## Lutz G. Arnold

Print publication date: 2002

Print ISBN-13: 9780199256815

Published to Oxford Scholarship Online: October 2011

DOI: 10.1093/acprof:oso/9780199256815.001.0001

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# (p.146) (p.147) Appendix 1 Stochastic Second-order Difference Equations

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This appendix analyses stochastic linear second-order difference equations. To begin with, we explain the definitions and the behaviour of the sine and cosine functions. Then it is shown that under appropriate parameter combinations the solution to a non-stochastic second-order difference equation displays damped sinusoidal oscillations. Finally, it is demonstrated that the solution to a stochastic second-order difference equation which displays damped oscillations in the absence of shocks exhibits variability, persistence, and reversion in the presence of recurrent shocks.

# Trigonometric functions

The sine and cosine functions are defined by

A first important result is

(A1.1)

for all x. Differentiating the left-hand side of (A1.1) with respect to x gives 2 (sin x cos x − cos x sin x) = 0. Hence, sin2 x + cos2 x = a for some constant a. Setting x = 0 yields a = 1, which proves (A1.1).

Next, consider two differentiable functions f (x) and g (x) satisfying

(A1.2)

By definition, these requirements are fulfilled by f (x) = sin x and g (x) = cos (x). But there are further examples, such as f (x) = cos (−x) and g (x) = sin (−x), and f (x) = sin (x + y) and g (x) = cos (x + y), where y is an (p.148) arbitrary real number. Consider the functions

Differentiating with respect to x and using (A1.2) yields cos xg (x)−sin xf (x)+ sin xf (x)−cos xg (x) = 0 and cos xf (x)+sin xg (x)−sin xg (x)−cos xf (x) = 0, respectively. Hence, there exist two real numbers a and b such that

Multiply the former equation by sin x and the latter by cos x and add. Then multiply the former by cos x and the latter by sin x and subtract the former from the latter. Using (A1.1), one obtains

(A1.3)
(A1.4)

(A1.3) and (A1.4) can be used to prove

(A1.5)

To do so, let f (x) = cos (−x) and g (x) = sin (−x), which, we know, satisfy (A1.2). Then (A1.3) and (A1.4) become

Setting x = 0 yields b = 0 and a = −1. Inserting this into the equations above proves (A1.5).

Next, we derive the addition formulas

(A1.6)

from (A1.3) and (A1.4). Let f (x) = sin (x + y) and g (x) = cos (x + y), where y is an arbitrary real number. Then,

Setting x = 0 yields b = cos y and a = − sin y, which yields (A1.6) upon substitution into the above pair of equations.

(p.149) Since sin2 x + cos2 x = 1, the sine and cosine curves range between −1 and +1. They display harmonic oscillations: there exists a real number π such that

(A1.7)
(A1.8)
(A1.9)
(A1.10)
(A1.11)
(A1.12)

1. (A1.7): This holds true by the definitions of sine and cosine.

2. (A1.8): At the origin, the sine curve is upward-sloping (sin′ 0 = cos 0 = 1). So the cosine curve is downward-sloping for x small: cos′ x = − sin′ x 〈 0. There exists an x (〉 0) such that sin x = 1 (and cos x = 0). Suppose this is not the case. Then cos x = sin′ x 〉 0 for all x. sin x 〈 1 and sin′ x 〉 0 imply that sin x converges to a constant a 〉 0. But this implies that cos′ x =− sin′ x converges to −a 〈 0. This contradicts cos x 〉 0 for all x. The smallest value x such that cos x = 0isdenotedby π/2.

3. (A1.9): This follows from the addition formula for the cosine function: cos π = cos (π/2 + π/2) = cos (π/2) cos (π/2) − sin (π/2) sin (π/2) = −1. From sin2 x + cos2 x = 1, it follows that sin π = 0.

4. (A1.10): Similarly, sin (3π/2) = sin (π + π/2) = sin π cos (π/2) + sin (π/2) cos π =−1 and cos (3π/2) = 0.

5. (A1.11): cos (2π) = cos (π + π) = cos π cos π − sin π sin π = 1 and sin (2π) = 0.

6. (A1.12): sin (x + 2π) = sin x cos (2π) + sin (2π)cos x = sin x and cos (x + 2π) = cos x cos (2π) − sin x sin (2π) = cos x. The sine and cosine functions take on the same value every 2π periods.

Finally, we prove DeMoivre’s theorem:

($i ≡ − 1$ denotes the imaginary unit). The proof is by induction. The validity for t = 1 is obvious. So it remains to show the validity for t − 1, (p.150) that is,

entails validity for t. Multiplying both sides of the equation by cos ω ± i sin ω gives

From the addition formulas (A1.6), cos ω cos ω(t − 1) − sin ω sin ω(t − 1) = cos ωt and sin ω cos ω(t − 1) + cos ω sin ω(t − 1) = sin ωt. It follows that

This completes the proof of DeMoivre’s theorem.

