Switching Regime Estimation

Transcription

1 Switching Regime Estimation Series de Tiempo BIrkbeck March 2013 Martin Sola (FE) Markov Switching models 01/13 1 / 52

2 The economy (the time series) often behaves very different in periods such as booms and recessions. Markov Switching models are a useful way of characterizing this phenomena. Uses of Markov Switching Models: Estimating models which are state dependent. Use the structure of the model to assess the probability that a state takes place, say a boom. Incorporate this feature in Rational Expectation models, pricing, derivatives, etc. STRUCTURE OF THE LECTURE Definition and Properties. Estimation and testing Multivariate extensions and a R.E. example Martin Sola (FE) Markov Switching models 01/13 2 / 52

3 Introduction Underlying assumption in all econometric models is that all observations have been drawn from the same distribution conditional on some constant parameter set The standard approach consists of trying to detect the existence of the regime changes and then imposing dummies. Models with too many dummies Martin Sola (FE) Markov Switching models 01/13 3 / 52

4 How should we model a change in the process? Suppose that the series under scrutiny has a break in its unconditional mean at time t 1. For data prior to t 1 we might use: y t µ 0 = φ(y t 1 µ 0 ) + ε t for t 1 < t and for data after t 1 y t µ 1 = φ(y t 1 µ 1 ) + ε t for t t 1 Even if this captures the break, it is not a satisfactory model: A complete time series model would include a description of the probability law governing the change from µ 0 to µ 1. Martin Sola (FE) Markov Switching models 01/13 4 / 52

5 We might consider the process to be influenced by an unobserved random variable x t x t is the state or regime In the example above, we could regard x t as: Thus, we could write x t = { 0 if the process has mean µ0 1 if the process has mean µ 1 y t µ xt = φ 1 (y t 1 µ xt 1 ) + ε t where µ xt = (1 x t )µ 0 + x t µ 1 Process for the unobserved variable: Markov chain Martin Sola (FE) Markov Switching models 01/13 5 / 52

6 Properties of the Markov Process Definition Let x t be a random variable that can take values 0 and 1. If the probability that x t takes a particular value at time t, only depends on its value at t 1, this variable is governed by a Markov process of order 1. P((x t = i x t 1 = j, x t 2 = k...) = P((x t = i x t 1 = j) Thus the process is summarized by the probabilities: P(x t = 0 x t 1 = 0) = q, P(x t = 1 x t 1 = 1) = p. Martin Sola (FE) Markov Switching models 01/13 6 / 52

7 The transition Matrix 0 1 (time t-1) 0 q (1 p) 1 (1 q) p (time t) Martin Sola (FE) Markov Switching models 01/13 7 / 52

8 Autoregressive Representation of Markov Process [ 1 xt+1 x t+1 ] [ = q (1 p) (1 q) p ] [ 1 xt x t ] + [ ζ1,t+1 ζ 2,t+1 ] The second row gives x t+1 = (1 q) + ( 1 + p + q)x t + ζ 2,t+1 This expression can be recognized as an AR(1) process with constant term 1 q and autoregressive coeffi cient ( 1 + p + q) Martin Sola (FE) Markov Switching models 01/13 8 / 52

9 Expected value of x t : or E (x t+1 ) = (1 q) + ( 1 + p + q)e (x t ) E (x t ) = 1 q 2 p q since E (x t+1 ) = E (x t ) for a stationary process. Martin Sola (FE) Markov Switching models 01/13 9 / 52

10 Unconditional probabilities of being in state 1 and 0. Notice that E (x t ) = 0P(x t = 0) + 1P(x t = 1) = 1 q 2 p q. Then, P(x t = 1) = P(x t = 0) = 1 q 2 p q, 1 p 2 p q. Martin Sola (FE) Markov Switching models 01/13 10 / 52

11 Conditional and Unconditional Probabilities of States 0 and 1 An alternative derivation It can be shown that the unconditional probabilities at time zero multiplied by the matrix of transition probabilities are equal to the unconditional probabilities at time one: [ P(x1 = 0) P(x 1 = 1) ] [ = q (1 p) (1 q) p ] [ P(x0 = 0) P(x 0 = 1) ] Martin Sola (FE) Markov Switching models 01/13 11 / 52

12 Equilibrium Probabilities If the process is stationary, there exist state probabilities {π 0, π 1 } that satisfy: Π = PΠ where P is the matrix of transition probabilities. [ ] [ ] [ ] π0 q (1 p) π0 = π 1 (1 q) p π 1 where π 0 = P(x t j = 0) and π 1 = P(x t j = 1) for all values of j. Martin Sola (FE) Markov Switching models 01/13 12 / 52

