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 Economic Dynamics

Dynamics is the study of the changes in the economy over time. This would also include the investigation of the growth rate of the economy and the characteristics of any cycle, such as its period. Dynamic models give the present performance of the economy in terms of aspects of the economy in previous time periods. For example, we might have an equation such as

#### Y t = 100 + 0.6Yt-1 + 0.16Yt-2.

This type of equation is known as a difference equation, but better names are iteration scheme and recursion relation. We can solve such an eqution iteratively if we know the two initial values. If Y 1 = 100 and Y 2 = 120, then Y3 = 100+72+16=188. Then it follows that Y4 = 100+112.8+19.2=232 and so on.

Often however we want a solution that enables us to compute Yt without computing all of the intermediate values.

Such a solution involves three steps:

• 1. Obtaining any particular solution to the equation; i.e., any solution no matter how special.
• 2. Obtaining the general solution to the corresponding homogeneous difference equation.
• 3. Finding values for the parameters in the general solution so the initial conditions are satisfied.

For the above example, we might look for a constant solution; i.e. Yt=Y. In order for Y to be a solution it must be that

Y = 100 + 0.6Y + 0.16Y.

This implies that Y = 100/0.34 = 294.12. This solution does not, however, satisfy the initial conditions that
Y 1 = 100 and Y 2 = 120.

We will need to add to the particular solution the general solution of the homogeneous difference equation. The homogeneous difference equation is the equation satisfied by yt=Yt-Y. Alternatively, it is the equation obtained by deleting the nonhomogeneous term 100; i.e.,

#### yt = 0.6yt-1 + 0.16yt-2.

We solve this equation by looking for solutions of the form

y t = ckt.

If yt = ckt then yt-1 = ckt-1 and yt-2 = ckt-2.

Thus

#### ckt = 0.6ckt-1 + 0.16ckt-2.

The coefficient c cancels out. If we divide by the lowest power of k we get

#### k2 = 0.6k + 0.16 or k2 - 0.6k - 0.16 = 0.

This is just a quadratic equation. Its roots are

#### k1 = 0.3 + 0.5 = 0.8 and k2 = 0.3 - 0.5 = - 0.2.

The general solution to the homogeneous difference equation is thus

#### c1(0.8)t + c2(-0.2)t

and therefore the general solution to the inhomogeneous difference equation is

#### 100/0.34 + c1(0.8)t + c2(-0.2)t

The coefficients c1 and c2 must be chosen so that Y1=100 and
Y 2=120; i.e.,

#### (100/0.34) + c1(0.8) + c2(-0.2) = 100 and (100/0.34) + c1(0.8)2 + c2(-0.2)2 = 120

Therefore c1 = -266.17625 and c2 = -94.1176. The complete solution is therefore

#### Yt = (100/0.34) -266.17625(0.8)t - 94.1176(-0.2)t

The solution asymptotically approaches (100.0/.34). The term -266.17625(0.8)t goes to zero monotonically as t increases without bound, whereas the term -94.1176(-0.2)t goes to zero cyclically switching sign.

Now suppose the difference equation is

#### k2 = 0.6k - 0.4, or k2 - 0.6k + 0.4 = 0.

The roots of this equation are:

These reduce to:

#### k1 = 0.3 + i0.5568 and k2 = 0.3 - i0.5568.

This pair of complex roots corresponds to a cyclical solution. A complex number a + bi can be expressed in polar form as

#### r*exp(iθ) where r=(a2 + b2)1/2and θ = tan-1(b/a)

In the above example,

Then

#### k 1 = (0.632)exp(i61.68) = and k 1t = (0.632)texp(i(61.68)t)

By DeMoivre's Theorem,

#### exp(iθt) = cos(θt) + i*sin(θt).

The general solution to the homogeneous difference equation is

#### c1k1t + c2k2t

where c1 and c2 may be complex. A particular solution of the inhomogeneous equation is Yt=Y, where must satisfy the condition

#### Y = 100 + 0.6Y - 0.4Y.

Therefore Y = 100/0.8 = 125. The general solution to the inhomogeneous difference equation is

#### Y t = 125 + c1k1t +c2k2t = 125 + (c1+c2)(0.632)tcos(61.68t) + i(c 1-c2)(0.632)tsin(61.68).

Although the coefficients c 1 and c2 may be complex the coefficients d1 = c1+c2 and d2 =i(c1-c2) must be real. Thus,

#### Yt = 125 + d1(0.632)tcos(61.68t) + d2(0.632)tsin(61.68t).

In order for Y1=100 and Y2=120 we must have

Thus

#### d10.3 + d20.556 = -25 d1(-0.22) + d20.334 = -5.

This means that d1 = -25.03 and d2 = -25.39. Thus the complete solution to the difference equation problem is

#### Yt = 125 + (0.632)t[-25.03cos(661.68t) - 25.39sin(61.68t)].

The solution goes therough a cycle when the argument of the cosine and sine functions goes through the interval 0 to 360o; i.e., 61.68T = 360 or T= 360/61.68 = 5.836 periods.

### The Accelerator Model

Production requires capital and labor so an increase in production in an economy operating at capacity will need an increase in capital; i.e., investment. Thus growth induces investment and the increased investment demand further stimulates production.

Let the capital stock K required to produce output Y be given by

#### K = σY.

The parameter σ is called the capital-output ratio. For the U.S. economy the value of σ seems to be about 3.

Investment is the net increase in capital stock in year t; i.e,

#### It = Kt - (1-δ)Kt-1

where δ is the depreciation rate for capital.

If the economy is operating at capacity production then

#### It = σYt - σ(1-δ)Yt-1.

Usually the accelerator model is represented as

#### It = γ(Yt-1 - Yt-2).

This assumes that δ=0 and that (Yt - Yt-1) is somehow related to (Yt-1 - Yt-2).

Thus when Ct = C0βYt-1, Gt = G0 and NXt = NX0

#### Yt = βYt-1 + γ(Yt-1 - Yt-2) + G0 + NX0

The equilibrium output Y satisfies the equation

#### Y = C0 + βY + γ(Y - Y) + G0 + NX0.

Subtracting this equation from the previous one gives:

#### Yt-Y = β(Yt-Y) + γ(Yt-1-Y) - γ(Yt-2-Y) or if yt = Yt-Y then yt = βyt-1 +γ(yt-1 - yt-2)

The solution of this homogeneous second-order difference equation involves the roots of the quadratic equation:

#### k2 - (β+γ)k + γ = 0

The roots of this equation are:

#### [(β+γ) ± ((β+γ)2-4γ)1/2]/2

The crucial thing for stability is the modulus of the roots, which for complex roots is:

#### γ

Thus the stability of this accelerator model is determined by the magnitude of the accelerator coefficient.

### The Time Scale and the Magnitude of the Accelerator Coefficient

It is clear that the dynamics of macroeconomic models depends crucially upon the magnitude of the accelerator coefficient; i.e. the coefficient of the growth of output in determining the level of investment. Likewise the capital/output ratio is important (and may be the same as the accelerator coefficient
. But the capital output ratio depends upon the time unit being used. When the output of the U.S. was \$6 trillion per year the capital stock was \$16 trillion. The capital/output ratio was therefore 3 years. If the output is measured per quarter then the output was \$1.5 trillion per quarter and the capital/output ratio was 12 quarters. It means the same thing but the magnitude is clearly different. In the case of the investment function

#### It = γ[Yt-1 - Yt-2]

If an increase in output of \$100 billion per year in output from 1990 to 1991 leads to investment of \$200 billion per year in 1992 then the accelerator coefficient γ is equal to 2 years. But the increase of \$100 billion per year from \$6000 billion per year in 1990 to \$6100 billion in 1991 is an increase from \$1500 billion per quarter in 1990I to \$1525 billion in 1991I, an increase of \$25 billion per quarter over a four quarter period. The level of investment that is induced in 1992I is \$50 billion per quarter. This corresponds to an accelerator coefficient of 2 quarters, which is not the same a 2 years even if the magnitudes are the same. If we take into account that the increase in output occured over four quarters the rate of increase is 25/4 or \$6.25 billion. In this case the accelerator coefficient would be 8 quarters, which means the same as 2 years but is not of the same magnitude. Clearly the models have to be formulated in a way that the time unit is irrelevant and the crucial parameters are dimensionless numbers.

