ES10006: Core skills for economists: Mathematics 2 Seminar 10 Problem set no. 7: Questions 6 and 7
Simultaneous difference equations The direct method The following system of n = 2 linear difference equations with constant coefficients and terms y t+1 = a 11 y t + a 12 x t + b 1 x t+1 = a 21 y t + a 22 x t + b 2 (1) can be written in matrix form as y t+1 = Ay t + b (2) where y t = [ y t x t ], A = [ a 11 a 12 a 21 a 22 ], and b = [ b 1 b 2 ]. 2/13
Simultaneous difference equations The direct method To find the homogeneous solution we use the trial solution y t = kr t where k is a 2 1 vector of arbitrary constants and r is a scalar. Substituting into the homogeneous system of equations (and excluding the trivial solution r = 0) 3/13 y t+1 = Ay t kr t+1 = Akr t Akr t rkr t = 0 (A ri)k = 0 (3) where I is the identity matrix and 0 is the zero-vector. This is a system of linear homogeneous equations; it yields only the trivial solution (k 1 = k 2 = 0) if its coefficient matrix is nonsingular. To rule out this possibility (and obtain instead an infinite number of solutions) we require the determinant of (A ri) to vanish, 1 that is A ri = 0 (4) Example (where the determinant of the first matrix is different from zero and that of the second matrix equals zero): [ 5 1 6 2 ] [k 1 k 2 ] = [ 0 0 ] k 1 = k 2 = 0 but [ 3 1 6 2 ] [k 1 k 2 ] = [ 0 0 ] k 2 = 3k 1 1 The solution outcomes for a linear-equation system are summarized in Chiang and Wainwright (2005, Table 5.1 on p. 106).
Simultaneous difference equations The direct method Equation (4) is the characteristic equation of matrix A. A nonzero vector, k i (i = 1, 2), which is a solution of equation (3) (or equivalently of Ak = rk) for a particular eigenvalue, r i, is called the eigenvector of matrix A corresponding to the eigenvalue r i. In what follows we have (n =) 2 equations and 2 roots. Assuming that there are two distinct real roots (as it is the case with Questions 6 and 7), the solution to the homogeneous form y t+1 = Ay t is the following linear combination of the 2 distinct solutions to the system: y h t = A 1 k 1 (r 1 ) t + A 2 k 2 (r 2 ) t (5) To find the particular solution, set y t = y t+1 = y. Then y = Ay + b (I A)y = b y = (I A) 1 b (6) and the general solution is y t = y h t + y = A 1 k 1 (r 1 ) t + A 2 k 2 (r 2 ) t + y (7) For the definite solution, initial conditions (y 0 and x 0 ) are needed to assign specific values to the constants A 1 and A 2. 4/13
Simultaneous difference equations Convergence of the time path The terms in solution (7) that contain (r 1 ) t and (r 2 ) t converge to zero as t if 1 < r 1 < 1 and 1 < r 2 < 1. In this case the system is stable: whatever the initial conditions, y t converges to y [the steady state given by (6)]. In general, A system of two linear difference equations with constant coefficients and terms [that is system (1)] is asymptotically stable if and only if the absolute value of both characteristic roots are less than unity (see Hoy et al. 2011, Theorem 24.10 on p. 835). If there is a single real root, r = r 1 = r 2, the system is stable if 1 < r < 1 (see Hoy et al. 2011, Example 24.21). If there are complex roots, r 1, r 2 = h ± ui, the system is stable if R = h 2 + u 2 < 1 (see Hoy et al. 2011, Example 24.22). 5/13
