜔 𝑡 Note that we simplify GBM to BM of 𝑉𝑡, and remove some constants. 18
- 19. Dynamic Programming 19 Consider a stage problem to minimize (or maximize) the accumulated costs over a system path. s0 s11 s12 time t = 0 t = 1 S21 S22 s23 t = 2 3 2 3 5 Cost = 𝑐3 + ∑ 𝑐𝑡 2 𝑡=0 s3 s12 5 4 3 1 5
- 20. Dynamic Programming Formulation State change: 𝑥 𝑘+1 = 𝑓𝑘 𝑥 𝑘, 𝑢 𝑘, 𝜔 𝑘 𝑘: time 𝑥 𝑘: state 𝑢 𝑘: control decision selected at time 𝑘 𝜔 𝑘: a random noise Cost: 𝑔 𝑁 𝑥 𝑁 + ∑ 𝑔 𝑘 𝑥 𝑘, 𝑢 𝑘, 𝜔 𝑘 𝑁−1 𝑘=0 Objective: minimize (maximize) the expected cost. We need to take expectation to account for the noise, 𝜔 𝑘. 20
- 21. Principle of Optimality Let 𝜋∗ = 𝜇0 ∗, 𝜇1 ∗, ⋯ , 𝜇 𝑁−1 ∗ be an optimal policy for the basic problem, and assume that when using 𝜋∗ , a give state 𝑥𝑖 occurs at time 𝑖 with positive probability. Consider the sub-problem whereby we are at 𝑥𝑖 at time 𝑖 and wish to minimize the “cost-to-go” from time 𝑖 to time 𝑁. 𝐸 𝑔 𝑁 𝑥 𝑁 + ∑ 𝑔 𝑘 𝑥 𝑘, 𝑢 𝑘, 𝜔 𝑘 𝑁−1 𝑘=𝑖 Then the truncated policy 𝜇𝑖 ∗, 𝜇𝑖+1 ∗, ⋯ , 𝜇 𝑁−1 ∗ is optimal for this sub-problem. 21
- 22. Dynamic Programming Algorithm For every initial state 𝑥0, the optimal cost 𝐽∗ 𝑥 𝑘 of the basic problem is equal to 𝐽0 𝑥0 , given by the last step of the following algorithm, which proceeds backward in time from period 𝑁 − 1 to period 0: 𝐽 𝑁 𝑥 𝑁 = 𝑔 𝑁 𝑥 𝑁 𝐽 𝑘 𝑥 𝑘 = min 𝑢 𝑘 𝐸 𝑔 𝑘 𝑥 𝑘, 𝑢 𝑘, 𝜔 𝑘 + 𝐽 𝑘+1 𝑓𝑘 𝑥 𝑘, 𝑢 𝑘, 𝜔 𝑘 22
- 23. Value function Terminal condition: 𝐺 𝑇, 𝑉, 𝑥 = 𝑉 𝛾 DP equation: 𝐺 𝑡, 𝑉𝑡, 𝑥𝑡 = m𝑎𝑎 ℎ 𝑡 𝐸 𝐺 𝑡 + 𝑑𝑑, 𝑉𝑡+𝑑𝑑, 𝑥𝑡+𝑑𝑑 𝐺 𝑡, 𝑉𝑡, 𝑥𝑡 = m𝑎𝑎 ℎ 𝑡 𝐸 𝐺 𝑡, 𝑉𝑡, 𝑥𝑡 + Δ𝐺 By Ito’s lemma: Δ𝐺 = 𝐺𝑡 𝑑𝑑 + 𝐺 𝑉 𝑑𝑑 + 𝐺 𝑥 𝑑𝑥 + 1 2 𝐺 𝑉𝑉 𝑑𝑑 2 + 1 2 𝐺 𝑥𝑥 𝑑𝑥 2 + 𝐺 𝑉𝑥 𝑑𝑉 𝑑𝑑 23
- 24. Hamilton-Jacobi-Bellman Equation Cancel 𝐺 𝑡, 𝑉𝑡, 𝑥 𝑡 on both LHS and RHS. Divide by time discretization, Δ𝑡. Take limit as Δ𝑡 → 0, hence Ito. 0 = m𝑎𝑎 ℎ 𝑡 𝐸 Δ𝐺 m𝑎𝑎 ℎ 𝑡 𝐸 𝐺𝑡 𝑑𝑑 + 𝐺 𝑉 𝑑𝑑 + 𝐺 𝑥 𝑑𝑑 + 1 2 𝐺 𝑉𝑉 𝑑𝑑 2 + 1 2 𝐺 𝑥𝑥 𝑑𝑑 2 + 𝐺 𝑉𝑉 𝑑𝑑 𝑑𝑑 = 0 The optimal portfolio position is ℎ 𝑡 ∗ . 24
- 25. HJB for Our Portfolio Value m𝑎𝑎 ℎ 𝑡 𝐸 𝐺𝑡 𝑑𝑑 + 𝐺 𝑉 𝑑𝑑 + 𝐺 𝑥 𝑑𝑑 + 1 2 𝐺 𝑉𝑉 𝑑𝑑 2 + 1 2 𝐺 𝑥𝑥 𝑑𝑑 2 + 𝐺 𝑉𝑉 𝑑𝑑 𝑑𝑑 = 0 m𝑎𝑎 ℎ 𝑡 𝐸 𝐺𝑡 𝑑𝑑 + 𝐺 𝑉 ℎ 𝑡 𝑘 𝜃 − 𝑥 𝑡 𝑑𝑑 + ℎ 𝑡 𝜂𝜂𝜔 𝑡 + 𝐺 𝑥 𝑑𝑑 + 1 2 𝐺 𝑉𝑉 ℎ 𝑡 𝑘 𝜃 − 𝑥 𝑡 𝑑𝑑 + ℎ 𝑡 𝜂𝜂𝜔 𝑡 2 + 1 2 𝐺 𝑥𝑥 𝑑𝑑 2 + 𝐺 𝑉𝑉 ℎ 𝑡 𝑘 𝜃 − 𝑥 𝑡 𝑑𝑑 + ℎ 𝑡 𝜂𝜂𝜔 𝑡 𝑑𝑑 = 0 m𝑎𝑎 ℎ 𝑡 𝐸 𝐺𝑡 𝑑𝑑 + 𝐺 𝑉 ℎ 𝑡 𝑘 𝜃 − 𝑥 𝑡 𝑑𝑑 + ℎ 𝑡 𝜂𝜂𝜔 𝑡 + 𝐺 𝑥 𝑘 𝜃 − 𝑥 𝑡 𝑑𝑑 + 𝜂𝜂𝜔 𝑡 + 1 2 𝐺 𝑉𝑉 ℎ 𝑡 𝑘 𝜃 − 𝑥 𝑡 𝑑𝑑 + ℎ 𝑡 𝜂𝜂𝜔 𝑡 2 + 1 2 𝐺 𝑥𝑥 𝑘 𝜃 − 𝑥 𝑡 𝑑𝑑 + 𝜂𝜂𝜔 𝑡 2 + 𝐺 𝑉𝑉 ℎ 𝑡 𝑘 𝜃 − 𝑥 𝑡 𝑑𝑑 + ℎ 𝑡 𝜂𝜂𝜔 𝑡 × 𝑘 𝜃 − 𝑥 𝑡 𝑑𝑑 + 𝜂𝜂𝜔 𝑡 = 0 25
- 26. Taking Expectation All 𝜂𝜂𝜔 𝑡 disapper because of the expectation operator. Only the deterministic 𝑑𝑑 terms remain. Divide LHR and RHS by 𝑑𝑑. m𝑎𝑎 ℎ 𝑡 𝐺𝑡 + 𝐺 𝑉 ℎ 𝑡 𝑘 𝜃 − 𝑥 𝑡 + 𝐺 𝑥 𝑘 𝜃 − 𝑥 𝑡 + 1 2 𝐺 𝑉𝑉 ℎ 𝑡 𝜂 2 + 1 2 𝐺 𝑥𝑥 𝜂2 +𝐺 𝑉𝑉 ℎ 𝑡 𝜂2 = 0 26
- 27. Dynamic Programming Solution Solve for the cost-to-go function, 𝐺𝑡. Assume that the optimal policy is ℎ 𝑡 ∗ . 27
