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Economics 440.602 Macroeconomic Theory Lecture 3
发布时间:2023-02-23
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Economics 440.602
Macroeconomic Theory
Lecture 3 - SEcoND 0RDER D1FFERENcE EoUAT1oNs
1 LEARN1Nc 0BJEcT1vEs
Students should understand:
● Second Order Difference Equation
● Distinct Real Roots
● Repeated Real Roots
● Complex and Imaginary Numbers
● Complex Roots
● Second Order Nonhomogeneous Systems
● Interpretation of the Second Order Systems
2 SEcoND 0RDER D1FFERENcE EoUAT1oN
1. The economic model being discussed may require the use of two lagged variables. We first discuss the nature of the general solution of a second-order homogenous difference equation. Given
yt = φ 1yto1 + φ2yto2 (1)
(a) where φ 1 and φ2 are arbitrary constants. We can form an auxiliary equation by letting
yt = λt
i. and substituting it and its lagged equivalents into (1) to obtain
λt = φ1 λto1 + φ2 λto2 (2)
ii. It is easy to see that this reduced into a quadratic equation which when solved gives two values of λ, say λ 1 and λ2 . This is, rewriting (2) we have
λt - φ 1 λto1 - φ2 λto2 = 0
iii. Dividing λto2 into both sides of the equation, we obtain λ2 - φ 1 λ - φ2 = 0
iv. The solution to (4) is, according to the quadratic formula,
(3)
(4)
φ 1 +^(-φ 1 )2 + 4φ2 φ 1 - ^(-φ 1 )2 + 4φ2
2 2
(b) Three distinct cases can occur, depending on the nature of the number appearing under the radical sign. (1) if (-φ 1 )2 + 4φ2 is real and positive, then λ 1 and λ2 are distinct real roots; (2) if (-φ 1 )2 + 4φ2 = 0, then λ 1 and λ2 are repeated real roots and λ 1 = λ2 ; and (3) if (-φ 1 )2 + 4φ2 < 0, then λ 1 and λ2 are complex roots. The methods of solution differ, depending on the type of case. Case (1) is first discussed, then Case (2). Case (3) will be treated after a discussion of complex roots and de Moivre’s theorem.
3 CAsE 1: D1sT1NcT REAL RooTs
2. If λ 1 and λ2 , discussed in the preceding section, are unequal roots satisfying the aux- iliary equation (3), these roots will each satisfy the difference equation (1). That is
yt = λ1(t)
yt = λ2(t)
(a) which each satisfy the difference equation
yt = φ 1yto1 + φ2yto2
i. For example, we can write the lagged equivalents
(5)
(6)
(7)
yto1 = λ1(t)o1
yto2 = λ1(t)o2
ii. Then,
yt = φ 1 λ1(t)o1 + φ2 λ1(t)o2
= λ1(t)
(b) Indeed, our choice of both λ 1 and λ2 is such that they satisfy the difference equation. However, it is usually the case that the initial conditions are not satisfied by (5) and (6). For instance, if the initial conditions of (7) are
y0 = a y1 = b (8)
i. then there is no general assurance that for t = 0 and 1 λ1(0) = a
ii. and
λ 1 = b;
iii. and similarly for yt = λ2(t) when t = 0, 1. Thus yt = λ1(t) and yt = λ2(t) are each insufficient to be the numerical solution to the difference equation consisting of (7) and (8). To obtain the solution it is necessary for another condition to be met.
(c) Theorem 3: If f1 (t) and f2 (t) both satisfy a difference equation, then a linear combination of f1 (t) and f2 (t), namely, C1 f1 (t) + C2 f2 (t), where C1 and C2 are arbitrary constants, also satisfies the difference equation.
i. Hence, the general solution to the system (7) and (8) is
yt = C1 λ1(t) + C2 λ2(t) (9)
ii. To obtain the numerical solution, we note that the initial conditions require that
y0 = a = C1 λ1(0) + C2 λ2(0) = C1 + C2 (10)
iii. and
y1 = b = C1 λ 1 + C2 λ2 (11)
iv. Since a, b, λ 1 , and λ2 are now known, C1 , and C2 can be obtained simulta- neously from (10) and (11). We next illustrate the solution procedure by an example.
(d) To find the numerical solution for
y0 = 3 y1 = 5
i. we first obtain the auxiliary equation of the difference equation. Let yt = λt
ii. then
λt = 3λto1 - 2λto2
iii. Therefore
λ2 - 3λ + 2 = 0
iv. and
λ 1 , λ2 = = = 2, 1
v. The general solution to the system (12) is
yt = C1 λ1(t) + C2 λ2(t)
= C1 (2)t + C2 (1)t
vi. To obtain the numerical solution, we first let
y0 = C1 (2)0 + C2 (1)0 = C1 + C2 = 3
y1 = C1 (2) + C2 (1) = 2C1 + C2 = 5
vii. Solving the these two equations simultaneously, we have
C1 = 2 C2 = 1
viii. We conclude, then, that the numerical solution is
yt = 2(2)t + 1(1)t = 2(2)t + 1 (13)
ix. We now check to see if (13) satisfies both the initial conditions and the dif- ference equation.
x. First, the initial conditions are satisfied, since
y0 = 20 + 2 = 3
y1 = 2 . 21 + 1 = 5
xi. Second, substituting
yto1 = 2(2)to1 + 1
yto2 = 2(2)to2 + 1
xii. into (12) we have
yt = 3[2(2)to1 + 1] - 2[2(2)to2 + 1]
= 3 . 2 . (2)to1 - 2 . 2 . (2)to2 + 1
= 3(2)to+ - (2)to+ + 1
= 3(2)t - (2)t + 1
= 2(2)t + 1
= yt
xiii. which is (13), thus satisfying the difference equation.