# Non-stochastic equations

Next, we examine second-order difference equations in the absence of stochastic disturbances:

(A1.13)

The first important thing to note is that if y 1,t and y 2,t are two particular solutions of (A1.13), then any linear combination y t = A 1 y 1,t + A 2 y 2,t of the two also satisfies (A1.13) (A 1 and A 2 are arbitrary, non-zero constants):

Suppose there exist numbers λ ≠ 0 such that y t = λ t are solutions to (A1.13). Then λ t + a 1 λ t−1 + a 2 λ t−2 = 0 or, dividing by λ t−2 (≠ 0),

This is the characteristic equation of (A1.13). Its solutions,

(p.151) are called the characteristic roots of (A1.13). Assume $Δ ≡ a 1 2 − 4 a 2 〈 0$. Then the characteristic roots are complex conjugates:

where α ≡−a 1/2 and $θ ≡ | Δ | / 2$. Since $λ 1 t$ and $λ 2 t$ are distinct solutions to (A1.13), the linear combination

also solves (A1.13). This equation is the general solution of (A1.13). In order for y t to be real for all t, A 1 and A 2 must be complex numbers. Let

(A1.14)

where A and e are real numbers. We proceed to show that the solution of (A1.13) is , where ω is a real number. This equation is derived in several steps, the non-self explanatory of which are explained below:

(A1.15)
(A1.16)
(A1.17)
(A1.18)
(A1.19)

1. (A1.16): Let ω and r be determined by cos ωα/r and sin ωθ/r. Then cos ω/sin ω = α/θ. Since, from (A1.7) and (A1.9), cos ω/sin ω equals ∞ for ω = 0 and −∞ for ω = π, there exists an ω and, hence, an r = α/cos ω which satisfy these equations. Substituting α = r cos ω and θ = r sin ω into (A1.15) gives (A1.16). Notice that α 2 + θ 2 = r 2(sin2 ω + cos2 ω) = r 2, hence $r = α 2 + θ 2$.

2. (A1.17): This is the crucial step in the proof: The time argument ‘wanders’ into the sine and cosine terms. We obtain the equation by applying DeMoivre’s theorem.

3. (p.152) (A1.18): In this step, the imaginary unit i disappears. Use is made of the fact that A 1 + A 2 = A cos e and (A 1A 2)i = (−iA sin e)i = − i 2 A sin e = A sin e, as implied by (A1.14) together with i 2 =−1.

4. (A1.19): This follows from (A1.5) and (A1.6):

The period of oscillation of y t is given by t′ − t where ωt′ − e = ωte + 2π. It is equal to t′ − t = 2π/ω.

# Random variables

Random variables are variables which take on different possible values with given probabilities. For our purposes, it is sufficient to consider continuous random variables, which can take on arbitrary real numbers y. Suppose the distribution of the variable can be described by means of the continuously differentiable distribution function H (y), where H (y) is the probability that the random variable takes on a value no greater than y. H (y) is non-decreasing with limy→−∞ H (y) = 0 and limy→∞ H (y) = 1. The probability that the random variable falls into the interval [y, y + dy] is H (y + dy) − H (y). As dy goes to zero, this probability approaches dH (y) and the average value of the random variable in this interval approaches y. The expectation of the random variable is obtained by ‘summing’ over the probability-weighted y-values:

Two random variables x and y are independent when the distribution functions G(x) and H (y) are independent of each other. Independent random variables satisfy E(xy) = Ex Ey:

# (p.153) Stochastic equations

We proceed to derive the equations concerned with the variance and autocorrelations of y t in the presence of shocks. Since (1 + a 1 + a 2)Ey = = 0, the expectation of y t is Ey t = 0. The variance of y t is $σ y 2 ≡ E y t 2$, the covariance between y t and y tj is E(y t y tj), and the correlation between y t and y tj is ρ jE(y t y tj)/σ 2. The covariance function satisfies

Hence, ρ j = ρ j. From y t + a 1 y t−1 + a 2 y t−2 = εt, we have

(A1.20)

Dividing by σ 2 and making use of the fact that εt is independent of y t−1 and y t−2 and that E(y t−1 y tj) = E[y t y t−(j−1)] and E(y t−2 y tj) = E[y t y t−(j−2)], one obtains

for all j 〉 0. Setting j = 1 and j = 2, it follows that

(A1.21)

where use is made of the symmetry property ρ j = ρ j. To calculate the variance σ 2 of y t, set j = 0 in equation (A1.20) and notice that $E ( ε t y t ) = σ ε 2$ because εt is independent of y t−1 and y t−2:

Substituting the expressions in (A1.21) for ρ 1 and ρ 2 yields the formula reported in the main text: