Mathematical Foundations -1- Difference equations. Systems of difference equations. Life cycle model 2. Phase diagram 4. Eigenvalue and eigenvector 5

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1 Mathematical Foundations -- Difference equations Systems of difference equations Life cycle model Phase diagram 4 Eigenvalue and eigenvector 5 The general two variable model 9 Complex eigenvalues 5 Stable and unstable systems 4 John Riley August 4, 0

2 Mathematical Foundations -- Difference equations Review: Life-cycle model: the evolution of the system is defined by two linear equations. c = t α( + + r) c, t FOC W = t ( + r)( W + t ct). wealth equation For the baseline case we make the following assumptions. Assumptions: ( + r) α > and α < Consumption sequence: If c t > 0 consumption is growing larger at a constant rate. If c t < 0 consumption is growing more negative at a constant rate. John Riley August 4, 0

3 Mathematical Foundations -3- Difference equations More analysis: ct ( + ) = ( + r) αct ( ) Wt ( + ) = ( + r)( Wt ( ) Ct ( )). Constant growth paths. x( t+ ) = λx( t) Consumption and wealth grow at the same rate if Wt ( + ) = ( + r) αwt ( ). Substituting, Wt ( + ) = ( + r)( Wt ( ) Ct ( )) = ( + r) αwt ( ). Then Wt () Ct () = αwt () and so Ct () = ( α) Wt (). Summarizing, if Ct () = ( α) Wt () both consumption and wealth grow at the rate λ = ( + r) α. Also if Ct () = 0, then Ct+ ( ) = 0and Wt ( + ) = ( + rwt ) ( ) so both grow at the same rate + r John Riley August 4, 0

4 Mathematical Foundations -4- Difference equations Phase Diagram We can express the FOC and wealth equation in matrix form as follows. Wt ( + ) + r ( + r) Wt ( ) = ct ( + ) 0 ( + r) α ct ( ) () This is a linear difference equation system. John Riley August 4, 0

5 Mathematical Foundations -5- Difference equations. Solution with a constant growth rate Wt ( + ) Wt ( ) = λ ct ( + ) ct ( ). Any λ for which this holds is called an eigenvalue (or characteristic root.) Any vector Wt () v ct () = v for which this holds is called an eigenvector. Appealing to () Wt ( + ) + r ( + r) Wt ( ) Wt ( ) Wt ( ) λ 0 Wt ( ) λ λ ct ( ) = 0 ( r) α ct ( ) = ct ( ) = I = + + ct ( ) 0 λ ct ( ) + r ( + r) Wt ( ) λ 0 Wt ( ) = 0 ( + r) α c( t) 0 λ c( t) John Riley August 4, 0

6 Mathematical Foundations -6- Difference equations Solving for the eigenvalues (constant growth rates) If you write out the equation system it will be clear that this equation can be rewritten as follows. + r λ ( + r) W( t) 0 = 0 ( + r) α λ c( t) 0. () It follows that the determinant of this matrix is zero. + r λ ( + r) = 0 0 ( + r) α λ that is (( + r) α λ)( + r λ) = 0 Thus the eigenvalues are λ = ( + r) α and λ = + r. John Riley August 4, 0

7 Mathematical Foundations -7- Difference equations Solving for the eigenvectors Let v = ( v, v ) be an eigenvector associated with the first eigenvalue λ. From (), + r λ ( + r) v 0 0 ( r) α λ = v 0. + Substituting for λ. ( + r)( α) ( + r) v = v 0. That is, ( + r)( α) ( + r) v = v 0. Solving, ( α)v = v. Thus v = (, α) is an eigenvector. John Riley August 4, 0

8 Mathematical Foundations -8- Difference equations v = (, α) is an eigenvector. Note that this vector is in the positive quadrant or negative quadrant if and only if α <. This is the case depicted. Exercise: Consider the dynamics if α >. Similarly, substituting for λ, v = (, 0) is an eigenvector. If If x() x = v, then () v, = then x( t+ ) = λ v x( t ) t t + = λ v. The eigenvectors and constant growth paths are depicted in a phase diagram John Riley August 4, 0

9 Mathematical Foundations -9- Difference equations c = t α( + + r) c t W = t ( + r)( W + t ct). r Hence W t W = t (( + + r)( Wt ct) + r John Riley August 4, 0