- 28. First Order Condition Differentiate w.r.t. ℎ 𝑡. 𝐺 𝑉 𝑘 𝜃 − 𝑥 𝑡 + ℎ 𝑡 ∗ 𝐺 𝑉𝑉 𝜂2 + 𝐺 𝑉𝑉 𝜂2 = 0 ℎ 𝑡 ∗ = − 𝐺 𝑉 𝑘 𝜃−𝑥 𝑡 +𝐺 𝑉𝑉 𝜂2 𝐺 𝑉𝑉 𝜂2 In order to determine the optimal position, ℎ 𝑡 ∗ , we need to solve for 𝐺 to get 𝐺 𝑉, 𝐺 𝑉𝑉, and 𝐺 𝑉𝑉. 28
- 29. The Partial Differential Equation (1) 𝐺𝑡 − 𝐺 𝑉 𝐺 𝑉 𝑘 𝜃−𝑥 𝑡 +𝐺 𝑉𝑉 𝜂2 𝐺 𝑉𝑉 𝜂2 𝑘 𝜃 − 𝑥 𝑡 + 𝐺 𝑥 𝑘 𝜃 − 𝑥 𝑡 + 1 2 𝐺 𝑉𝑉 𝜂2 𝐺 𝑉 𝑘 𝜃−𝑥 𝑡 +𝐺 𝑉𝑉 𝜂2 𝐺 𝑉𝑉 𝜂2 2 + 1 2 𝐺 𝑥𝑥 𝜂2 − 𝐺 𝑉𝑉 𝜂2 𝐺 𝑉 𝑘 𝜃−𝑥 𝑡 +𝐺 𝑉𝑉 𝜂2 𝐺 𝑉𝑉 𝜂2 = 0 29
- 30. The Partial Differential Equation (2) 𝐺𝑡 − 𝐺 𝑉 𝑘 𝜃 − 𝑥 𝑡 𝐺 𝑉 𝑘 𝜃−𝑥 𝑡 +𝐺 𝑉𝑉 𝜂2 𝐺 𝑉𝑉 𝜂2 + 𝐺 𝑥 𝑘 𝜃 − 𝑥 𝑡 + 1 2 𝐺 𝑉 𝑘 𝜃−𝑥 𝑡 +𝐺 𝑉𝑉 𝜂2 2 𝐺 𝑉𝑉 𝜂2 + 1 2 𝐺 𝑥𝑥 𝜂2 − 𝐺 𝑉𝑉 𝐺 𝑉 𝑘 𝜃−𝑥 𝑡 +𝐺 𝑉𝑉 𝜂2 𝐺 𝑉𝑉 = 0 30
- 31. Dis-equilibrium Let 𝑏 = 𝑘 𝜃 − 𝑥 𝑡 . Rewrite: 𝐺𝑡 − 𝐺 𝑉 𝑏 𝐺 𝑉 𝑏+𝐺 𝑉𝑉 𝜂2 𝐺 𝑉𝑉 𝜂2 + 𝐺 𝑥 𝑏 + 1 2 𝐺 𝑉 𝑏+𝐺 𝑉𝑉 𝜂2 2 𝐺 𝑉𝑉 𝜂2 + 1 2 𝐺 𝑥𝑥 𝜂2 − 𝐺 𝑉𝑉 𝐺 𝑉 𝑏+𝐺 𝑉𝑉 𝜂2 𝐺 𝑉𝑉 = 0 Multiply by 𝐺 𝑉𝑉 𝜂2 . 𝐺𝑡 𝐺 𝑉𝑉 𝜂2 − 𝐺 𝑉 𝑏 𝐺 𝑉 𝑏 + 𝐺 𝑉𝑉 𝜂2 + 𝐺 𝑥 𝑏𝐺 𝑉𝑉 𝜂2 + 1 2 𝐺 𝑉 𝑏 + 𝐺 𝑉𝑉 𝜂2 2 + 1 2 𝐺 𝑥𝑥 𝐺 𝑉𝑉 𝜂4 − 𝐺 𝑉𝑉 𝜂2 𝐺 𝑉 𝑏 + 𝐺 𝑉𝑉 𝜂2 = 0 31
- 32. Simplification Note that −𝐺 𝑉 𝑏 𝐺 𝑉 𝑏 + 𝐺 𝑉𝑉 𝜂2 + 1 2 𝐺 𝑉 𝑏 + 𝐺 𝑉𝑉 𝜂2 2 − 𝐺 𝑉𝑉 𝜂2 𝐺 𝑉 𝑏 + 𝐺 𝑉𝑉 𝜂2 = − 1 2 𝐺 𝑉 𝑏 + 𝐺 𝑉𝑉 𝜂2 2 The PDE becomes 𝐺𝑡 𝐺 𝑉𝑉 𝜂2 + 𝐺 𝑥 𝑏𝐺 𝑉𝑉 𝜂2 + 1 2 𝐺 𝑥𝑥 𝐺 𝑉𝑉 𝜂4 − 1 2 𝐺 𝑉 𝑏 + 𝐺 𝑉𝑉 𝜂2 2 = 0 32
- 33. The Partial Differential Equation (3) 𝐺𝑡 𝐺 𝑉𝑉 𝜂2 + 𝐺 𝑥 𝑏𝐺 𝑉𝑉 𝜂2 + 1 2 𝐺 𝑥𝑥 𝐺 𝑉𝑉 𝜂4 − 1 2 𝐺 𝑉 𝑏 + 𝐺 𝑉𝑉 𝜂2 2 = 0 33
- 34. Ansatz for G 𝐺 𝑡, 𝑉, 𝑥 = 𝑓 𝑡, 𝑥 𝑉 𝛾 𝐺 𝑇, 𝑉, 𝑥 = 𝑉 𝛾 𝑓 𝑇, 𝑥 = 1 𝐺𝑡 = 𝑉 𝛾 𝑓𝑡 𝐺 𝑉 = 𝛾𝑉 𝛾−1 𝑓 𝐺 𝑉𝑉 = 𝛾 𝛾 − 1 𝑉 𝛾−2 𝑓 𝐺 𝑥 = 𝑉 𝛾 𝑓𝑥 𝐺 𝑉𝑥 = 𝛾𝑉 𝛾−1 𝑓𝑥 𝐺 𝑥𝑥 = 𝑉 𝛾 𝑓𝑥𝑥 34
