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Solução de Série Temporais - Walter Enders
Tipologia: Exercícios
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Semester Project 137
Questions and Exercises 42
APPENDIX 1.1 Imaginary Roots and de Moivre’s Theorem 44 APPENDIX 1.2 Characteristic Roots in Higher-Order Equations 46
Nearly all students will have some familiarity the concepts developed in the chapter. A first course in integral calculus makes reference to convergent versus divergent solutions. I draw the analogy between the particular solution to a difference equation and indefinite integrals. It is important to stress the distinction between convergent and divergent solutions. Be sure to emphasize the relationship between characteristic roots and the convergence or divergence of a sequence. Much of the current time-series literature focuses on the issue of unit roots. It is wise to introduce students to the properties of difference equations with unitary characteristic roots at this early stage in the course. Question 5 at the end of this chapter is designed to preview this important issue. The tools to emphasize are the method of undetermined coefficients and lag operators. Few students will have been exposed to these methods in other classes. I use overheads to show the students several data series and ask them to discuss the type of difference equation model that might capture the properties of each. The figure below shows three of the real exchange rate series used in Chapter 4. Some students see a tendency for the series to revert to a long-run mean value. The classroom discussion might center on the appropriate way to model the tendency for the levels to meander. At this stage, the precise models are not important. The objective is for students to conceptualize economic data in terms of difference equations.
60
80
100
120
140
160
180
1980 1983 1986 1989 1992 1995 1998 2001
(1996 = 100)
U.S. Canada Germany
yt = a 0 + a 1 yt - = a 0 + a 1 [ a 0 + a 1 yt -2 ] = a 0 + a 0 a 1 + a 0 ( a 1 )^2 + .... + a 0 ( a 1 ) t -1^ + ( a 1 ) ty 0
Bill added the homogeneous and particular solutions to obtain: yt = a 0 /(1 - a 1 ) + ( a 1 ) t [ y 0 - a 0 /(1 - a 1 )].
A. Show that the two solutions are identical for a 1 < 1.
Answer : The key is to demonstrate:
a 0 + a 0 a 1 + a 0 ( a 1 )^2 + .... + a 0 ( a 1 ) t -1^ + ( a 1 ) ty 0 = a 0 /(1 - a 1 ) + ( a 1 ) t [ y 0 - a 0 /(1 - a 1 )]
First, cancel ( a 1 ) ty 0 from each side and then divide by a 0. The two sides of the equation are identical if:
1 + a 1 + ( a 1 )^2 + .... + ( a 1 ) t -1^ = 1/(1 - a 1 ) - ( a 1 ) t /(1 - a 1 )
A. Find the homogeneous solution for p * t
Answer : Form the homogeneous equation p * t - (1 - α) pt *− 1 = 0. The homogeneous solution is:
p (^) t = A (1- α) t
where A is an arbitrary constant and (1- α) is the characteristic root.
B. Use lag operators to find the particular solution. Check your answer by substituting your answer into the original difference equation.
Answer : The particular solution can be written as
[ 1 - (1- α) L ] p * t = α pt -
or p * t = α pt -1/[ 1 - (1- α) L ] so that
p t = α[ pt -1 + (1- α) pt -2 + (1- α) (^2) p t -3 + ... ]
To check the answer, substitute the particular solution into the original difference equation
α[ pt -1 + (1- α) pt -2 + (1- α)^2 pt -3 + ... ] = α pt -1 + (1- α) α[ pt -2 + (1- α) pt -3 + (1- α)^2 pt -4 + ... ]
It should be clear that the equation holds as an identity.
A. Show that it is possible to express mt + n in terms of the known value mt and the sequence { ε t +1, ε t +2, ... , ε t + n ).