10 Mathematical Foundations -0- Difference equations Consider any other x (). Since v and v are linearly independent we can write x () as some linear combination x() = α v + α v Then x() = Ax() = A( α v + α v ) = α Av + α A v = α λv + α λ v. Similarly x(3) = Ax() = A( αλv + αλv ) = ( αλav + αλa v ) = ( αλ v + αλ v ) Repeating this... x( t+ ) = α λ v + α λ v. t t Suppose λ > λ Then x( t+ ) = λ ( α ( λ ) v + α v ) λ α v t t t λ Thus the solution converges to the constant growth path with the higher eigenvector. John Riley August 4, 0

11 Mathematical Foundations -- Difference equations SYSTEMS OF LINEAR DIFFERENCE EQUATIONS The general two variable model x( t+ ) = Ax( t) = λx( t) = λi x( t). Rearranging, x( t+ ) λx( t) = Ax( t) λix( t) = ( A λ I ) x( t) = 0. For a stationary state this must hold with λ =, that is ( A I ) x( t) = 0. We assume that A I 0, so that the matrix A-I is invertible. Then the unique solution to the equation system ( A I ) x( t) = 0 is x() t = 0. John Riley August 4, 0

12 Mathematical Foundations -- Difference equations Next consider λ. For a constant growth path the equation system ( A λ I) x( t) = 0must have a non-zero solution. Arguing as above, this is not possible if A λ I is invertible. Hence for constant growth, a λ a A I = = ( + ) + A = 0. () λ λ a a λ a a λ This equation is known as the characteristic equation of the matrix A. The two roots λ, λ of the quadratic equation are known as the characteristic roots or eigenvalues. Associated with each, is an initial value of the state vector v and eigenvalues differ, the two initial state vectors are independent. v. These are known as the eigenvectors. As long as the two John Riley August 4, 0

13 Mathematical Foundations -3- Difference equations Real eigenvectors: We have seen that λ ( a + a ) λ + A = 0 () We can also write the characteristic equation as follows. ( λ λ )( λ λ ) = λ ( λ + λ ) λ λλ = 0. () Comparing () and (), the sum of the roots is the sum of the diagonal elements of A and the product of the roots is the determinant of A. Solving the quadratic characteristic equation yields the two roots λ λ = ( a+ a) + 4( a+ a) A = ( a+ a) 4( a+ a) A, (3) Thus there are two distinct real roots if and only if ( a + a ) 4 A = ( a a ) + 4a a > 0. Thus a sufficient condition for two distinct real roots is that aa > 0. John Riley August 4, 0

14 Mathematical Foundations -4- Difference equations Suppose this condition holds. Arguing exactly as in the example, any initial state vector x () can be expressed as a linear combination of v and v, that is x() = α v + α v. i i Since A v = λ v, i =,, it follows that A t v i = λ t v i, i =,. Then i i x( t+ ) = α λ v + α λ v. t t Thus the long run dynamics are determined by the eigenvalue with the larger absolute value. John Riley August 4, 0

15 Mathematical Foundations -5- Difference equations Complex eigenvalues a λ a A I = = ( + ) + A = ( )( ) = 0. (4) λ λ a a λ λ λ λ λ a a λ Consider the solution to the characteristic equation. λ λ = ( a+ a) + 4( a+ a) A = ( a+ a) 4( a+ a) A, (5) Suppose that the expression under the square root in (5) is negative. Then define α = ( a + a ) and β = A ( a + a ). 4 Then (5) can be rewritten as follows. λ = α + β λ = α β John Riley August 4, 0

16 Mathematical Foundations -6- Difference equations λ = α + β λ = α β It follows that there is no solution in terms of real numbers and so there can be no constant growth paths. We employ a remarkable mathematical sleight of hand and introduce complex numbers. We define i =. Then λ = α + iβ and λ = α iβ John Riley August 4, 0

17 Mathematical Foundations -7- Difference equations Example: Suppose that A =. The solution for a particular starting value is depicted. λ / A λi = = ( λ) + 4 = 0. / λ While there is no real root we note that i = and so ( λ) = i. 4 Taking the square root, λ = ± i John Riley August 4, 0