- 35. Another PDE (1) 𝑉 𝛾 𝑓𝑡 𝛾 𝛾 − 1 𝑉 𝛾−2 𝑓𝜂2 + 𝑉 𝛾 𝑓𝑥 𝑏𝛾 𝛾 − 1 𝑉 𝛾−2 𝑓𝜂2 + 1 2 𝑉 𝛾 𝑓𝑥𝑥 𝛾 𝛾 − 1 𝑉 𝛾−2 𝑓𝜂4 − 1 2 𝛾𝑉 𝛾−1 𝑓𝑏 + 𝛾𝑉 𝛾−1 𝑓𝑥 𝜂2 2 = 0 Divide by 𝛾 𝛾 − 1 𝜂2 𝑉2𝛾−2. 𝑓𝑡 𝑓 + 𝑓𝑥 𝑏𝑓 + 1 2 𝑓𝑥𝑥 𝑓𝜂2 − 𝛾 2 𝛾−1 𝑓 𝑏 𝜂 + 𝑓𝑥 𝜂 2 = 0 35
- 36. Ansatz for 𝑓 𝑓𝑓𝑡 + 𝑏𝑏𝑓𝑥 + 1 2 𝜂2 𝑓𝑓𝑥𝑥 − 𝛾 2 𝛾−1 𝑏 𝜂 𝑓 + 𝜂𝑓𝑥 2 = 0 𝑓 𝑡, 𝑥 = 𝑔 𝑡 exp 𝑥𝑥 𝑡 + 𝑥2 𝛼 𝑡 = 𝑔 exp 𝑥𝑥 + 𝑥2 𝛼 𝑓𝑡 = 𝑔𝑡 exp 𝑥𝑥 + 𝑥2 𝛼 + 𝑔 exp 𝑥𝑥 + 𝑥2 𝛼 𝑥𝛽𝑡 + 𝑥2 𝛼 𝑡 𝑓𝑥 = 𝑔 exp 𝑥𝑥 + 𝑥2 𝛼 𝛽 + 2𝑥𝑥 𝑓𝑥𝑥 = 𝑔 exp 𝑥𝑥 + 𝑥2 𝛼 𝛽 + 2𝑥𝑥 2 + 𝑔 exp 𝑥𝑥 + 𝑥2 𝛼 2𝛼 𝑓𝑥 𝑓 = 𝛽 + 2𝛼𝛼 36
- 37. Boundary Conditions 𝑓 𝑇, 𝑥 = 𝑔 𝑇 exp 𝑥𝑥 𝑇 + 𝑥2 𝛼 𝑇 = 1 𝑔 𝑇 = 1 𝛼 𝑇 = 0 𝛽 𝑇 = 0 37
- 38. Yet Another PDE (1) 𝑔 exp 𝑥𝑥 + 𝑥2 𝛼 𝑔𝑡 exp 𝑥𝑥 + 𝑥2 𝛼 + 𝑔 exp 𝑥𝑥 + 𝑥2 𝛼 𝑥𝛽𝑡 + 𝑥2 𝛼 𝑡 + 𝑏𝑏 exp 𝑥𝑥 + 𝑥2 𝛼 𝑔 exp 𝑥𝑥 + 𝑥2 𝛼 𝛽 + 2𝑥𝑥 + 1 2 𝜂2 𝑔 exp 𝑥𝑥 + 𝑥2 𝛼 𝑔 exp 𝑥𝑥 + 𝑥2 𝛼 𝛽 + 2𝑥𝑥 2 + 𝑔 exp 𝑥𝑥 + 𝑥2 𝛼 2𝛼 − 𝛾 2 𝛾−1 𝑏 𝜂 𝑔 exp 𝑥𝑥 + 𝑥2 𝛼 + 𝜂𝜂 exp 𝑥𝑥 + 𝑥2 𝛼 𝛽 + 2𝑥𝑥 2 = 0 Divide by g exp 𝑥𝑥 + 𝑥2 𝛼 exp 𝑥𝑥 + 𝑥2 𝛼 . 𝑔𝑡 + 𝑔 𝑥𝛽𝑡 + 𝑥2 𝛼 𝑡 + 𝑏𝑏 𝛽 + 2𝑥𝑥 + 1 2 𝜂2 𝑔 𝛽 + 2𝑥𝑥 2 + 𝑔 2𝛼 − 𝛾 2 𝛾−1 𝑔 𝑏 𝜂 + 𝜂 𝛽 + 2𝑥𝑥 2 = 0 𝑔𝑡 + 𝑔 𝑥𝛽𝑡 + 𝑥2 𝛼 𝑡 + 𝑏𝑏 𝛽 + 2𝑥𝑥 + 1 2 𝜂2 𝑔 𝛽 + 2𝑥𝑥 2 + 𝜂2 𝑔𝑔 − 𝛾 2 𝛾−1 𝑔 𝑏 𝜂 + 𝜂 𝛽 + 2𝑥𝑥 2 = 0 38
- 39. Yet Another PDE (2) 𝜆 = 𝛾 2 𝛾−1 𝑔𝑡 + 𝑔 𝑥𝛽𝑡 + 𝑥2 𝛼 𝑡 + 𝑏𝑏 𝛽 + 2𝑥𝑥 + 1 2 𝜂2 𝑔 𝛽 + 2𝑥𝑥 2 + 𝜂2 𝑔𝑔 − 𝜆𝑔 𝑏 𝜂 + 𝜂 𝛽 + 2𝑥𝑥 2 = 0 39
- 40. Expansion in 𝑥 𝑔𝑡 + 𝑔𝑔𝛽𝑡 + 𝑔𝑥2 𝛼 𝑡 + 𝑏𝑏𝛽 + 2𝑥𝑥𝑏𝑏 + 1 2 𝜂2 𝑔 𝛽2 + 4𝑥2 𝛼2 + 4𝑥𝑥𝑥 + 𝜂2 𝑔𝑔 − 𝜆𝑔 𝑏2 𝜂2 + 𝜂2 𝛽2 + 4𝜂2 𝑥2 𝛼2 + 2𝑏𝑏 + 4𝑏𝑏𝑏 + 4𝜂2 𝑥𝑥𝑥 = 0 𝑔𝑡 + 𝑔𝑔𝛽𝑡 + 𝑔𝑥2 𝛼 𝑡 + 𝑘 𝜃 − 𝑥 𝑔𝛽 + 2𝑥𝑥𝑘 𝜃 − 𝑥 𝑔 + 1 2 𝜂2 𝑔 𝛽2 + 4𝑥2 𝛼2 + 4𝑥𝑥𝑥 + 𝜂2 𝑔𝑔 − 𝜆𝑔 𝑘2 𝜃−𝑥 2 𝜂2 + 𝜂2 𝛽2 + 4𝜂2 𝑥2 𝛼2 + 2𝑘 𝜃 − 𝑥 𝛽 + 4𝑘 𝜃 − 𝑥 𝑥𝑥 + 4𝜂2 𝑥𝑥𝑥 = 0 𝑔𝑡 + 𝑔𝑔𝛽𝑡 + 𝑔𝑥2 𝛼 𝑡 + 𝑘𝑘𝛽𝛽 − 𝑘𝑘𝛽𝑥 + 2𝑥𝑥𝑘𝑘𝜃 − 2𝛼𝑘𝑘𝑥2 + 1 2 𝜂2 𝑔𝛽2 + 2𝜂2 𝑔𝑥2 𝛼2 + 2𝜂2 𝑔𝑔𝑔𝑔 + 𝜂2 𝑔𝑔 − 𝜆𝑔 𝜂2 𝑘2 𝜃2 + 2 𝜆𝑔 𝜂2 𝑘2 𝜃𝑥 − 𝜆𝑔 𝜂2 𝑘2 𝑥2 − 𝜆𝑔𝜂2 𝛽2 − 4𝜆𝑔𝜂2 𝑥2 𝛼2 − 2𝜆𝑔𝑔𝛽𝛽 + 2𝜆𝑔𝑔𝛽𝑥 − 4𝜆𝑔𝑔𝜃𝜃𝜃 + 4𝜆𝑔𝑔𝑥2 𝛼 − 4𝜆𝑔𝜂2 𝑥𝑥𝑥 = 0 40
- 41. Grouping in 𝑥 𝑔𝑡 + 𝑘𝑘𝛽𝛽 + 1 2 𝜂2 𝑔𝛽2 + 𝜂2 𝑔𝑔 − 𝜆𝑔 𝜂2 𝑘2 𝜃2 − 𝜆𝑔𝜂2 𝛽2 − 2𝜆𝑔𝑔𝛽𝛽 + 𝑔𝛽𝑡 − 𝑘𝑘𝛽 + 2𝛼𝑘𝑘𝜃 + 2𝜂2 𝑔𝑔𝑔 + 2 𝜆𝑔 𝜂2 𝑘2 𝜃 + 2𝜆𝑔𝑔𝛽 − 4𝜆𝑔𝑔𝜃𝜃 − 4𝜆𝑔𝜂2 𝛼𝛼 𝑥 + 𝑔𝛼 𝑡 − 2𝛼𝑘𝑘 + 2𝜂2 𝑔𝛼2 − 𝜆𝑔 𝜂2 𝑘2 − 4𝜆𝑔𝜂2 𝛼2 + 4𝜆𝑔𝑔𝛼 𝑥2 = 0 41
- 42. The Three PDE’s (1) 𝑔𝛼 𝑡 − 2𝛼𝑘𝑘 + 2𝜂2 𝑔𝛼2 − 𝜆𝑔 𝜂2 𝑘2 − 4𝜆𝑔𝜂2 𝛼2 + 4𝜆𝑔𝑔𝛼 = 0 𝑔𝛽𝑡 − 𝑘𝑘𝛽 + 2𝛼𝑘𝑘𝜃 + 2𝜂2 𝑔𝑔𝑔 + 2 𝜆𝑔 𝜂2 𝑘2 𝜃 + 2𝜆𝑔𝑔𝛽 − 4𝜆𝑔𝑔𝜃𝜃 − 4𝜆𝑔𝜂2 𝛼𝛼 = 0 𝑔𝑡 + 𝑘𝑘𝛽𝛽 + 1 2 𝜂2 𝑔𝛽2 + 𝜂2 𝑔𝑔 − 𝜆𝑔 𝜂2 𝑘2 𝜃2 − 𝜆𝑔𝜂2 𝛽2 − 2𝜆𝑔𝑔𝛽𝛽 = 0 42
- 43. PDE in 𝛼 𝛼 𝑡 + 2𝜂2 − 4𝜆𝜆2 𝛼2 + 4𝜆𝑘 − 2𝑘 𝛼 − 𝜆 𝜂2 𝑘2 = 0 𝛼 𝑡 = 𝜆 𝜂2 𝑘2 + 2𝑘 1 − 2𝜆 𝛼 + 2𝜂2 2𝜆 − 1 𝛼2 43
- 44. PDE in 𝛽, 𝛼 𝛽𝑡 − 𝑘𝛽 + 2𝜂2 𝛼𝛼 + 2𝜆𝑘𝛽 − 4𝜆𝜆2 𝛼𝛼 − 4𝜆𝑘𝜃𝜃 + 2 𝜆 𝜂2 𝑘2 𝜃 + 2𝛼𝑘𝜃 = 0 𝛽𝑡 = 𝑘 − 2𝜂2 𝛼 − 2𝜆𝑘 + 4𝜆𝜆2 𝛼 𝛽 + 4𝜆𝑘𝜃𝜃 − 2 𝜆 𝜂2 𝑘2 𝜃 − 2𝛼𝑘𝜃 44
- 45. PDE in 𝛽, 𝛼, 𝑔 𝑔𝑡 + 𝑘𝑘𝛽𝛽 + 1 2 𝜂2 𝑔𝛽2 + 𝜂2 𝑔𝑔 − 𝜆𝑔 𝜂2 𝑘2 𝜃2 − 𝜆𝑔𝜂2 𝛽2 − 2𝜆𝑔𝑔𝛽𝛽 = 0 𝑔𝑡 = −𝑘𝑘𝛽𝛽 − 1 2 𝜂2 𝑔𝛽2 − 𝜂2 𝑔𝑔 + 𝜆𝑔 𝜂2 𝑘2 𝜃2 + 𝜆𝑔𝜂2 𝛽2 + 2𝜆𝑔𝑔𝛽𝛽 𝑔𝑡 = 𝑔 −𝑘𝛽𝛽 − 1 2 𝜂2 𝛽2 − 𝜂2 𝛼 + 𝜆 𝜂2 𝑘2 𝜃2 + 𝜆𝜆2 𝛽2 + 2𝜆𝑘𝛽𝛽 45
- 46. Riccati Equation A Riccati equation is any ordinary differential equation that is quadratic in the unknown function. 𝛼 𝑡 = 𝜆 𝜂2 𝑘2 + 2𝑘 1 − 2𝜆 𝛼 + 2𝜂2 2𝜆 − 1 𝛼2 𝛼 𝑡 = 𝐴0 + 𝐴1 𝛼 + 𝐴2 𝛼2 46
- 47. Solving a Riccati Equation by Integration Suppose a particular solution, 𝛼1, can be found. 𝛼 = 𝛼1 + 1 𝑧 is the general solution, subject to some boundary condition. 47
- 48. Particular Solution Either 𝛼1 or 𝛼2 is a particular solution to the ODE. This can be verified by mere substitution. 𝛼1,2 = −𝐴1± 𝐴1 2 −4𝐴2 𝐴0 2𝐴2 48
- 49. 𝑧 Substitution Suppose 𝛼 = 𝛼1 + 1 𝑧 . 1 𝑧 ̇ = 𝐴0 + 𝐴1 𝛼1 + 1 𝑧 + 𝐴2 𝛼1 + 1 𝑧 2 = 𝐴0 + 𝐴1 𝛼1 + 𝐴1 1 𝑧 + 𝐴2 𝛼1 2 + 𝐴2 1 𝑧2 + 2𝐴2 𝛼1 𝑧 = 𝐴0 + 𝐴1 𝛼1 + 𝐴1 1 𝑧 + 𝐴2 𝛼1 2 + 𝐴2 1 𝑧2 + 2𝐴2 𝛼1 𝑧 = 𝐴0 + 𝐴1 𝛼1 + 𝐴2 𝛼1 2 + 𝐴1+2𝛼1 𝐴2 𝑧 + 𝐴2 𝑧2 goes to 0 by the definition of 𝛼1 49
- 50. Solving 𝑧 1 𝑧 ̇ = 𝐴1+2𝛼1 𝐴2 𝑧 + 𝐴2 𝑧2 − 1 𝑧2 𝑧̇ = 𝐴1+2𝛼1 𝐴2 𝑧 + 𝐴2 𝑧2 1st order linear ODE 𝑧̇ + 𝐴1 + 2𝛼1 𝐴2 𝑧 = −𝐴2 𝑧 𝑡 = −𝐴2 𝐴1+2𝛼1 𝐴2 + 𝐶exp − 𝐴1 + 2𝛼1 𝐴2 𝑡 50
- 51. Solving for 𝛼 𝛼 = 𝛼1 + 1 −𝐴2 𝐴1+2𝛼1 𝐴2 +𝐶exp− 𝐴1+2𝛼1 𝐴2 𝑡 boundary condition: 𝛼 𝑇 = 0 𝛼1 + 1 −𝐴2 𝐴1+2𝛼1 𝐴2 +𝐶exp− 𝐴1+2𝛼1 𝐴2 𝑇 = 0 𝐶exp − 𝐴1 + 2𝛼1 𝐴2 𝑇 = − 1 𝛼1 + 𝐴2 𝐴1+2𝛼1 𝐴2 𝐶 = exp 𝐴1 + 2𝛼1 𝐴2 𝑇 𝐴2 𝐴1+2𝛼1 𝐴2 − 1 𝛼1 51
- 52. 𝛼 Solution (1) 𝛼 = 𝛼1 + 1 −𝐴2 𝐴1+2𝛼1 𝐴2 +𝐶exp− 𝐴1+2𝛼1 𝐴2 𝑡 = 𝛼1 + 1 −𝐴2 𝐴1+2𝛼1 𝐴2 +exp 𝐴1+2𝛼1 𝐴2 𝑇 𝐴2 𝐴1+2𝛼1 𝐴2 − 1 𝛼1 exp − 𝐴1+2𝛼1 𝐴2 𝑡 = 𝛼1 + 1 −𝐴2 𝐴1+2𝛼1 𝐴2 +exp 𝐴1+2𝛼1 𝐴2 𝑇−𝑡 𝐴2 𝐴1+2𝛼1 𝐴2 − 1 𝛼1 = 𝛼1 + 𝛼1 𝐴1+2𝛼1 𝐴2 −𝛼1 𝐴2+exp 𝐴1+2𝛼1 𝐴2 𝑇−𝑡 𝛼1 𝐴2−𝐴1−2𝛼1 𝐴2 52