13 Using π 0 + π 1 = 1: π 0 = (1 p) (2 p q) and π 1 = (1 q) (2 p q) where π 0 and π 1 are the equilibrium unconditional probabilities. Martin Sola (FE) Markov Switching models 01/13 13 / 52

14 Given the following initial values: [ p 0 (0) p 0 (1) ] [ π0 = π 1 ] It can be shown (by multiplying n times by the transition probability matrix) that the unconditional probability vector at time n is: [ p n ] [ ] (0) π0 p n = (1) π 1 Therefore, the distribution does not change with time and the stochastic process is always in equilibrium. Martin Sola (FE) Markov Switching models 01/13 14 / 52

15 Forecasts for a Markov Chain A n- period ahead forecast for a Markov chain can be obtained simply by multiplying n times by the transition probability: [ ] [ ] n [ ] P(xt+n = 0) q (1 p) P(xt = 0) = P(x t+n = 1) (1 q) p P(x t = 1) Martin Sola (FE) Markov Switching models 01/13 15 / 52

16 P n is derived in the following way: 1) Find the eigenvalues of the transition probability Matrix. λ 1 = 1, λ 2 = 1 + p + q, 2) Find the associated eigenvectors. (1 p) (2 p q) (1 q) (2 p q), [ 1 1 ] Martin Sola (FE) Markov Switching models 01/13 16 / 52

17 3) Express P as T ΛT 1, where T = (1 p) (2 p q) (1 q) (2 p q) 1 is the matrix of eigenvectors and [ ] 1 0 Λ = p + q a diagonal matrix of eigenvalues. 1 Martin Sola (FE) Markov Switching models 01/13 17 / 52

18 4) Use the result that or P n = T Λ n T 1 P n = [ 1 p 1 (2 p q) 1 q 2 p q 1 ] [ p + q ] n [ 1 p 1 (2 p q) 1 q 2 p q 1 ] 1 Martin Sola (FE) Markov Switching models 01/13 18 / 52

19 0 1 ( t+n) 0 1 ( t) (1 p) (2 p q) + (1 q)(p+q 1)n (2 p q) (1 q) (2 p q) (1 q)(p+q 1)n (2 p q) (1 p) (2 p q) (1 p)(p+q 1)n (2 p q) (1 q) (2 p q) + (1 p)(p+q 1)n (2 p q) Note that by making n = 1 in the above matrix we end with the matrix of transition probabilities, P. Martin Sola (FE) Markov Switching models 01/13 19 / 52

20 In addition, we can derive the conditional expectations: (1 q) E (x t+n x t = 0) = (2 p q) E (x t+n x t = 1) = (1 q)(p + q 1)n (2 p q) (1 q) (1 p)(p + q 1)n + (2 p q) (2 p q) Martin Sola (FE) Markov Switching models 01/13 20 / 52

21 The expected value of x t at time n conditional on s at time zero is: E (x t+n x t = x t ) = (1 q) (2 p q) + (x (1 q) t )(p + q 1)n (2 p q) This result is derived assuming that x t is observed (conditional on x t ) Martin Sola (FE) Markov Switching models 01/13 21 / 52

22 A brief description of Hamilton s non linear filter The procedure assumes that discrete states of the economy are not known: inferred from the data. States follow a discrete Markov process. Martin Sola (FE) Markov Switching models 01/13 22 / 52

23 A brief description of Hamilton s non linear filter The observed variable, y t, is assumed to follow an AR(m): y t µ xt = φ 1 (y t 1 µ xt 1 ) + φ 2 (y t 2 µ xt 2 ) +.. +φ m (y t m µ xt m ) + σ xt ε t ε t is distributed N(0,1) and µ xt is parameterized as α 0 + α 1 x t and σ xt as w 0 + w 1 x t Martin Sola (FE) Markov Switching models 01/13 23 / 52

24 y t µ xt = φ 1 (y t 1 µ xt 1 ) + φ 2 (y t 2 µ xt 2 ) +.. +φ m (y t m µ xt m ) + σ xt ε t The error ε t is assumed to be independent of all x t j 0 Martin Sola (FE) Markov Switching models 01/13 24 / 52

25 Hamilton s Filter Step 1 Step 1 Calculate the joint density of the m past states and the current state conditional on the information included in y t 1 and all past values of y, where y is the variable that is observed: p(x t, x t 1,,..., x t m y t 1,.., y 0 ) = p(x t x t 1 )p(x t 1,,..., x t m y t 1,..., y 0 ) Martin Sola (FE) Markov Switching models 01/13 25 / 52

26 Hamilton s Filter Step 1 Step 1 As in all the subsequent steps the second term on the right-hand-side is obtained from the preceding step of the filter. In this case, p(x t 1, x t 2,..., x t m y t 1, y t 2,..., y 0 ) is known from the input to the filter, which in turn represents the result of the iteration at date t 1 (from step 5). Martin Sola (FE) Markov Switching models 01/13 26 / 52

27 Hamilton s Filter Step 1 Step 1 To begin with the iteration, it is necessary to assign some initial values to the parameters, and to impose some initial conditions on the Markov process. The unconditional distribution p(x m 1, x m 2,...x 0 ) has been chosen for the first observation: p(x m 1, x m 2,...x 0 ) = p(x m 1 x m 2 )...p(x 1 x 0 )p(x 0 ) where p(x 0 ) are the equilibrium unconditional probabilities as defined above. Martin Sola (FE) Markov Switching models 01/13 27 / 52

28 Hamilton s Filter Step 2 Step 2 Calculate the joint conditional distribution of y t and (x t, x t 1,..., x t m ). = p(y t, x t, x t 1,..., x t m y t 1, y t 2,..., y 0 ) p(y t x t, x t 1,..., x t m, y t 1, y t 2,..., y 0 ). p(x t, x t 1, x t 2,..., x t m y t 1, y t 2,..., y 0 ) Martin Sola (FE) Markov Switching models 01/13 28 / 52

29 Hamilton s Filter Step 2 We assume that = p(y t x t, x t 1,., x t m, y t 1, y t 2,., y 0 ) 1 2π(ω0 + ω 1 x t ) exp[ 1 2[ω 0 + ω 1 x t ] 2 ((y t α 1 x t α 0 ) φ 1 (y t 1 α 1 x t 1 α 0 ).. φ m (y t m α 1 x t m α 0 )) 2 ] Martin Sola (FE) Markov Switching models 01/13 29 / 52

30 Hamilton s Filter Step 3 Step 3 Marginalize the previous joint density with respect to the states giving the conditional density, from which the (conditional) likelihood function is calculated. = p(y t y t 1, y t 2,., y 0 ) p(y t, x t, x t 1,., x t m y t 1, y t 2,., y 0 ) x t =0 x t 1 =0 x t m =0 Martin Sola (FE) Markov Switching models 01/13 30 / 52

31 Hamilton s Filter Step 4 Step 4 Combine the results from steps 2 and 3 to calculate the joint density of the state conditional on the observed current and past realizations of y p(x t, x t 1,., x t m y t, y t 1, y t 2,., y 0 ) = p(y t, x t, x t 1,., x t m y t 1, y t 2,., y 0 ) p(y t y t 1, y t 2,., y 0 ) Martin Sola (FE) Markov Switching models 01/13 31 / 52

32 Hamilton s Filter Step 5 Step 5 The desired output is then obtained from = p(x t, x t 1,., x t m+1 y t, y t 1, y t 2.., y 0 ) 1 p(x t, x t 1,., x t m y t, y t 1, y t 2,., y 0 ) x t m =0 The output of step 5 is used as an input to the filter in the next iteration. Martin Sola (FE) Markov Switching models 01/13 32 / 52

33 Hamilton s Filter Step 5 Note that to iterate, estimates of the parameters are required. Maximum likelihood estimates can be obtained numerically from Step 3 as a by-product of the filter ln p(y t, y t 1, y t 2,., y m y m 1,., y 0 ) = T ln p(y t y t 1,., y 0 ). t=m which can be maximized numerically with respect to the unknown parameters (α 1, α 0, p, q, ω 0, ω 1, φ 1, φ 2...φ m ). Martin Sola (FE) Markov Switching models 01/13 33 / 52

34 Comments Notice that p and q, the parameters of the transition matrix, are also estimated by maximum likelihood. Hamilton s filter requires the numerical optimization of a very complicated non-linear function. Martin Sola (FE) Markov Switching models 01/13 34 / 52

35 Specification Tests Specification tests based on the properties of the standardized residuals, ε t = y t E (y t I t 1 ) σ t. First calculate the conditional expectation of y t given information at time t 1: E (y t I t 1 ) = α 0 + α 1 E (x t I t 1 ) + φ 1 (y t 1 α 1 E (x t 1 I t 1 ) α 0 ) where E (x t I t 1 ) = φ m (y t m α 1 E (x t m I t 1 ) α 0 ), (1 q) (2 p q) + (P(x (1 q) t 1 I t 1 ) )(p + q 1), (2 p q) E (x t m I t 1 ) = P(x t m = 1 I t 1 ), m 1. P(x t m I t 1 ), for m > 1, are called smoothing probabilities They can be calculated from the filtering probabilities. Martin Sola (FE) Markov Switching models 01/13 35 / 52

36 Computing the standard deviation Step 1 First make use of the autoregressive representation of the Markov process x t = (1 q) + ( 1 + p + q)x t 1 + ζ 2,t For this process the error, conditional on x t 1 = 1, can be characterized as ζ 2,t = and conditional on x t 1 = 0 (1 p) with probability p p with probability 1 p ζ 2,t = (1 q) with probability q q with probability 1 q Martin Sola (FE) Markov Switching models 01/13 36 / 52

37 Computing the standard deviation Step 2 Calculate the variance of the error term, ζ 2,t, conditional on the state at t 1. E (ζ 2 2,t x t 1 = 1) = (1 p) 2 p + p 2 (1 p) = p(1 p) E (ζ 2 2,t x t 1 = 0) = (1 q) 2 q + q 2 (1 q) = q(1 q) Martin Sola (FE) Markov Switching models 01/13 37 / 52

38 Computing the standard deviation Step 3 Calculate the conditional variance (conditional on I t 1 = {y t 1,...y 0 }) We start by calculating the state dependent variance σ 2 x t function of the Markov switching parameters. as a Martin Sola (FE) Markov Switching models 01/13 38 / 52

39 Computing the standard deviation Step 3 Conditional on x t 1 = 1, the switching variance can be written as: σ 2 x t = E (σ 2 x t x t 1 = 1) + V (µ xt x t 1 = 1) where E (σ 2 x t x t 1 = 1) = (E (σ xt x t 1 = 1)) 2 + Var((σ xt x t 1 = 1)) (since σ xt is a random variable) and Var((σ xt x t 1 = 1)) = E (σ 2 x t x t 1 = 1) (E (σ xt x t 1 = 1)) 2 Martin Sola (FE) Markov Switching models 01/13 39 / 52

40 Computing the standard deviation Step 3 Then using that (E (σ xt x t 1 = 1)) 2 = (w 0 + w 1 E (x t x t 1 = 1) 2 = (w 0 + w 1 p) 2 Var((σ xt x t 1 = 1)) = Var(w 0 + w 1 x t x t 1 = 1) = w1 2 p(1 p). V (µ xt x t 1 = 1) = Var(α 0 + α 1 x t x t 1 = 1) = α 2 1p(1 p) Martin Sola (FE) Markov Switching models 01/13 40 / 52

41 Computing the standard deviation Step 3 Collecting all these terms we can see that σ 2 x t = (w 0 + w 1 p) 2 + w1 2 p(1 p) + α2 1p(1 p) We can obtain a similar formulae for the variance conditional on x t 1 = 0. σ 2 x t = E (σ 2 x t x t 1 = 0) + V (µ xt x t 1 = 0) and doing the same transformations for state 0 we obtain σ 2 x t = (w 0 + w 1 (1 q)) 2 + w1 2 q(1 q) + α2 1q(1 q). Martin Sola (FE) Markov Switching models 01/13 41 / 52

42 Computing the standard deviation Step 3 Clearly the state is not observed at time t 1 but we can use the filtered probabilities to make an inference of the unobserved state. Then the conditional variance (on information on time t 1) is σ 2 t = ((w 0 + w 1 p) 2 + w1 2 p(1 p) +α 2 1 p(1 p))p(x t 1 = 1 I t 1 )+ ((w 0 + w 1 (1 q)) 2 + w1 2 q(1 q)+ α 2 1 q(1 q))(1 P(x t 1 = 1 I t 1 )). Martin Sola (FE) Markov Switching models 01/13 42 / 52

43 Computing the standard deviation Step 3 Then the standardized residuals are simply, v t = ε t /σ t and we may conduct standard specification tests for these residuals. Martin Sola (FE) Markov Switching models 01/13 43 / 52

44 Number of tests and specification tests Crucial: rightly identify the number of states or regimes. Hamilton proposes to use simple specification tests as a mean of assessing whether the estimated equation contains the right number of states. Martin Sola (FE) Markov Switching models 01/13 44 / 52

45 A Bivariate model with regime switching We consider a VAR process in two variables, with 1 lag, with the feature that the means of each equation and the variance-covariance matrix are allowed to switch endogenously between two possible states. The two equations that define the VAR are influenced by the same state variable. The state is not observed and has to be inferred from a filter. S t = ΦS t 1 + ψd t 1 + (ω 0 + ω 1 x t )ν t D t = ϕs t 1 + ΩD t 1 + (τ 0 + τ 1 x t )ε t Martin Sola (FE) Markov Switching models 01/13 45 / 52

46 A Bivariate model with regime switching The centered variables are defined by the two following equations: S t = S t α 0 α 1 x t D t = D t β 0 β 1 x t A prime ( ) is used to denote centred variables in the remainder of the presentation x t denotes the unobserved state of the system and takes values 0 and 1. Martin Sola (FE) Markov Switching models 01/13 46 / 52

47 A Bivariate model with regime switching x t is governed by a Markov process P(x t = 0 x t 1 = 0) = q P(x t = 1 x t 1 = 1) = p The errors ν t, ε t are assumed to be independent of all x t j. j 0. Martin Sola (FE) Markov Switching models 01/13 47 / 52

48 A Bivariate model with regime switching Substituting the centered variables into the VAR and rearranging terms, we obtain the following expression for S t and D t S t = α 0 (1 Φ) β 0 ψ + ΦS t 1 + ψd t 1 +α 1 (x t Φx t 1 ) β 1 ψx t 1 + (ω 0 + ω 1 x t )ν t D t = α 0 ϕ + β 0 (1 Ω) + ϕs t 1 + ΩD t 1 α 1 ϕx t 1 + β 1 (x t Ωx t 1 ) + (τ 0 + τ 1 x t )ε t Martin Sola (FE) Markov Switching models 01/13 48 / 52

49 Testing The Term Structure of Interest Rates The process that drives the spread and the short-term interest rate difference is the VAR of equation described above in which D t denotes the first difference of the three month rate, R 1t R 1t 1 and S t denotes the yield spread R 2t R 1t. The expectations hypothesis of the term structure of the interest rates can be written as S t = (1/2)E t D t+1 + θ + u t The restrictions imposed by the expectations model are presented below. Both an unrestricted and a restricted VAR can be estimated, and the restrictions tested using a likelihood ratio test. Martin Sola (FE) Markov Switching models 01/13 49 / 52

50 Derivation of the restrictions in the regime-shifting VAR. [ ] [ ] [ ] [ S t Φ ψ S D t = t 1 (ω0 + ω ϕ Ω D t x t )ν t (τ 0 + τ 1 x t )ε t To find the restrictions we condition on information available at t 1 on both sides of the term structure equation. E (S t I t 1 ) = (1/2)E (D t+1 I t 1 ) + θ ] were I t 1 = {S t 1, S t 2,., D t 1, D t 2,., x t 1,., }. Martin Sola (FE) Markov Switching models 01/13 50 / 52

51 Then, we need to calculate expected values E [D t+1 It 1 ] and E [S t It 1 ]. These can be calculated in the following way: where E [D t+1 I t 1 ] = β 0 + β 1 E (x t+1 I t 1 ) + E (D t+1 I t 1 ), where ρ = 1 q 2 p q E [D t+1 I t 1 ] = [0, 1] 2 Z t 1, E (x t+1 I t 1 ) = [ρ + (x t 1 ρ)λ 2 ] and λ = (p + q 1). where E [S t I t 1 ] = α 0 + α 1 E (x t I t 1 ) + E (S t I t 1 ) E (S t I t 1 ) = [ 1 0 ] Z t 1 E (x t I t 1 ) = [ρ + (x t 1 ρ)λ] Martin Sola (FE) Markov Switching models 01/13 51 / 52

52 Finally, substituting the expected (difference of the) short-term interest rates and the expected spread, the term structure of interest rates relationship can be expressed as; α 0 + α 1 [ρ + (x t 1 ρ)λ] + [1, 0] Z t 1 ((A4)) = 1 2 [β 0 + β 1 [ρ + (x t 1 ρ)λ 2 ] + [0, 1] 2 Z t 1 ] + θ The restrictions written out in full are: α 1 = 1 2 β 1 λ Φ = ψ = ϕω 2 ϕ Ω 2 2 ϕ Martin Sola (FE) Markov Switching models 01/13 52 / 52