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Unformatted text preview: Res Ec 797A: Forecasting Order Of Topics And Readings For Lecture 10 (October 5, 2011) Source Pagegs) Top'c Announcements Exam 1 will be either on Wednesday evening, October 26 or Monday evening, October 31 or
Wednesday evening, November 2. Starting time will be 6:00 pm. Next Monday, October 10 is a holiday. We do not meet.
Tuesday, October 11 follows Monday’s schedule. We meet. As a reminder, start looking at: Enders Chapter 1 Difference equations: Pages 114; 14—17; 2238
(Start reading this material. Expect it to be tough). Continuing With DifferenceEquation Material:
Please look at the pages at the front end before class on October 5.
I will not get through all of these pages on October 5. BM 4548 Examples of solving ﬁrst—order difference equations by the
method of iteration (These pages were distributed as Lecture 9
material). BM 49 Solving two firstorder difference equations BM 50 A general solution and a homogeneous solution for a ﬁrst order difference equation BM 5157 Second—order difference equations: general solution;
homogeneous solution; and particular solution Will I Get To Any Of The Following? If Not, Please Look It Over Before Next Class. BM 5859 Examples: Applying the conditions for determining stability
BM 59—60 Introducing a stochastic component to the difference equation
BM 61 Determining the stability of a difference equation using: . rules pertaining to the equation’s coefficients
. the equation’s characteristic roots BM 62 Stability and Stationarity Example: Solve the difference equation Answer: a =
b = Solution: Is this equation stable? Why? What is the equilibrium state?
Do we converge to equilibrium?
Why? Example: Solve the difference equation Answer: a =
b = Solution: Is this equation stable?
Why?
What is the equilibrium state? Do we converge to equilibrium? Why? 49 Additional terminology: Return to our original ﬁrstorder, nonhomogeneous difference equation:
yl = ay‘_l + b. Bring it one period forward
y”. = ayt + b. and make it homogeneous: yl+l = ayl + 0. or y”l  ay‘ = 0.
We know that the general solution for yt = ay‘_l + b is
13‘
= ‘ + b __
y] a YO [1 a : So, the homogeneous solution for yt = ayl_l +b is yr : alyo A downtoearth explanation of a homogeneous solution is that it is that part of the solution which
is common among all y. Notice: y1 = a‘yo
y”. = a"'yo
yﬁz = aUZYO
y”; = a"3yo We now replace yo with the (more general) multiplicative constant B. We have: 3'. = a‘B
yt+l atllB The reason for this should become clear when dealing with secondorder difference equations. 50 SecondOrder Difference Equations Background Notation: $First difference: Ay = yvyl‘l 1 $ Second difference: $ deﬁnition: the ﬁrst difference of the ﬁrst difference A(Ayt) = My,  y,_.)
= Ayi Ayn =(y1_ y1])(y(. y1.2)
= yr  23m + ya = Ali/l Recall that an equation involving lags for the periods tl and t2 would be a second order
difference equation. A typical secondorder, linear, nonhomogeneous difference equation: yr + yll " 2y12 =12 Suppose we carry the notation two periods forward: y1+2 + y1+ _ 2Y1 = and we seek a solution (or solutions) to this equation. How do we proceed? This becomes quite a chore if we decide to use repeated substitution. Likewise coming up with a
general solution similar to that in the ﬁrstdifference situation requires another approach due to
the presence of the additional 2nddifference term. Start withageneral example: yl+2 + alyl+l + 32yl = b
and write its homogeneous form: yt+2 + alym + alyt = 0 Notice that we are looking for solutions for yt+2, yt+1 and y, to be expressed in terms of
coefﬁcients a] and a2. To get started, think of the purpose of the homogeneous solution. We know, from the previous
page, that y = aty0= atB plays a role in the general solution of a difference equation. 51 So, yt+1 = amB, and yt+2 = at+2B. Substitute these results into the previous equation:
Yt+2 + alYt+l + a2Yt = 0
i i
Ba“2 + alBat+l + azBat = 0
Now, cancel the common factor Bat to get:
a2 + ala + a2 = 0
Sorry about limiting myself to Aa=s@ only, but it should be apparent that this is a quadratic equationalso known as the characteristic equationwhich possesses two quadratic rootsor two
characteristic roots solved as follows mli— asolution la asolulion 2 _
2( 1) Three possibilities exist regarding these roots. It all hinges on the square root expression in the previous equation; that is, on Jaf  4(l)(a2) (1) When a? > 4a2 , the square root is a real number, and the two characteristic
roots will be real and distinct. (2) When 3,2 = 4212 , the square root vanishes, and the characteristic roots will be
repeated. (3) When 312 < 4212 , the characteristic roots will be conjugate complex. We are now in a position to put all of this previous background infonnation to work for us.
Making use of homogeneous solutions of our secondorder difference equation to ﬁnd the characteristic roots of the characteristic equation associated with this difference equation lets
us: (1) ﬁnd the solution for y; i.e., the time path of y; (2) detennine if the time path of y moves toward an equilibrium value.
Effectively, the previous set of notes is meant to explain Enders pages 114 so that we can pick
apart Enders pages 1417 and pages 223 8. Pages 19—21 provide economic examples that will interest some people. Pages 3841 introduces notation for lag operations that will be useful later
in the course. So, for now, focus your efforts on pages 1417 and 2238. 52 Before proceeding with concrete examples, let=s put some old terminology into perspective, and
let=s deﬁne some new terminology. On pages 14—15, Enders relates the concepts of homogeneous solution, particular solution, and
general solution. To appreciate these, 1et=s return to our original ﬁrstorder, nonhomogeneous difference
equation: y: = aym + b
Recall the general solution to this difference equation:
yt = awei) + i for a¢1
la la Recall that the homogeneous solution to the difference equation was deﬁned as a‘B. B is a
multiplicative constant. Now, look at the general solution directly above. Let B = (y0~—b—) so that a‘B = a‘(yOi) is the homogeneous solution. la la
Return to the original difference equation: Yt = aytl + b 
Suppose your task is to ﬁnd a valid solutiona particular solution—a solution that stabilizes or
becomes constant for different t. Proceed as follows. Let this desirable constant value of y be called k. Insert k for y and yH into the equation above, and solve for k in terms of a and b.