- 53. 𝛼 Solution (2) 𝛼 = 𝛼1 + 𝛼1 𝐴1+2𝛼1 𝐴2 −𝛼1 𝐴2+exp 𝐴1+2𝛼1 𝐴2 𝑇−𝑡 −𝐴1−𝛼1 𝐴2 = 𝛼1 1 − 𝐴1+2𝛼1 𝐴2 𝐴2+exp 𝐴1+2𝛼1 𝐴2 𝑇−𝑡 𝐴1 𝛼1 +𝐴2 = 𝛼1 1 − 𝐴1 𝐴2 +2𝛼1 1+ 𝐴1 𝛼1 𝐴2 +1 exp 𝐴1+2𝛼1 𝐴2 𝑇−𝑡 53
- 54. 𝛼 Solution (3) 𝛼 𝑡 = 𝑘 2𝜂2 1 − 1 − 𝛾 + 2 1−𝛾 1+ 1− 2 1− 1−𝛾 exp 2𝑘 1−𝛾 𝑇−𝑡 54
- 55. Solving 𝛽 𝛽𝑡 = 𝑘 − 2𝜂2 𝛼 − 2𝜆𝑘 + 4𝜆𝜆2 𝛼 𝛽 + 4𝜆𝑘𝜃𝜃 − 2 𝜆 𝜂2 𝑘2 𝜃 − 2𝛼𝑘𝜃 Let 𝜏 = 𝑇 − 𝑡 𝛽̂ 𝜏 = 𝛽 𝑇 − 𝑡 𝛽̂ 𝜏 𝜏 = −𝛽𝑡 𝜏 −𝛽̂ 𝜏 𝜏 = 𝛽𝑡 𝜏 = 𝑘 − 2𝜂2 𝛼 𝜏 − 2𝜆𝑘 + 4𝜆𝜆2 𝛼 𝜏 𝛽 𝜏 + 4𝜆𝑘𝜃𝜃 𝜏 − 2 𝜆 𝜂2 𝑘2 𝜃 − 2𝛼 𝜏 𝑘𝜃 𝛽̂ 𝜏 𝜏 = −𝑘 + 2𝜂2 𝛼� + 2𝜆𝑘 − 4𝜆𝜆2 𝛼� 𝛽̂ + −4𝜆𝑘𝜃𝛼� + 2 𝜆 𝜂2 𝑘2 𝜃 + 2𝛼�𝑘𝜃 𝛽̂ 𝜏 𝜏 = 2𝜆 − 1 𝑘 + 2𝜂2 𝛼� 1 − 2𝜆 𝛽̂ + 2𝛼�𝑘𝜃 1 − 2𝜆 + 2 𝜆 𝜂2 𝑘2 𝜃 55
- 56. First Order Non-Constant Coefficients 𝛽̂ 𝜏 = 𝐵1 𝛽̂ + 𝐵2 𝐵1 𝜏 = 2𝜆 − 1 𝑘 + 2𝜂2 𝛼� 1 − 2𝜆 𝐵2 𝜏 = 2𝛼�𝑘𝜃 1 − 2𝜆 + 2 𝜆 𝜂2 𝑘2 𝜃 56
- 57. Integrating Factor (1) 𝛽̂ 𝜏 − 𝐵1 𝛽̂ = 𝐵2 We try to find an integrating factor 𝜇 = 𝜇 𝜏 s.t. 𝑑 𝑑𝜏 𝜇𝛽̂ = 𝜇 𝑑𝛽� 𝑑𝜏 + 𝛽̂ 𝑑𝜇 𝑑𝜏 = 𝜇𝐵2 Divide LHS and RHS by 𝜇𝛽̂. 1 𝛽� 𝑑𝛽� 𝑑𝑑 + 1 𝜇 𝑑𝜇 𝑑𝜏 = 𝐵2 𝛽� By comparison, −𝐵1 = 1 𝜇 𝑑𝜇 𝑑𝜏 57
- 58. Integrating Factor (2) ∫ −𝐵1 𝑑𝜏 = ∫ 𝑑𝜇 𝜇 = log 𝜇 + 𝐶 𝜇 = exp ∫ −𝐵1 𝑑𝜏 𝜇𝛽̂ = ∫ 𝜇𝐵2 𝑑𝜏 + 𝐶 𝛽̂ = ∫ 𝜇𝐵2 𝑑𝜏+𝐶 𝜇 𝛽̂ = ∫ exp ∫ −𝐵1 𝑑𝑢 𝐵2 𝑑𝜏+𝐶 exp ∫ −𝐵1 𝑑𝜏 58
- 59. 𝛽̂ Solution 𝛽̂ = ∫ exp ∫ −𝐵1 𝑢 𝑑𝑑 𝜏 0 𝐵2 𝑠 𝑑𝑠 𝜏 0 exp ∫ −𝐵1 𝑢 𝑑𝑢 𝜏 0 + 𝐶 𝛽̂ 𝜏 = exp ∫ 𝐵1 𝑢 𝑑𝑑 𝜏 0 ∫ exp ∫ −𝐵1 𝑢 𝑑𝑑 𝑠 0 𝐵2 𝑠 𝑑𝑑 𝜏 0 + 𝐶 59
- 60. 𝐵1, 𝐵2 ∫ 𝐵1 𝑠 𝑑𝑑 𝜏 0 = ∫ 2𝜆 − 1 𝑘 + 2𝜂2 1 − 2𝜆 𝛼 𝑠 𝑑𝑑 𝜏 0 𝐼2 = ∫ exp ∫ −𝐵1 𝑢 𝑑𝑑 𝑠 0 𝐵2 𝑠 𝑑𝑑 𝜏 0 60
- 61. 𝛽 Solution 𝛽 𝑡 = 𝑘𝜃 𝜂2 1 + 1 − 𝛾 exp 2𝑘 1−𝛾 𝑇−𝑡 −1 1+ 1− 2 1− 1−𝛾 exp 2𝑘 1−𝛾 𝑇−𝑡 61
