Introduction to Algorithmic Trading Strategies Lecture 3

Transcription

1 Introduction to Algorithmic Trading Strategies Lecture 3 Trend Following Haksun Li haksun.li@numericalmethod.com

2 References Introduction to Stochastic Calculus with Applications. Fima C Klebaner. nd Edition. Estimating continuous time transition matrices from discretely observed data. Inamura, Yasunari. April 006. Optimal Trend Following Trading Rules. Dai, Min and Zhang, Qing and Zhu, Qiji Jim. 011.

3 3 Stochastic Calculus

4 Brownian Motion Independence of increments. dd = B t B s is independent of any history up to s. Normality of increments. dd is normally distributed with mean 0 and variance t s. Continuity of paths. B t is a continuous function of t. 4

5 Stochastic vs. Newtonian Calculus Newtonian calculus: when you zoom in a function enough, the little segment looks like a straight line, hence the approximation: dd = f dd Derivative exists. The function is differentiable. Stochastic calculus: no matter how much you zoom in or how small dd is, the function still looks very zig-zag and random. It is nothing like a straight line. For Brownian motion, however much you zoom in, how small the segment is, it still just looks very much like a Brownian motion. The function is therefore no where differentiable. 5

6 Examples not a trend not mean reversion 6

7 dbdb dd = dddd = dd t 0 db t n i=1 Z n,i Z n,i is of N 0, t n t db 0 t for all i. sum of variances of Z n,i t Making n or dd 0, we have t db 0 t = t, convergence in probability db t = dd, in differential form dddt = 0 dtdd = 0 7

8 Asset Price Model Want to model asset price movement. Change in price = dd Change in price is not too meaningful as $1 change in a penny stock is more significant than $1 change in GOOG. Return = dd S Model return using two parts. Deterministic: μdd, the predictable part. E.g., fixed deposit interest rate. Random/stochastic: σdd, where σ is the volatility of returns and db is a sample from a probability distribution, e.g., Normal. 8

9 Geometric Brownian Motion Asset price: dd S = μμμ + σσb db: normally distributed E db = 0 Variance = dd. It is intuitive that db should be scaled by dd otherwise the (random) return drawn would be too big from any Normal distribution for dd 0. db = φ dd, where φ is a standard Normal distribution. Reasonably good model for stocks and indices. Real data have more big rises and falls than this model predicts, i.e., extreme events. 9

10 GBM Properties Markov property: the distribution of the next price S + dd depend only on the current price S. E ds = E μsdd + σsdb = E μμμμ + E σσσb = μμμμ + σσ E db = μμμμ Var dd = E dd dbdd = 0 dddd = 0 dbdb = dd E dd = σ S dd 10

11 Stochastic Differential Equation Both μ and σ can be as simple as constants, deterministic functions of t and S, or as complicated as stochastic functions adapted to the filtration generated by S t. ds t = μμμ + σσb t ds t = μ t, S t dd + σ t, S t db t 11

12 Univariate Ito s Lemma Assume dx t = μ t dd + σ t db t f t, X t is twice differentiable of two real variables We have dd t, X t = + μ t + σ t f x dd + σ t db t 1

13 Proof Taylor series df t, X = f t dd + f X dx + 1 f ttdddd + f tx dddx + f XX dxdt + f XX dddx = f t dd + f X dd + 1 f XXdddd = f t dd + f X μdd + σdb + 1 f XX μdd + σdd = f t dd + f X μdd + σdd + 1 f XX μdd + σdd = f t dd + f X μdd + σdd + 1 f XX μ dd + σ db + μddσdd = f t dd + f X μdd + σdd + 1 f XXσ dd = f t dd + f X μdd + σdd + 1 f XXσ dd = f t + μf X + 1 σ f XX dd + σf X dd 13

14 Log Example For G.B.M., dx t = μx t dd + σx t dz t, d log X t =? f x = log x f t = 0 f x = 1 x f x = 1 x d log X t = μx t 1 + σx t X t 1 X t dd + σx t 1 X t db t = μ σ dd + σdb t 14

15 Exp Example dx t = μdd + σdz t, de X t =? f x = e x f t = 0 f x = ex f x = ex de X t = μe X t + σ ex t dd + σe X tdb t = e X t dex t e X t μ + σ = μ + σ dd + σdb t dd + σdb t 15

16 Stopping Time A stopping time is a random time that some event is known to have occurred. Examples: when a stock hits a target price when you run out of money when a moving average crossover occurs When a stock has bottomed out is NOT a stopping time because you cannot tell when it has reached its minimum without future information. 16

17 17 Optimal Trend Following Strategy

18 Asset Model Two state Markov model for a stock s prices: BULL and BEAR. ds r = S r μ αr dd + σσb r, t r T < The trading period is between time t, T. α r = 1, are the two Markov states that indicates the BULL and BEAR markets. μ 1 > 0 μ < 0 Q = λ 1 λ 1, the generator matrix for the Markov λ λ chain. 18

19 Generator Matrix From transition matrix to generator matrix: P t, t + Δt = I + Δtt + o Δt We can estimate Q by estimating P. P t, s = P t, t + mδt I + Δtt m = P t, t + Δt = exp Δtt P t exp ΔtQ Q 1 Δt log P t E.g., Δt = 1 5 I + s t m Q m = exp s t Q 19

20 Optimal Stopping Times i = 0, Λ 0 = τ 1, ν 1, τ, ν, no position, bank the $ no position, LONG LONG LONG bank the $ no position, bank the $ t τ 1 ν 1 τ ν τ n ν n BUY SELL BUY SELL BUY SELL T i = 0, Λ 0 = ν 1, τ, ν, τ 3, 0

21 Parameters ρ 0, interest free rate S t = S, the initial stock price i = 0,1, the net position K b, the transaction cost for BUYs in percentage K s, the transaction cost for SELLs in percentage 1

22 Expected Return (starting flat) When i = 0, E 0,t R t = n=1 E t e ρ τ 1 t S νn S τ n 1 K s 1+K b I τ n<t e ρ τ n+1 υ n You are long between τ n and ν n and the return is determined by the price change discounted by the commissions. You are flat between ν n and τ n+1 and the money grows at the risk free rate.

23 Expected Return (starting long) When i = 1, E 1,t R t = E t S ν 1 S eρ τ ν 1 1 K s S νn n= You sell the long position at ν 1. S τ n 1 K s 1+K b I τ n<t e ρ τ n+1 υ n 3

24 Value Functions It is easier to work with the log of the returns. n=1 J 0 S, α, t, Λ 0 = E t ρ τ 1 t + log S νn S τ n + I τn <T log 1 K s 1+K b + ρ τ n+1 υ n J 1 S, α, t, Λ 1 = E t log S ν1 + ρ τ S ν 1 + log 1 K s + log S νn n= S τ n + I τn <T log 1 K s 1+K b + ρ τ n+1 υ n 4

25 Conditional Probability of Bull Market p r = P α r = 1 σ S u 0 u r p r is therefore a random process driven by the same Brownian motion that drives S u. dp r = λ 1 + λ p r + λ dd + μ 1 μ p r 1 p r db r = d log S r μ 1 μ p r +μ σ σ dd p t+1 = p t + g p t Δt + μ 1 μ p t 1 p t σ log S t+1 S t σ db r g p = λ 1 + λ p + λ μ 1 μ p 1 p μ 1 μ p+μ σ σ Make p t+1 between 0 and 1 if it falls outside the bounds. 5

26 Objective Find an optimal trading sequence (the stopping times) so that the value functions are maximized. V i p, t = sup Λi J i S, p, t, Λ i V i : the maximum amount of expected returns 6

27 Realistic Expectation of Returns Lower bounds: V 0 p, t ρ T t V 1 p, t ρ T t + log 1 K s Upper bound: V i p, t μ 1 σ No trading zones: T t One should never buy when ρ μ 1 σ 7

28 Principal of Optimality Given an optimal trading sequence, Λ i, starting from time t, the truncated sequence will also be optimal for the same trading problem starting from any stopping times τ n or υ n. 8

29 Coupled Value Functions V 0 p, t = sup E t ρ τ 1 t log 1 + K b + V 1 p τ1, τ 1 τ 1 V 1 p, t = sup ν 1 E t log S ν1 S t + log 1 K s + V 0 p ν1, ν 1 9

30 Hamilton-Jacobi-Bellman Equations min LV 0 ρ, V 0 V 1 + log 1 + K b = 0 min LV 1 f ρ, V 1 V 0 log 1 K s = 0 V with terminal conditions: 0 p, T = 0 V 1 p, T = log 1 K s L = t + 1 μ 1 μ p 1 p σ pp + λ 1 + λ p + λ p 30

31 Penalty Formulation Approach The HJB formulation is equivalent to this system of PDEs. LV 0 h ρ = ξ V 1 V 0 log 1 + K + b LV 1 f ρ = ξ V 0 V 1 + log 1 K + s ξ is a penalization factor such that ξ. h ρ = ρ 31

32 Trading Boundaries BB = p 0,1 [0, T) p p b t SR = p 0,1 [0, T) p p s t 3

33 SP500 33

34 SSE 34

35 Problems The buy entry signals are always delayed (as expected). The market therefore needs to bull long enough for the strategy to make profit. If the bear market comes in too fast and hard, the exit entry may not come in soon enough. A proper stoploss from max drawdown may help. 35

36 Finding the Boundaries At any time t, we need to find p b and p s to determine whether we buy or sell or go flat. p b is determined by V 1 p b, t V 0 p b, t p s is determined by = log 1 + K b V 1 p s, t V 0 p s, t = log 1 K s Need to compute V 0 p, t and V 1 p, t for all p and t. Solve the coupled PDEs. 36

37 Discretization of L ρ, t 0,1 [0,1) ρ, t as ρ j, t n j 0,, M ρ 0 = 0 ρ M = 0 n 1,, N t 1 = 0 t N = 1 V i V n+1 n i,j Vi,j t t 37