Consider a continuous time version of the accelerator model. Let τ1 be a time lag such that

#### C(t) = C0 + βY(t-τ1).

Likewise let τ2 be such that

#### I(t) = γ[dY(t-τ2)/dt]

This leads to the delay differential equation:

#### Y(t) = C0 + βY(t-τ1) + γ[dY(t-τ2)/dt] + G0 + NX0

The corresponding homogeneous equation for the deviations from the equilibrium is then:

#### y(t) = βy(t-τ1) + γ[dy(t-τ2)/dt]

It is worthwhile to check the units of this equation. The variable y, as the variable Y, is in units of dollars per year, \$/T. The marginal propensity to consume is dimensionless. The derivative dy/dt has the dimensions of dollars per year per year or \$/T2. But the accelerator coefficient has dimensions of years so the product correctly has the dimensions of \$/T. The time lags of course have the dimensions of years. A technique that is used in the physical sciences when dealing with differential equations is to put such equations into the form in which the terms are dimensionless. The obvious choices to use in this case to produce a dimensionless equation are the time lags, the equilibrium level and the initial level of output. If we divide through by the equilibrium output Y and divide the accelerator coefficient by τ2 the result is:

#### y(t)/Y = β(y(t-τ1)/Y) + (γ/τ2) τ2d(y(t-τ2)/Y)/dt

The time variable can be taken to be t/τ2. Let this variable be denoted as τ and let y/Y be denoted as z. Then the above equation takes the form:

#### z(τ) = βz(τ-s) + g(dz(τ-1)/dτ)

where s = τ12 and g=γ/τ2.

If we look for solutions of the form exp(kτ) then k must be such that

#### 1 = βexp(-ks) + gk*exp(-k)

This is a transcendental equation and usually the solutions cannot be found analytically and must be approximated numerically. And of course, the solutions for k may be complex numbers.

### Fourier Analysis

When econoists want to determine whether cycles exist in the economy and if so what are their characteristics such cycle period they use a technique called Fourier Analysis. This method is an outgrowth of the discovery in the early nineteenth century by the Frech mathematician, Jean Baptiste Fourier. Fourier discovered that any periodic function can be represented as the sum of sinusoidal functions of various frequencies. Another way of saying this is that any periodic function can be discomposed into a sum of sinusoidal functions.

Suppose there is a possibly periodic function of time f(t) and one wants to determine whether f(t) includes a component with cycle period of L. The way to do this is to first construct the values of the sinusoidal functions, sin(2πt/L) and cos(2πt/L); i.e., the sine and cosine functions with a period L. If the periodic function f(t) is available for continous time then the component of f which have period L is found by evaluating the following integrals:

#### a(L) = (1/π)∫f(t)cos(2πt/L)dt b(L) = (1/π)∫f(t)sin(2πt/L)dt

In principle the integral should be from -∞ (negative infinity) to +∞ (positive infinity) but, in practice, the limits are over some finite interval such as 0 to T.

Both a(L) and b(L) represent the presence in f(t) of a cycle of period L. It is reasonable to combine the two into a single value. The measure that is used is the sum of their squares; i.e.,

#### c(L)2 = a(L)2 + b(L)2

The value of c(L)2 for all values of L is known as the spectrum of the function f. It is often more convenient to use the frequency of a cycle rather that its cycle period as the variable for analysis. The frequency is just the reciprocal of the cycle period length; i.e., the frequency s of a cycle with time period L is just 1/L, s=1/L. The spectrum of a function is more commonly c(s)2, a function of the frequency s.

There is a very beautiful theorem, called De Moivre Theorem, which says

#### exp(iz) = cos(z) + isin(z)

Thus the spectrum coefficient c(s) can be computed by the formula

#### c(s) = (1/2π)∫f(t)exp(2πst)dt

In this case, c(s) is a complex number and the spectrum is the squared magnitude of c(s), c(s) times its complex conjugate. There is a one-to-one correspondence between a function f(t) and its Fourier representation c(s). Sometimes f(t) is said to be the time domain representation of the function and c(s) is its frequency domain representation. The function c(s) is said to be the Fourier transform of the function f(t). If the transform process is applied to c(s) the result is f(t); i.e., c(s) is the Fourier transform of f(t) and f(t) is the Fourier transform of c(s).

For determining the dynamics of a model what is needed is not so much the solution for the model but instead the Fourier transform of the solution.

### Differential Equation Models in Macroeconomics

The natural format for economic relations is difference equations, but there are times where it is convenient to consider models expressed in the form of differential equations. Differential equations have the time lags left out of the analysis and so are simpler to analyze. Consider the following simple macroeconomic model with an accelerator model of investment:

• Y(t) = C(t) + I(t) + G(t)
• C(t) = C0 + bY(t)
• I(t) = γdY/dt

This model reduces to the equation

#### Y(t) = (1/(1-b))(C0 + G(t)) + (γ/(1-b))dY/dt or, equivalently, dY/dt = ((1-b)Y(t) - C0 - G(t))/γ

This equation may be solved by means of what is known as an integrating factor. First, it simplifies things if (1-b)/γ is expressed as ν. Incidentally (1-b) is the marginal propensity to save and γ should be the capital-output ratio so ν is the growth rate given by the Harrod-Domar Growth Model.

When the terms involving Y are moved to the left side of the equation and the whole equation multiplied by exp(-νt) the left side of the equation is an exact differivation of the product of Y(t) and exp(-νt); i.e.,

#### dY/dt - νY(t) = - (C0 + G(t)) exp(-νt)dY/dt - Y(t)νt)exp(-νt) = - exp(-νt)(C0 + G(t)) (d/dt)(exp(-νt)Y(t)) = - exp(-νt)(C0 + G(t))

This means the equation can be integrated from, say, 0 to z. The integration of the left side gives

#### exp(-νz)Y(z) - Y(0)

The integration of the right side of the equation gives

#### - ∫z0exp(-νt))(C0 + G(t))dt

The left and right side are equal and the resulting equation can easily be solved for Y(z); i.e.,

#### Y(z) = exp(νz)Y(0) - exp(νz)∫z0exp(-νt)(C0 + G(t))dt

This solution can be put into the following form by letting t-z=s and reversing the direction of integration:

#### Y(z) = exp(νz)Y(0) + ∫z0exp(-νs)(C0 + G(t-s))ds

The solution consists of two parts. One part involves constant exponential growth at a rate of ν. The other part depends upon the autonomous elements of aggregate demand, C0+G(t), The form of the term involving the autonomous damands is called a convolution integral. The convolution integral is the convolution of the exponential function and autonomous demand. The convulution tends to average out cyclical variations in G(t).

### Transform Methods for the Solution of Differential Equations

Transform methods convert the problem of finding a solution to a differential equation into an algebraic problem. Two commonly used transform methods are the Laplace Transforms and Fourier Transforms.

One of the important properties of the transforms is that the transform of the derivative of a function is simply related to the transform of the function; i.e.,

### Distributed Lags

The preceding presumed fixed, difinite lags in the relationships, but a better, more flexible way to express lagged relationships is in terms of a distributed lag. For example, instead of expressing a lagged relationship between current consumption and past income in the form

#### C(t) = βY(t-τ)

we can express it in the general form

#### C(t) = β(∫0Th(s)Y(t-s)ds)

where we require h(z) to satisfy the normalization condition

#### ∫0Th(s)ds = 1.

The average lag could be defined as:

#### τ = ∫0Tzh(z)dz

For a specific lag of τ the h(z) function has a spike at z=τ.

For the accelerator model we could have

#### I(t) = γ(∫0TH(s)Y'(t-s)ds)

where Y'(t-s) is the derivative of Y with respect to time at time (t-s). The normalization condition for H(z) would apply here as well.

The dynamics of an economy with such distributed lags would be given by the equation:

#### Y(t) = β(∫0Th(s)Y(t-s)ds) + γ(∫0TH(s)Y'(t-s)ds) + G(t) + X(t) - M(t).

The corresponding homogeneous equation would be:

#### y(t) = β(∫0Th(s)y(t-s)ds) + γ(∫0TH(s)y'(t-s)ds)

where y(t) is the differernce between Y(t) and a particular solution to the equation.

The solution to this type of equation requires what is known as Transform methods. Two important transfroms are the Laplace transform and the Fourier transform. What makes the laplace or Fourier transform methods convenient is that the integrals involved in the equation can be treated as the convolution of two functions. The transform of the convolution of two functions is equal to the product of the transforms of the individual functions. The integrals in the equation can be considered as the convolution of the functions if the distributed lag functions, h() and H(), are extended from T to infinity with zero values. Let T[f] stand for the transfrom, either Laplace or Fourier, of the function f. Then the transform of the above equation gives:

#### T[y](s) = βT[y](s)T[h](s) + γT[dy/dt](s)T[H](s)

where s is the variable for the transform functions.

One of the important properties of the transforms is that the transform of the derivative of a function is simply related to the transform of the function; i.e.,

#### T[dy/dt](s) = sT[y] - y(0)

Thus the equation satisfied by the transform of the function y(t) is

#### T[y](s) = βT[y](s)T[h](s) + γ(sT[y](s)-y(0))T[H](s)

Solving for T[y](s) yields