4 CAsE 2: REPEATED REAL RooTs
3. Before proceeding, we state the following theorem.
(a) Theorem 4: If double roots are involved in the auxiliary equation to a second- order difference equation (that is, the roots are such that λ 1 = λ2 ), and Yt = λ1(t) satisfies the difference equation, then yt = t . λ1(t) also satisfies the difference equa- tion.
i. When this theorem is combined with Theorem 3, we can say that for the system
yt = φ 1yto1 + φ2yto2
y0 = c y1 = d
ii. (where the difference equation involves double roots)
yt = C1 λ1(t) + C2 . t . λ1(t)
yt = (C1 + C2 . t)λ1(t)
iii. will satisfy the difference equation. Then, given the initial conditions, it is easy to find the values of the coefficients C1 and C2 .
(b) We illustrate the procedure by solving the following system:
yt = 6yto1 - 9yto2
y0 = 8 y1 = 10
i. The auxiliary equation for yt = λt is
λt = 6λto1 - 9λto2
ii. Then,
λ2 - 6λto1 + 9 = 0
iii. so that
-(-6) · ^(-6)2 - 4 . (1) . 9 6 ·^36 - 36 6
2 2 2
iv. Now yt = 3t satisfies the difference equation, and from the proceeding para- graph, we know that
yt = C1 (3)t + C2t(3)t (14)
(15)
v. also satisfies the difference equation. Hence C1 and C2 in (14) can be deter- mined by referring to the initial conditions. That is,
y0 = C1 = 8
y1 = 3C1 + 3(1)C2 = 10
vi. It then follows that
C1 = 8 C2 = -
vii. Therefore the numerical solution is
yt = 8(3)t - . t . (3)t
= (3)t ╱8 - . t、
viii. We now check to see if the numeric solution satisfies both the initial condi- tions and the difference equation.
ix. First, the initial conditions are satisfied, since
y0 = (3)0 ╱8 - . 0、= 8
y1 = (3)1 ╱8 - . 1、= . = 10
x. Second, substituting
yto1 = (3)to1 ╱8 - . (t - 1)、
yto1 = (3)to2 ╱8 - . (t - 2)、
xi. into the original second-order difference equation, we have
yt = 6 ┌ (3)to1 ╱8 - 、┐ - 9 ┌ (3)to2 ╱8 - 、┐ = 6 ┌ (3)to1 ╱ 、┐ - 9 ┌ (3)to2 ╱ 、┐ = 6 ┌ (3)to1 ╱ 、┐ - 9 ┌ (3)to2 ╱ 、┐
= 2 . 3 ┌ (3)to1 ╱ 、┐ - 3 . 3 ┌ (3)to2 ╱ 、┐ = 2 ┌ (3)to+ ╱ 、┐ - ┌ (3)to+ ╱ 、┐
= (3)t ┌ 2 . ╱ 、┐ - ┌╱ 、┐
= (3)t ┌ - ┐
= (3)t ┌ ┐
= (3)t ┌ ┐
= (3)t ╱8 - . t、
yt = yt
xii. which indeed satisfies the difference equation.
(c) We next discuss some forms of complex numbers before proceeding to a discussion of the case of complex roots.
5 CoMPLEx AND IMAc1NARY NUMBERs
4. In the discussion of the number system, a complex number is given by
a + bi (16)
(a) where a and b are real numbers and i is the imaginary unit. Now, in solving quadratic equations it is quite often the case that the roots are such numbers that b 0 in (15). Here the roots occur in a conjugate pair, or complex conjugates, meaning that if a + bi is one of the roots, then the other root must be a - bi. Recall also the rectangular coordinate system in which the first member and the second member of an ordered pair of numbers (a, b) were respectively measured along the horizontal axis and the vertical axis. Now, any complex number can be represented on a rectangular system if we agree to measure the imaginary part, or the number multiplied by the imaginary unit, along the vertical axis. This is done in Figure 1 for three examples as follows:
1. 2 + 4i is represented by the point P
2. 3 - 2i is represented by the point Q
3. -2 + 5i is represented by the point R
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Imaginary part
Figure 1
(b) Indeed, there is a one-to-one correspondence between an ordered pair of numbers (a, b) and the complex number a = bi, and hence a one-to-one correspondence between the point (a, b) on the X - Y plane and the complex number a + bi. Thus a + bi is called the rectangular form. Now, from geometry and trigonometry the rectangular form of a complex number can be represented in the polar form.
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