38 The Grid ρ M =1 ρ j V i ρ j, t n ρ 0 =0 t 1 =0 t 1 t n t N =1 38

39 Crank-Nicolson Scheme V i ρ 1 n+1 n+1 V i,j+1 Vi,j 1 ρ + V n n i,j+1 V i,j 1 ρ V i ρ 1 n+1 n+1 n+1 V i,j+1 Vi,j +Vi,j 1 ρ + V i,j+1 n V i,j n +V i,j 1 n ρ 39

40 RHS ξ V 1 V 0 log 1 + K b + = n+1 n+1 n+1 V 1,j V0,j ξ 0,j k b = log 1 + K b + V 1,j n n V 0,j ξ V 0 V 1 + log 1 K s + = n+1 V 0,j ξ 1,j n+1 n+1 V1,j k s = log 1 K s + V 0,j n n V 1,j k b + k s 40

41 Penalty n+1 ξ 0,j = ξ t, if V 1,j n+1 ξ 1,j = ξ t, if V 0,j n+1 n+1 V0,j n+1 n+1 V1,j + V 1,j n n V 0,j 0, otherwise n n V 1,j + V 0,j 0, otherwise k b > 0 + k s > 0 41

42 The Discretized PDEs a 1,j V n+1 0,j+1 + a 0,j V n+1 0,j + a 1,j V n+1 0,j 1 = c n 0,j n+1 n+1 n+1 V 1,j V0,j + ξ 0,j + V 1,j n n V 0,j k b a 1,j V n+1 1,j+1 + a 0,j V n+1 1,j + a 1,j V n+1 1,j 1 = c n 1,j n+1 n+1 n+1 V 0,j V1,j + ξ 1,j j 1,, M 1 p = 0, p = 1 are excluded. + V 0,j n n V 1,j + k s 4

43 The Coefficients a 1,j = 1 σ p a 0,j = 1 + σ p t p + 1 μ p t p t t p a 1,j = 1 σ p 1 μ p p p c n n 0,j = a 1,j V 0,j+1 + a 0,j V n n 0,j a 1,j V 0,j 1 + h j p t c n n 1,j = a 1,j V 1,j+1 + a 0,j V n n 1,j a 1,j V 1,j 1 + f j p t t 43

44 Boundary Conditions j = 0, p = 0 a 0,0 =1 a 1,0 = 0 c n 0,0 = V n 0,1 c n 1,0 = V n 1,1 j = M, p = 1 a 0,M =1 a 1,M = 0 n c 0,M n c 1,M n = V 0,M 1 n = V 1,M 1 + h 0 t + f 0 t + h 1 t + f 1 t n = 0, t = 0, for all j 0,, M V 0,j = 0 V 1,j = k s 44

45 System in Matrix Form An V 0 n+1 = c 0 n A n V 1 n+1 = c 1 n A n = a 0,0 a 1, a 0,1 0 a 1,1 a 1,M 1 0 a 0,M a 1,M 1 a 0,M 45

46 Solving the System of Equations We already know the values of V 0 0 and V 1 0. Starting from n = 0, using A n, we iteratively solve for V 0 n+1 and V 1 n+1 until n = M 1. There are totally M systems of equations to solve. The equations are linear in the unknowns V n+1 0,j and V n+1 n+1 1,j if and only if ξ 0,j = 0. When the equations are linear, we can use Thomas algorithm to solve for a tri-diagonal system of linear equations. Otherwise, we use an iterative scheme. 46

47 Iterative Scheme 1 st step, initialization, k = 0 n+1 Assume ξ i,j k th step 0 = 0. Solve for V i n+1 0. Using V n+1 n+1 i k 1, update ξ i,j n+1 When ξ i,j Repeat until convergence. k. k > 0, adjust the A n k and c i n k. 47

48 Adjustments (A) A n k = a 0,0 + ξδt a 1, a 0,1 + ξδt 0 a 1,1 a 1,M 1 0 a 0,M 1 + ξδt a 1,M 1 a 0,M + ξδt 48

49 Adjustments (c) c 0 n k = n c 0,M ξδt V 1,0 n+1 k 1 c n 0,1 + ξδt V 1,1 n+1 k 1 + ξδt n+1 V 1,M 1 k 1 + V 1,0 n n V 0,0 k b + V 1,1 n n V 0,1 k b + V n n 1,M 1 V 0,M 1 k b 1 + ξδt V 1,M n+1 k 1 + V 1,M n n V 0,M k b c 1 n k = n c 1,M ξδt V 0,0 n+1 k 1 c n 1,1 + ξδt V 0,1 n+1 k 1 + ξδt n+1 V 0,M 1 k 1 + V 0,0 n n V 1,0 + k s + V 0,1 n n V 1,1 + k s + V n n 0,M 1 V 1,M 1 + k s 1 + ξδt n+1 V 0,M k 1 + V 0,M n n V 1,M + k s 49

50 Convergence V i n+1 k V i n+1 k 1 1 < ε 50

51 51 Miscellaenous

52 Model Estimation Estimate the transition probabilities in the HMM by EM. Estimate μ and σ. The log of the prices are Gaussian. σ : sample variance σ = σ Δt σ = σ 1 Δt m i : sample mean m i = μ i σ μ i = 1 Δt m i + σ Δt 5