Answer : One method is to use forward iteration. Updating the money supply process one period yields mt +1 = m + ρ mt + ε t +1. Update again to obtain
mt +2 = m + ρ mt +1 + ε t + = m + ρ[ m + ρ mt + ε t +1] + ε t +2 = m + ρ m + ε t +2 + ρε t +1 + ρ^2 mt
Repeating the process for mt +
mt +3 = m + ρ mt +2 + ε t +
= m + ε t +3 + ρ[ m + ρ m + ε t +2 + ρε t +1 + ρ^2 mt ]
For any period t+n , the solution is
mt + n = m (1 + ρ + ρ^2 + ρ^3 + ... + ρ n -1) + ε t + n + ρε t + n -1 + ... + ρ n -1 ε t +1 + ρ nmt
B. Suppose that all values of ε t + i for i > 0 have a mean value of zero. Explain how you could use your result in part A to forecast the money supply n -periods into the future.
Answer : The expectation of ε t +1 through ε t + n is equal to zero. Hence, the expectation of the money supply n periods into the future is
m (1 + ρ + ρ^2 + ρ^3 + ... + ρ n -1) + ρ nmt
As n → ∞, the forecast approaches m /(1- ρ).
Answer : Assuming a 1 < 1, you can use lag operators to write the equation as (1 - a 1 L ) yt = ε t + β 1 ε t -1. Hence, yt = ( ε t + β 1 ε t -1)/(1 - a 1 L ).
Now apply the expression (1 - a 1 L )-1^ to each term in the numerator so that
yt = ε t + a 1 ε t -1 + ( a 1 )^2 ε t -2 + ( a 1 )^3 ε t -3 + ... + β 1 [ ε t -1 + a 1 ε t -2 + ( a 1 )^2 ε t -3 + ...]
yt = ε t + ( a 1 + β 1 ) ε t -1 + a 1 ( a 1 + β 1 ) ε t -2 + ( a 1 )^2 ( a 1 + β 1 ) ε t -3 + ( a 1 )^3 ( a 1 + β 1 ) ε t -4 + ...
If a 1 = 1, the improper form of the particular solution is:
ε β ε t-i i=
∞
1
where : an initial condition is needed to eliminate the constant b 0 and the non-convergent sequence.
ii. yt = a 1 yt -1 + ε 1 t + βε 2 t
Answer : Write the equation as yt = ε 1 t /(1- a 1 L ) + βε 2 t /(1- a 1 L ). Now, apply (1 - a 1 L )-1^ to each term in the numerator so that
A 1 + A 2 (-1) t
iii) yt = a 0 + 2 yt -1 - yt -2 + ε t Answer : The homogeneous equation is yt -2 yt -1 + yt -2 = 0. The homogeneous solution always takes the form yt = A α t. To form the characteristic equation, first substitute this challenge solution into the homogeneous equation to obtain
A α t^ - 2 A α t -1^ + A α t -2^ = 0
Next, divide by A α t -2^ to obtain the characteristic equation
α^2 - 2 α + 1 = 0
The two characteristic roots are α 1 = 1, and α 2 = 1; hence there is a repeated root. Thelinear combination of the two homogeneous solutions is also a solution. Letting A 1 and A 2 be two arbitrary constants, the complete homogeneous solution is
A 1 + A 2 t
iv) yt = a 0 + yt -1 + 0.25 yt -2 - 0.25 yt -3 + ε t Answer : The homogeneous equation is yt - yt -1 - 0.25 yt -2 + 0.25 yt -3 = 0. The homogeneous solution always takes the form yt = A α t. To form the characteristic equation, first substitute this challenge solution into the homogeneous equation to obtain
A α t^ - A α t -1^ -0.25 A α t -2^ + 0.25 A α t -3^ = 0
Next, divide by A α t -3^ to obtain the characteristic equation
α^3 - α^2 -0.25 α + 0.25 = 0
The three characteristic roots are α 1 = 1, α 2 = 0.5, and α 3 = -0.5. The linear combination of the three homogeneous solutions is also a solution. Hence, letting A 1 , A 2 and A 3 be three arbitrary constants, the complete homogeneous solution is A 1 + A 2 (0.5) t^ + A 3 (-0.5) t
B. Show that each of the backward-looking particular solutions is not convergent.
i) yt = a 0 + 1.5 yt -1 - .5 yt -2 + ε t Answer : Using lag operators, write the equation as (1 - 1.5 L + 0.5 L^2 ) yt = a 0 + ε t. Factoring the polynomial yields (1 - L )(1 - 0.5 L ) yt = a 0 + ε t. Although the expression ( a 0 + ε t )/(1 - 0.5 L ) is convergent, ( a 0 + ε t )/(1 - L ) does not converge.
ii) yt = a 0 + yt -2 + ε t Answer : Using lag operators, write the equation as (1 - L )(1 + L ) yt = a 0 + ε t. It is clear that neither ( a 0 + ε t )/(1 - L ) nor ( a 0 + ε t )/(1 + L ) converges.
iii) yt = a 0 + 2 yt -1 - yt -2 + ε t Answer : Using lag operators, write the equation as (1 - L )(1 - L ) yt = a 0 + ε t. Here there are two characteristic roots that equal unity. Dividing ( a 0 + ε t ) by either of the (1 - L ) expressions does not lead to a convergent result.
iv) yt = a 0 + yt -1 + 0.25 yt -2 - 0.25 yt -3 + ε t Answer : Using lag operators, write the equation as (1 - L )(1 - 0.5 L )(1 + 0.5 L ) yt = a 0 + ε t. The expressions ( a 0 + ε t )/(1 + 0.5 L ) and ( a 0 + ε t )/(1 - 0.5 L ) are convergent, but the expression ( a 0 + ε t )/(1 - L ) is not convergent.
C. Show that equation (i) can be written entirely in first-differences; i.e., ∆ yt = a 0 + .5∆ yt -1 + ε t. Find
the particular solution for ∆ yt. [HINT: Define y t *= ∆ yt so that y t *= a 0 - 0.5 yt *− 1 + ε t. Find the
particular solution for y t *in terms of the { ε t } sequence.]
Answer : Subtract yt -1 from each side of yt = a 0 + 1.5 yt -1 - .5 yt -2 + ε t to obtain
yt - yt -1 = a 0 + 0.5 yt -1 - .5 yt -2 + ε t so that ∆ yt = a 0 + 0.5 yt -1 - 0.5 yt -2 + ε t = a 0 + 0.5∆ yt -1 + ε t
The particular solution for y * t = a 0 + 0.5 yt *− 1 + ε t is given by
y (^) t = ( a 0 + ε t )/(1 - 0.5 L ) so that
y t = 2 a 0 + ε t + 0.5 ε t -1 + 0.25 ε t -2 + 0.125 ε t -3 + ....
D. Similarly transform the other equations into their first-difference form. Find the backward- looking particular solution, if it exists, for the transformed equations.
ii) yt = a 0 + yt -2 + ε t , Answer : Subtract yt -1 from each side to form yt - yt -1 = a 0 - yt -1 + yt -2 + ε t or ∆ yt = a 0 - ∆ yt -1 + ε t so that
y t = a 0 -
y (^) t − 1 + ε t
Note that the first difference ∆ yt has characteristic root that is equal to -1. The proper form of the backward-looking solution does not exist for this equation. If you attempt
Answer : You can use iteration or the Method of Undetermined Coefficients to verify that the solution is
[ 1 - (-1) ] 2
1
t t i
t i+t
i=
t y= + y +^ a
Using the iterative method, y 1 = a 0 + ε 1 - y 0 and y 2 = a 0 + ε 2 - y 1 so that
y 2 = a 0 + ε 2 - a 0 - ε 1 + y 0 = ε 2 - ε 1 + y 0
Since y 3 = a 0 + ε 3 - y 2 , it follows that y 3 = a 0 + ε 3 - ε 2 + ε 1 - y 0. Continuing in this fashion yields
y 4 = a 0 + ε 4 - y 3 = a 0 + ε 4 - a 0 - ε 3 + ε 2 - ε 1 + y 0 = ε 4 - ε 3 + ε 2 - ε 1 + y 0
To confirm the solution for yt note that (-1) i + t^ is positive for even values of ( i + t ) and negative for odd values of ( i + t ), (-1) t^ is positive for even values of t , and ( a 0 /2)[1 - (-1) t ] equals zero when t is even and a 0 when t is odd.
π t = -.05 + 0.7 π t -1 + 0.6 π t -2 + ε t
A. Suppose that in periods 0 and 1, the inflation rate was 10% and 11%, respectively. Find the homogeneous, particular, and general solutions for the inflation rate.
Answer : The homogeneous equation is π t - 0.7 π t -1 - 0.6 π t -2 = 0. Try the challenge solution π t = A α t , so that the characteristic equation is
A α t^ - 0.7 A α t -1^ - 0.6 A α t -2^ = 0 or α^2 - 0.7 α - 0.6 = 0
The characteristic roots are: α 1 = 1.2, α 2 = -0.5. Thus, the homogeneous solution is
π t = A 1 (1.2) t^ + A 2 (-0.5) t
The backward-looking particular solution is explosive. Try the challenge solution: π t = b + Σ bi ε t - i. For this to be a solution, it must satisfy
b + b 0 ε t + b 1 ε t -1 + b 2 ε t -2 + b 3 ε t -3 + ... = -0.05 + 0.7( b + b 0 ε t -1 + b 1 ε t -2 + b 2 ε t -3 + b 3 ε t -4 + ...)
Matching coefficients on like terms yields:
b = -0.05 + 0.7 b + 0.6 b ⇒ b = 1/ b 0 = 1 b 1 = 0.7 b 0 ⇒ b 1 = 0. b 2 = 0.7 b 1 + 0.6 b 0 ⇒ b 2 = 0.49 + 0.6 = 1.
All successive values for bi satisfy the explosive difference equation
bi = 0.7 bi -1 + 0.6 bi -
If you continue in this fashion, the successive values of the bi are: b 3 =1.183; b 4 = 1.4821; b 5 = 1.74727; b 6 = 2.11235; b 7 = 2.527007...
Note that the forward-looking solution is not satisfactory here unless you are willing to assume perfect foresight. However, this is inconsistent with the presence of the error term. (After all, the regression would not have to be estimated if everyone had perfect foresight.) The point is that the forward-looking solution expresses the current inflation rate in terms of the future values of the { ε t } sequence. If { ε t } is assumed to be a white-noise process, it does not make economic sense to posit that the current inflation rate is determined by the future realizations of ε t + i.
Although the backward-looking particular solution is not convergent, imposing the initial conditions on the particular solution yields finite values for all π t (as long as t is finite). Consider the general solution
π t = 1/6 + ε t + 0.7 ε t -1 + b 2 ε t -2 + ... + bt- 2 ε 2 + bt -1 ε 1 + bt ε 0 + bt +1 ε-1 + ... + A 1 (1.2) t^ + A 2 (-0.5) t
For t = 0 and t = 1:
0.10 = 1/6 + ε 0 + 0.7 ε-1 + b 2 ε-2 + ... + A 1 + A 2 0.11 = 1/6 + ε 1 + 0.7 ε 0 + b 2 ε-1 + ... + A 1 (1.2) + A 2 (-0.5)
These last two equations define A 1 and A 2. Inserting the solutions for A 1 and A 2 into the general solution for π t eliminates the arbitrary constants.
b 0 = 1 b 1 = 0 b 2 = a 2 b 0 ⇒ b 2 = a 2 b 3 = a 2 b 1 ⇒ b 3 = 0 (since b 1 = 0)
Continuing in this fashion, it follows that
bi = ( a 2 ) i /2^ if i is even and 0 if i is odd.
Answer : Using lag operators, rewrite the equation as β pt +1 - (1 + β) pt = α - mt. Combining terms yields [1 - (1 + 1/ β) L ] pt +1 = ( α - mt )/ β so that lagging by one period results in
[1 - (1 + 1/ β) L ] pt = ( α - mt -1)/ β
Since β is assumed to be positive, the expression (1 + 1/ β) is greater than unity. Hence, the backward-looking solution for pt is divergent.
