Difinitions And Concepts

\ diffrrtnce equation expresses a relationship between a dependent variable and a lagged link|vtideni variable (or variables) which changes at discrete intervals of time, for example. /, - f[ Y, i >. where / and Y are measured at the end of each year. The order of a difference equation i* determined by the greatest number of periods lagged. A Jim-order difference equation expresses a time lag «»I one period: a second-order, two periods: etc. The change in y as i changes from t to /+ I b c ilkd the first difference of y. It is written

where A is .in operator replacing dldi that is used to measure continuous change in differential equations. The»olunon of a difference equation defines y for every value of i and does not contain a difference expression See Examples I and 2.

EXAMPLE 1. I of the following is a difference equation of the order indicated.

Q, ~ a + />/*,-! order t y.. i - 9y„j + 2y,., + 6y, = 8 order 3

EXAMPLE 2 Of. en that the initial value of y is y®. in the difference equation y,.i = by, (17.2)

a solution is found as follows. By successive substitutions of t = 0, 1, 2, 3, etc. in (77.2), y 1 = byo y3 = by2 = b(b2y0) = b3y0

This method is called the iterative method. Since y0 is a constant, notice the crucial role b plays in determining values for y as t changes.


Given a first-order difference equation which is linear (i.e., all the variables are raised to the first power and there are no cross products), yt = byt^ + a (17.3)

where b and a are constants, the general formula for a definite solution is yt={y°~i^b)bt + T^b when b*1 {17A)

If no initial condition is given, an arbitrary constant A is used for y0 — a/(I — b) in (17.4) and for yQ in (17.4a). This is called a general solution. See Example 3 and Problems 17.1 to 17.13.

EXAMPLE 3. Consider the difference equation yt = ~lyt-\ + 16 and yQ = 5. In the equation, b = — 1 and a = 16. Since b 1, it is solved by using (17.4), as follows:

To check the answer, substitute t = 0 and t = 1 in (17.5).

yo = 3(-7)° + 2 = 5 since (-7)° = 1 yi = 3(—7)1 + 2 = -19

Substituting y1 = -19 for yt and y0 = 5 for yt-x in the original equation,


Equation (17.4) can be expressed in the general form yt = Ab* + c (17.6)

where A = y0- a/( 1 - b) and c = al( 1 - b). Here AU is called the complementary function and c is the particular solution. The particular solution expresses the intertemporal equilibrium level of y\ the complementary function represents the deviations from that equilibrium. Equation (17.6) will be dynamically stable, therefore, only if the complementary function Ab1-* 0, as t—All depends on the base b. Assuming A = 1 and c = 0 for the moment, the exponential expression bc will generate seven different time paths depending on the value of b, as illustrated in Example 4. As seen there, if \b \> 1, tha txxaa ^atix^pVotfte, MíraXhex iarth^i xwsy iiom equWíoiram; ii\b\ < 1. the time path will be damped and move toward equilibrium. Ií b < 0, the time path will oscillate betw---positive and negative values; if b > 0, the time path will be nonoscillating. If A ^ 1, the value of th=

Fig. 17-1 Time path of br multiplicative constant will scale up or down the magnitude of b\ but will not change the basic pattern of movement. If A = —1, a mirror image of the time path of bl with respect to the horizontal axis will be produced. If c 0, the vertical intercept of the graph is affected, and the graph shifts up or down accordingly. See Examples 4 and 5 and Problems 17.1 to 17.13.

EXAMPLE 4. In the equation yt = b\ b can range from -oo to oo. Seven different time paths can be generated, each of which is explained below and graphed in Fig. 17-1.

1. If b > 1, b' increases at an increasing rate as t increases, thus moving farther and farther away from the horizontal axis. This is illustrated in Fig. 17-1 (a), which is a step function representing changes at discrete intervals of time, not a continuous function. Assume b = 3. Then as t goes from 0 to 4, b* = 1, 3, 9, 27, 81.

2. If b = 1, bl = 1 for all values of t. This is represented by a horizontal line in Fig. 17-1(&).

3. If 0 < b < 1, then b is a positive fraction and bl decreases as t increases, drawing closer and closer to the horizontal axis, but always remaining positive, as illustrated in Fig. 17-1 (c). Assume b = Then as t goes from 0 to 4, b' = l,i ifj, h,

4. If b = 0, then b* = 0 for all values of t. See Fig. 17-1(d).

5. If -l<b <0, then b is a negative fraction; b' will alternate in sign and draw closer and closer to the horizontal axis as t increases. See Fig. 17-l(e). Assume b = Then as t goes from 0 to 4, bl = 1, —5, 5,

6. If b = — 1, then b' oscillates between +1 and — 1. See Fig. 17-1 (/).

7. If b < — 1, then will oscillate and move farther and farther away from the horizontal axis, as illustrated in Fig. 17-1 (g). Assume b = -3. Then bc = 1, -3, 9, -27, 81, as t goes from 0 to 4.

In short, if \b\>l the time path explodes

| b | < 1 the time path converges b > 0 the time path is nonoscillating b < 0 the time path oscillates

EXAMPLE 5. In the equation yt = 6(—J)' + 6, since b = —\ < 0, the time path oscillates. Since | b | < 1, the time path converges.

When yt = 5(6)' + 9 and b = 6>0, there is no oscillation. With \ b\>l, the time path explodes.

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