18 Mathematical Foundations -8- Difference equations As the first step in solving for the general solution, note that the eigenvalue λ and associated eigenvector v must satisfy the constant growth condition: ( A λ I ) v = 0, That is ( a λ ) v + av = 0. Without loss of generality we may choose v = then v = ( a λ)/ a. Since λ is a complex number, so is v. We will write this more succinctly as v = k+ ik. John Riley August 4, 0

19 Mathematical Foundations -9- Difference equations It proves extremely useful to express the vector of parameters ( α, β ) in polar coordinates. See Fig. B.4-4 below. Note that λ = α + iβ and λ = α iβ Then we can rewrite the eigenvalue as r = α + β = A and tan θ = β / α. λ = r(cosθ + isin θ), λ = r(cosθ isin θ) Fig. B.4-4: Polar coordinates John Riley August 4, 0

20 Mathematical Foundations -0- Difference equations iz Mathematical aside: For any z, cos z+ isin z = e. To see this consider the Taylor expansion around z = df d f d f 3 d f 4 f( z) = f(0) + (0) z+ (0) z + (0) z + (0) z dz! dz 3! dz 4! dz 3 sin z = sin0 + cos(0) z sin(0) z cos(0) z +... so 3! 3 4 cos z = cos(0) sin(0) z cos(0) z + sin(0) z + sin(0) z +... so! 3! 4! 3 sin z = z z ! Summing: 3 4 iz z iz z cos z+ isin z = !! 3! 4! 3 sin z = z z ! 4 z z cos z = ! 4! iz Finally Taylor s Expansion of e. iz i0 i0 i0 3 i0 3 4 i0 4 e = e + ie z+ i e z + i e z + i e z so! 3! 4! 3 4 iz iz z iz z e = !! 3! 4! John Riley August 4, 0

21 Mathematical Foundations -- Difference equations iθ cosθ + isinθ = e = cosθ + isinθ t iθ t Then (cosθ + isin θ) = e = cosθt+ isinθt And so if x() = v= k ik, and λ = r(cosθ + isin θ) + t t x( t+ ) = Ax( t) = A v=λ v t iθ t = re k ik + t = r (cos θ t+ isin θ t) k ik + John Riley August 4, 0

22 Mathematical Foundations -- Difference equations Collecting real and complex terms, Define cos sin t θt t θt xt ( + ) = r + ir k cosθt k sinθt k sinθt+ k cosθt. (6) cosθt sinθt. t t v () t = r and v () t = r kcosθt ksinθt ksinθt+ kcosθt We now argue that both v t iv t (( ) + ( ) is a solution, v () t and v () t are solutions to the difference equation system. Since x( t+ ) = v ( t+ ) + iv ( t+ ) = Ax( t) = Av ( t) + ia v ( t). John Riley August 4, 0

23 Mathematical Foundations -3- Difference equations Collecting real and complex terms, v t+ = v t v t+ = v t ( ) A ( ) and ( ) A ( ). Choose α and α so that combination. Then α v () + α v () = x(). Since v () t and v () t are solutions, so is any linear x() t = α v () t + α v () t is the general solution, given the initial state x (). John Riley August 4, 0

24 Mathematical Foundations -4- Difference equations Stable and unstable systems From (3) = + i = ( a+ a) + 4( a+ a) A λ α β = i = ( a+ a) 4( a+ a) A λ α β From Fig. B.4-4, we can rewrite these eigenvalues in polar coordinates as follows. where λ = r(cosθ + isin θ), λ = r(cosθ isin θ) r = α + β and Substituting for α and β, r (, ) ( ( a a), A 4( a a) ) αβ = + +. = A. From () the amplitude of the oscillation is t r at time t. Thus the amplitude of the oscillations is increasing if r > and decreasing if r <. Therefore the cycles of an oscillating system with complex eigenvalues are damped if and only if A <. John Riley August 4, 0

25 Mathematical Foundations -5- Difference equations Example (continued) If A =, we have seen that the eigenvalues are λ = + i and λ = i. Transforming the eigenvalues into polar coordinates, 5 λ = (cos θt+ i sin θt ) and Fig. B.4-5: Polar coordinates 5 (cos sin ) λ = θt i θt, where tanθ =. Note that, since r = A > the amplitude of the oscillations is increasing. Thus the dynamic system is unstable. John Riley August 4, 0