B. Obtain the forward-looking particular solution for pt in terms of the { mt } sequence. In forming the general solution, why is it necessary to assume that the money market is in long-run equilibrium?
Answer : Solving for β pt yields
β pt = [ α - mt -1]/[1 - (1 + 1/ β) L ]
The expression (- mt -1)/[1 - (1 + 1/ β) L ] can be obtained using Property 6 of lag operators. Let (1 + 1/ β) correspond to the term a and apply Property 6 so that
(- mt -1)/[1 - aL ] = ( aL )-1[( aL )^0 + ( aL )-1^ + ( aL )-2^ + ( aL )-3^ + ... ] mt - = a -1[( aL )^0 + ( aL )-1^ + ( aL )-2^ + ( aL )-3^ + ... ] mt = a -1[ mt + a -1 mt +1 + a -2 mt +2 + a -3 mt +3 + ... ]
Since (1 + 1/ β) = a , it follows that a -1^ = β/(1+ β). Also note that α/[1 - (1 + 1/ β) L ] = - αβ. Thus, the solution for pt is
_m
p =- + t+i
i
i=
t
∞ β
β β
α 1 1
0
C. Find the impact multiplier. How does an increase in mt +2 affect pt? Provide an intuitive explanation of the shape of the entire impulse response function.
Answer : The impact multiplier is the effect of mt on pt. As such ∂ pt /∂ mt = 1/(1+ β). For any i , the effect of mt + i on pt is [1/(1+ β)][ β/(1+ β)] i. Since [ β/(1+ β)] i^ decreases as i increases, future values of the money supply have a smaller effect on the current price level than the current value. Notice that a permanent increase in the money supply, such that ∆ mt = ∆ mt +1 = ... has a proportional effect on pt since [1/(1+ β)]Σ[( β/(1+ β)] i^ = 1.
Answer : Substitute each posited solution into the original difference. A. Since yt = c and yt -1 = c , it immediately follows that c - c = 0. B. Since yt -1 = c + a 0 ( t -1), it follows that c + a 0 t - c - a 0 ( t -1) = a 0. C. The issue is whether c + c 0 (-1) t^ - c - c 0 (-1) t -2^ = 0? Since (-1) t^ = (-1) t -2, the posited solution is correct. D. Does c + c 0 (-1) t^ + ε t + ε t -2 + ε t -4 + ... - c - c 0 (-1) t -2^ - ε t -2 - ε t -4 - ε t -6 - ... = ε t? Since c 0 (-1) t^ = c 0 (-1) t -2, the posited solution is correct.
A. yt - 1.2 yt -1 + .2 yt -2 B. yt - 1.2 yt -1 + .4 yt - C. yt - 1.2 yt -1 - 1.2 yt -2 D. yt + 1.2 yt - E. yt - 0.7 yt -1 - 0.25 yt -2 + 0.175 yt -3 = 0 [Hint: ( x - 0.5)( x + 0.5)( x - 0.7) = x^3 - 0.7 x^2 - 0.25 x + 0.175]
Answers : A. The characteristic equation α^2 - 1.2 α + 0.2 = 0 has roots α 1 = 1 and α 2 = 0.2. The unit root means that the { yt } sequence is not convergent. B. The characteristic equation α^2 - 1.2 α + 0.4 = 0 has roots α 1 , α 2 = 0.6 ± 0.2 i. The roots are imaginary. The { yt } sequence exhibits damped wave-like oscillations. C. The characteristic equation α^2 - 1.2 α - 1.2 = 0 has roots α 1 = 1.85 and α 2 = -0.65. One of the roots is outside the unit circle so that the { yt } sequence is explosive. D. The characteristic equation α + 1.2 = 0 has the root α = -1.2. The { yt } sequence has explosive oscillations. E. The characteristic equation α^3 - 0.7 α^2 - 0.25 α + 0.175 = 0 has roots α 1 = 0.7, α 2 = 0.5 and α 3 = -0.5. Although all roots are real, there are damped oscillations due to the presence of the
