1 First-order di erence equation

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1 References Hamilton, J. D., Time Series Analysis. Princeton University Press. (Chapter 1,2) The task facing the modern time-series econometrician is to develop reasonably simple and intuitive models capable of forecasting, interpreting and hypothesis testing regarding economic data. In this respect, the time-series econometrician is often concerned with the estimation of difference equations containing stochastic components. 1 First-order di erence equation The model: Suppose we are given the dynamic equation: y t = φ 1 y t 1 + w t (1) where y t is the value of variable at period t, w t is the value of variable at period t, t = 1,..., T. Difference equation: equation relating a variable y t to its previous (lagged) values. First order difference equation: only the first lag appears on the RHS of the equation. For now, the input variable, {w 1, w 2, w 3,...}, will simply be regarded as a sequence of deterministic numbers. Later on, we will assume that they are stochastic. 1.1 Solving a Di erence Equation by Recursive Substitution Assume that we know the starting value of y 1, called the initial condition. The, note the following y 0 = φ 1 y 1 + w 0 y 1 = φ 1 y 0 + w 1 = φ 1 (φ 1 y 1 + w 0 ) + w 1 = φ 2 1y 1 + φ 1 w 0 + w 1 y 2 = φ 1 y 1 + w 2 = φ 1 (φ 2 1y 1 + φ 1 w 0 + w 1 ) + w 2 = φ 3 1y 1 + φ 2 1w 0 + φ 1 w 1 + w y t = φ 1 y t 1 + w t = φ t+1 1 y 1 + φ t 1w 0 + φ t 1 1 w φ 1 w t 1 + w t (2) The procedure to express y t in term of the past values of w and the starting value y 1 is known as recursive substitution. 1

2 1.2 Dynamic multipliers 1. Computation of dynamic multipliers Assume that we start with y t 1 instead of y 1, i.e. we observe the value of y t 1 at period t 1. Can we say something about y t+j? y t = φ 1 y t 1 + w t y t+1 = φ 1 y t + w t+1 = φ 1 (φ 1 y t 1 + w t ) + w t+1 = φ 2 1y t 1 + φ 1 w t + w t y t+j = φ j+1 1 y t 1 + φ j 1 w t + φ j 1 1 w t φ 1 w t+j 1 + w t+j (3) The dynamic multiplier: y t+j = φ j 1 (4) Note that the multiplier does not depend on t. The dynamic multiplier y t the impact multiplier. 2. Impulse response function is called sometimes We can plot dynamic multipliers as a function of lag j, i.e plot { y t+j } J j=1. Because dynamic multipliers calculate the response of y to a single impulse in w, it is also referred to as the impulseresponse function. This function has many important applications in time-series analysis because it shows how the entire path of a variable is effected by a stochastic shock. Obviously, the dynamic of impulse response function depends on the value of φ 1 : (a) 0 < φ 1 < 1, the impulse response converges to zero; the system is stable (b) 1 < φ 1 < 0, the impulse response oscillates but converges to zero; the system is stable (c) φ 1 > 1, the impulse response is explosive; the system is unstable (d) φ 1 < 1, the impulse response is explosive; the system is unstable Impulse response functions for all possible cases are presented in Figure Permanent change in w In calculating dynamic multipliers in Figure 1, we were asking what would happen if w t were to increase by one unit with w t+1, w t+2,..., w t+j unaffected. We were finding the effect of a purely transitory change in w. 2

3 Impulse responses Impulse responses φ 1 = φ 1 = φ 1 = 0.5 φ 1 = φ 1 = Impulse responses Impulse responses φ 1 = φ 1 = Figure 1: Example of impulse response functions for first order difference equations. 3

4 Permanent change in w means that w t, w t+1,..., w t+j would all increase by one unit. The effect on y t+j of a permanent change in w beginning in period t is then given by y t+j + y t+j +1 + y t+j y t+j +j (5) 4

5 1.3 pth-order Di erence Equations 1. The model: A linear pth-order difference equation has the following form: y t = φ 1 y t 1 + φ 2 y t φ p y t p + w t (6) 2. Properties of the pth order di erence equation To illustrate the properties of pth order difference equation we consider the second order difference equation: so that p = 2. We write equation (7) as a first order vector difference equation. y t = φ 1 y t 1 + φ 2 y t 2 + w t (7) where ξ t = F ξ t 1 + v t (8) ξ = [ yt y t 1 ] [ ] φ1 φ 2, F =, v t = 1 0 [ wt 0 ] (9) If the starting value ξ t 1 is known we use the recursive substitution to obtain: ξ t = F t+1 ξ 1 + F t v 0 + F t 1 v 1 + F t 2 v F v t 1 + v t (10) Or if the starting value ξ t is known we use the recursive substitution to obtain: ξ t+j = F j+1 ξ t 1 + F j v t + F j 1 v t+1 + F j 2 v j F v t+j 1 + v t+j (11) 3. Computation of dynamic multipliers Recall the rules of vector and matrix differentiation: ξ t+j v t = F j v t v t = F j (12) This follows from the following rule. If x(β) = (x 1 (β),..., x m (β)) is an m 1 vector that depends on the n 1 vector β = (β 1,..., β n ), then 5

6 x β = In our the equation (12) x = F j v t and β = v t. x 1 β 1 x 1 β n... x m β 1 x m β n (13) Since the first element in ξ t+j is y t+j and the first element in v t is w t then the element (1,1) in the matrix F j is y t+j, i.e. the dynamic multiplier. For larger values of j, an easy way to obtain a numerical values for dynamic multiplier y t+j simulate the system. This is done as follows. Set y 1 = y 2 =... = y p = 0 and w 0 = 1, and set the values of w for all other dates to 0. Then use (6) to calculate the value for y t, t = 0, 1, 2,..., J. is to 4. Illustration of dynamic multipliers for the pth-order di erence equation. Example 1: Consider the second order difference equation. (a) Let the dynamic equation be as follows: so that F = Figure 2(a). y t = 0.6y t y t 2 + w t (14) [ ] The impulse-response function for this example is presented in 1 0 (b) Let the dynamic equation be as follows: so that F = Figure 2(b). y t = 0.8y t y t 2 + w t (15) [ ] The impulse-response function for this example is presented in 1 0 (c) Let the dynamic equation be as follows: so that F = Figure 2(c). y t = 0.9y t 1 0.5y t 2 + w t (16) [ ] The impulse-response function for this example is presented in 1 0 (d) Let the dynamic equation be as follows: y t = 0.5y t 1 1.5y t 2 + w t (17) 6

7 [ ] so that F =. The impulse-response function for this example is presented in 1 0 Figure 2(d). 5. The role of eigenvalue of the matrix F Similar to the first order difference equation, the impulse response function for the p-th order difference equation can be explosive or converge to zero. What determines the dynamics of impulse response function? The eigenvalues of matrix F determine whether the impulse response oscillates, converges or is explosive. The eigenvalues of a p p matrix F are those numbers λ for which F λi p = 0 (18) Proposition 1.1 The eigenvalues of the general matrix F defined in 9 are the values of λ that satisfy: λ p φ 1 λ p 1 φ 2 λ p 2... φ p λ φ p = 0 (19) If the eigenvalues are real but at least one eigenvalue is greater than unity in absolute value, the system is explosive. Example 1 cont. (a) Eigenvalues in the example 1(a): λ 2 0.6λ 0.2 = 0 (20) λ 1 = 0.838, λ 2 = 0.238, i.e. λ i < 1 and the system is stable. (b) Eigenvalues in the example 1(b) λ 2 0.8λ 0.4 = 0 (21) λ 1 = 1.148, λ 2 = 0.348, i.e. λ 1 > 1 and the system is unstable. (c) Eigenvalues in the example 1(c) λ λ = 0 (22) λ 1 = i, λ 2 = i, i.e. λ 1 = ( 0.45) = < 1 and the system is stable. Since eigenvalues are complex, the impulse response function oscillates. 7

8 (a) Impulse responses (b) Impulse responses φ 1 = 0.8, φ 2 = φ 1 = 0.6, φ 2 = (c) Impulse responses (d) Impulse responses φ 1 = -0.9, φ 2 = φ 1 = -0.5, φ 2 = Figure 2: Example of impulse response functions for second order difference equation. 8

9 (d) Eigenvalues in the example 1(d) λ λ = 0 (23) λ 1 = i, λ 2 = i, i.e. λ 1 = ( 0.25) = > 1 and the system is unstable. Since eigenvalues are complex, the impulse response function oscillates. 6. Why do eigenvalues determine the dynamics of dynamic multipliers? Recall that if the eigenvalues of p p matrix are distinct, there exists a nonsingular p p matrix T such that: F = T ΛT 1 (24) where Λ is a p p matrix with the eigenvalues of F on the principal diagonal and zeros elsewhere. Then, F j = T Λ j T 1 (25) and the value of eigenvalues in Λ determines whether the elements of F j explode or not. Recall from (12) that the dynamic multiplier is equal to ξ t+j v t determines whether the system is stable or not. = F j. Therefore, the size of eigenvalues 2 Lag operators Now we develop some of the results using time series operators, which are used in time-series econometrics very often. 1. Lag operator Generally, a time series operator transforms one times series or a group of time series into a new time series. A highly useful operator is the lag operator, represented by the symbol L. Faced with a time series defined in terms of compound operators, one may use the standard cummutative, associative and distributive laws for multiplication and addition. The main properties of the lag operator are as follows: Lx t = x t 1 9

10 L k x t = x t k L(βx t ) = β(lx t ) L(x t + w t ) = Lx t + Lw t Lc = c, c is a constant L i y t = y t+i (1 + φ 1 L + φ 2 1L )y t = y t 1 φ 1 L, a < First-order di erence equation The first-order difference equation has the form: y t = φ 1 y t 1 + w t y t = φ 1 (Ly t ) + w t (1 φ 1 L)y t = w t (26) Multiplying LHS and RHS of the equation 26 by (1 + φ 1 L + φ 2 L 2 + φ 3 L φ t L t ) we obtain: (1 φ t+1 1 L t+1 )y t = (1 + φ 1 L + φ 2 L 2 + φ 3 L φ t L t )w t y t = φ t+1 1 y 1 + w t + φ 1 w t 1 + φ 2 1w t φ t 1w 0 (27) Note that equation in (27) is the same as in equation (2). A sequence {y t } t= is said to be bounded if there exists a finite number ȳ such that y t < ȳ for all t. When φ 1 < 1, a sequence is bounded. 2.2 Second-order di erence equation A second-order difference equation is: (1 φ 1 L φ 2 L 2 )y t = w t y t = φ 1 y t 1 + φ 2 y t 2 + w t (1 λ 1 L)(1 λ 2 L)y t = w t (28) How do we find the values of λ 1 and λ 2? Step 1: Find the scalars z 1 and z 2 such that (1 φ 1 z φ 2 z 2 ) = 0 Step 2: Set λ 1 = z 1 1, λ 2 = z

11 Proposition 2.1 Factoring the polynomial (1 φ 1 L φ 2 L 2 ) as (1 φ 1 L φ 2 L) = (1 λ 1 L)(1 λ 2 L) (29) is the same calculation as finding eigenvalues of the matrix F in (9). The eigenvalues λ 1 and λ 2 of F are the same as the parameters λ 1 and λ 2 in (29). Given that the second-order difference equation is stable, with eigenvalues λ 1 and λ 2 distinct and both lie inside the unit circle the equation (28) can be written as (1 λ 1 L)(1 λ 2 L)y t = w t y t = (1 λ 1 L) 1 (1 λ 2 L) 1 w t (30) 2.3 pth-order Di erence Equations These techniques generalize in a straightforward way to a pth-order difference equation of the form: (1 φ 1 L φ 2 L 2... φ p L p )y t = w t y t = φ 1 y t 1 + φ 2 y t φ p y t p + w t (1 λ 1 L)(1 λ 2 L)...(1 λ p L)y t = w t (31) 11