Question 6 Two rival retailers I and II each adopt a price setting strategy of setting price today 10% lower than its rival s price yesterday. Write out the system of difference equation for the two prices, calling the price for retailer I, y and the price for retailer II, x. Solve the system assuming that y 0 = x 0. Show that prices converge to zero. Assuming Y t = [ y t x t ], we have the following system of difference equations { y t+1 = 0.9x t x t+1 = 0.9y t or (in matrix form) Y t+1 = AY t = [ 0 0.9 0.9 0 ] Y t We use equation (4) to find the characteristic roots: r 0.9 A ri = 0 0.9 r = 0 r2 0.9 2 = 0 r 1 = 0.9 and r 2 = 0.9 6/13
Question 6 From equation (3) we have Ak = rk. For r 1 = 0.9, the eigenvector is the solution to [ 0 0.9 0.9 0 ] [k 1 ] = 0.9 [ k 1 ] { k 1 = 1 k 2 k 2 k 2 = 1 For r 2 = 0.9, the eigenvector is the solution to [ 0 0.9 0.9 0 ] [k 1 k 2 ] = 0.9 [ k 1 k 2 ] { k 1 = 1 k 2 = 1 Hence, from equation (5), the general homogeneous solution to the system of equations is: Y t = A 1 [ 1 1 ] (0.9)t + A 2 [ 1 1 ] ( 0.9)t { y t = A 1 (0.9) t + A 2 ( 0.9) t x t = A 1 (0.9) t A 2 ( 0.9) t Finally, y 0 = x 0 A 1 + A 2 = A 1 A 2 A 2 = A 2 A 2 = 0 and consequently y 0 = A 1 + A 2 y 0 = A 1 and x 0 = A 1 A 2 x 0 = A 1. With 1 < 0.9 < 1, y t = y 0 (0.9) t and x t = x 0 (0.9) t tend to zero as t. 7/13
Question 6 The following graphs show convergence to Y(= b) = [ 0 ] as t, starting 0 from x 0 = y 0 on the left panel and x 0 y 0 on the right panel. On the right panel, given the initial conditions, A 1 = 15 and A 2 = 5, implying y(t) = 15(0.9) t 5( 0.9) t and x(t) = 15(0.9) t + 5( 0.9) t. 8/13
Question 7 Consider a duopoly model in which two firms (Firms 1 and Firm 2) share the market for the same product. Each firm chooses a strategy of producing an amount given by the following reaction functions: y t+1 = 1 4 x t + 60 x t+1 = 1 4 y t + 60 where y is Firm 1 s output and x is Firm 2 s output. Write the equations in matrix form. Solve the system of difference equations. Assuming Y t = [ y t x t ] we have the following system of difference equations y t+1 = 1 4 x t + 60 x t+1 = 1 4 y t + 60 or (in matrix form) Y t+1 = AY t + b = 0 1 / 4 1 Y t + [ 60 / 4 0 60 ] 9/13
Question 7 We use equation (4) to find the characteristic roots: A ri = 0 r 1 / 4 1 / 4 r = 0 r2 ( 1 4 ) 2 = 0 r 1 = 1 4 and r 2 = 1 4 From equation (3) we have Ak = rk. For r 1 = 1/4, the eigenvector is the solution to [ 0 1 / 4 1 / 4 0 ] [k 1 k 2 ] = 1 4 [k 1 k 2 ] { k 1 = 1 k 2 = 1 For r 2 = 1/4, the eigenvector is the solution to [ 0 1 / 4 1 / 4 0 ] [k 1 ] = 1 k 2 4 [k 1 ] { k 1 = 1 k 2 k 2 = 1 Hence, from equation (5), the general homogeneous solution to the system of equations is Yt h = A 1 [ 1 1 ] ( 1 t 4 ) + A 2 [ 1 1 ] ( 1 t 4 ) y t = A 1 ( 1 4 )t + A 2 ( 1 4 )t x t = A 1 ( 1 4 )t + A 2 ( 1 4 )t 10/13
Question 7 We use equation (6) to find the particular solution. 16 Y = (I A) 1 b 15 4 15 and the general solution is: 4 15 [ 60 16 60 ] = [48 48 ] 15 Y t = Y h t + Y = A 1 [ 1 1 ] ( 1 4 ) t + A 2 [ 1 1 ] ( 1 4 ) t + [ 48 48 ] y t = A 1 ( 1 4 )t + A 2 ( 1 4 )t + 48 x t = A 1 ( 1 4 )t + A 2 ( 1 4 )t + 48 Since r 1 < 1 and r 2 < 1, Y h t 0 and Y t Y as t. 11/13
Question 7 Whatever the initial conditions, the system converges to Y = [ 48 48 ]. 12/13
References Chiang, A. C. and Wainwright, K. (2005), Fundamental Methods of Mathematical Economics, 4th edn, McGraw-Hill. Hoy, M., Livernois, J., McKenna, C., Rees, R. and Stengos, R. (2011), Mathematics for Economics, 3rd edn, MIT Press. 13/13