- 62. Solving 𝑔 𝑔𝑡 = 𝑔 −𝑘𝛽𝛽 − 1 2 𝜂2 𝛽2 − 𝜂2 𝛼 + 𝜆 𝜂2 𝑘2 𝜃2 + 𝜆𝜆2 𝛽2 + 2𝜆𝑘𝛽𝛽 With𝛼 and 𝛽 solved, we are now ready to solve 𝑔𝑡. 𝑔 𝑡 𝑔 = 𝐺 𝑡 𝑑 𝑑𝑠 log 𝑔𝑡 = 𝑔 𝑡 𝑔 = 𝐺 log 𝑔𝑡 = ∫ 𝐺𝐺𝐺 + 𝐶 𝑔𝑡 = 𝐶exp ∫ 𝐺𝑑𝑑 = exp − ∫ 𝐺𝐺𝐺 𝑇 0 exp ∫ 𝐺𝐺𝐺 𝑡 0 𝑔𝑡 = exp − ∫ 𝐺𝐺𝐺 𝑇 𝑡 62
- 63. Computing the Optimal Position ℎ 𝑡 ∗ = − 𝐺 𝑉 𝑘 𝜃−𝑥 𝑡 +𝐺 𝑉𝑉 𝜂2 𝐺 𝑉𝑉 𝜂2 = − 𝛾𝑉 𝛾−1 𝑓 𝑘 𝜃−𝑥 𝑡 +𝛾𝑉 𝛾−1 𝑓𝑥 𝜂2 𝛾 𝛾−1 𝑉 𝛾−2 𝑓𝜂2 = − 𝑉𝑓 𝑘 𝜃−𝑥 𝑡 +𝑉𝑓𝑥 𝜂2 𝛾−1 𝑓𝜂2 = − 𝑉 𝛾−1 𝜂2 𝑓 𝑘 𝜃−𝑥 𝑡 +𝑓𝑥 𝜂2 𝑓 = − 𝑉 𝛾−1 𝜂2 𝑘 𝜃 − 𝑥𝑡 + 𝑓𝑥 𝑓 𝜂2 = 𝑉 1−𝛾 𝜂2 𝑘 𝜃 − 𝑥𝑡 + 𝜂2 𝛽 + 2𝛼𝛼 = 𝑉 1−𝛾 − 𝑘 𝜂2 𝑥𝑡 − 𝜃 + 2𝛼𝛼 + 𝛽 63
- 64. The Optimal Position ℎ 𝑡 ∗ = 𝑉𝑡 1−𝛾 − 𝑘 𝜂2 𝑥 𝑡 − 𝜃 + 2𝛼 𝑡 𝑥 𝑡 + 𝛽 𝑡 ℎ 𝑡 ∗~ − 𝑘 𝜂2 𝑥 𝑡 − 𝜃 64
- 65. P&L for Simulated Data The portfolio increases from $1000 to $4625 in one year. 65
- 66. Parameter Estimation Can be done using Maximum Likelihood. Evaluation of parameter sensitivity can be done by Monte Carlo simulation. In real trading, it is better to be conservative about the parameters. Better underestimate the mean-reverting speed Better overestimate the noise To account for parameter regime changes, we can use: a hidden Markov chain model moving calibration window 66
- 68. Two-State Markov Model 68 𝑑𝑆𝑟 = 𝑆𝑟 𝜇 𝛼 𝑟 𝑑𝑑 + 𝜎𝜎𝐵𝑟 𝛼 𝑟 = 0 DOWN TREND 𝛼 𝑟 = 1 UP TREND q p 1-q 1-p
- 69. Buying and Selling Decisions 69 𝑡 ≤ 𝜏1 0 ≤ 𝜈1 0 ≤ 𝜏2 0 ≤ 𝜈2 0 ≤ ⋯ ≤ 𝜏 𝑛 0 ≤ 𝜈 𝑛 0 ≤ ⋯ ≤ 𝑇
- 70. Optimal Trend Following Strategy 70
- 71. Trading SSE 2001 – 2011 71