What do you get? (Also, see Enders, page 30). k = ak + b
kak = b
k(1a) = b
k = i.
1a Where have you seen this particular solution before? Summary: general solution
FR
b b
y1 = a‘(yo_ ) +
1a la
a—/ W; homogeneous particular
solution solution Shortly, we will apply this result to a secondorder difference equation example. 53 Enders: top of page 17: The general solution to a difference equation is deﬁned to be a particular solution plus
all homogeneous solutions. Why the phrase Aall homogeneous solutions@? Don=t we have just one homogeneous solution?
Answer: Increasing the order of the difference equation increases the order of the polynomial
requiring a solution. These deﬁnitions provide a ﬁne perspective for what it is we are trying to accomplish. Recall
that the general solution to a difference equation gives us what we want; i.e., the time path of y.
This was pretty easy to do when the process generating the data abided by a ﬁrst order
difference equation. What if the data generating process abides by higher order differences?
From what we did previously with a second order difference equation, we saw that generating
solutions was a bit sticky. In fact, up to now, you may be quite unsure what to do with the roots
that we derived. Again, the logic is this: Step 1: For the given difference equation, form its homogeneous version and ﬁnd all of
the homogeneous solutions. Step 2: Find a particular solution. Step 3: Obtain the general solution as the sum of the particular solution and a linear
combination of all homogeneous solutions. Step 4: Impose any initial condition(s) on the general solution. Result: You have made an honest attempt to discover the time path of y.
(Notice that these are Ender's= steps presented on page 15). Example: middle of page 17: ﬁnd the time path of y, given: y‘ = 0.9y‘_l0.2yl_2 + 3 : homogeneous solutions —> 2 of them Rewrite above as: yt  0.9yH + 0.2y1_2 = 3 homogeneous equation: y‘  0.9yt_I + 0.2yl_2 0 carrytwo periods forward: y”2  0.9y“l + 0.2yt = O 54 Recall the homogeneous solution: yt = a‘y0 = a‘B; y‘+1 a'HB; etc. Substitute items above into previous equation: a”2 B  0.9a‘+1 B + 0.23t B = 0
Divide through by a‘B: a2  0.9a + 0.2 = O This is a second degree polynomial. Solve using the quadratic formula.
(It is also the characteristic equation and has two characteristic roots). 0.9i‘/0.814(1)(0.2) 0.9i\/0.810.80 0.9i\/0.01 The two roots forAa@: —— = — = — 0.9i0.1 1.0 08 2 7 You could verify that these solutions work by plugging them, one at a time, into the
homogeneous difference equation: yt+2  O.9yt+1+ 0.2yt = O .
Recall deﬁnition 6 a‘+zB 0.9 a‘HB + 0.2 alB= o
Dividethroughbya‘B 6 a2  0.9 a1 + 0.2 = o
Fora=0.5 6 (0.5)2 0.9(0.5) + 0.2 —— o
Fora=0.4 6 (0.4)2 0.9(0.4) + 0.2 = 0 Now, use these roots to ﬁnd the (two) homogeneous solutions.
(Look at the ﬁrst line at the top of this page). 11: homogeneous h: homogeneous
y'l'l = (0.5)‘ ijl = (0.4)‘
Subscript 1: Solution 1 Subscript 2: Solution 2 Enders: bottom p. 16: "Now, suppose that there are two separate solutions to the homogeneous
equation denoted by y:“ and y'z‘t . It is straightforward to show that for any two constants A] and A2, the linear combination Al yll‘t + A2 ygl is also a solution to the homogeneous equation. "  Enders shows this directly below this statement on page 16  55 : Find a particular solution: Take the nonhomogeneous equation:
yt  0.9 yl_l + 0.2 yt_2 = 3 Deﬁne the particular constant value of y as k. Insert into equation kO.9k+0.2k= 3
0.3k= 3
k=lO. This is the particular solution for y. : Sum the particular solution and the linear combination of the two
homogeneous solutions to get the general solution: general solution yt = A1(0.5)‘ + A2(o.4)t + 10 homogeneous particular
solutions solution Step 4 : Impose any initial condition(s) on the general solution. Suppose yo = 13 and yl = 11.3. By substituting each of these into the general solution above, we can solve
for A1 and A2 simultaneously:
yo = 13 = A,(0.5)° + A2(0.4)° +10} (1)
yl = 11.3 = Al(0.5)' + A2(0.4)' +10
Simplifying the two LN
II + 7
equations above Al A” (2)
1.3 = 0.5,AH + 0.4A2 multiply 1st equation by 0.5 and 1.5 = O.5,AH  0.5A2 (3)
subtract it from 2nd equation 1.3 = 0.5 A, + 0.4 A2 —o.2 = — 0.1 A2 56 So, A2 = 2.0 and A. = 1.0, and the general solution or time path ofyis:
yt = (0.5)1 + 2(O.4)l + 10 In this example, notice that the characteristic roots were real and distinct. Enders provides an
example of a situation where these roots are repeated (Case 2, pages 2324) and where the roots
are imaginary (Case 3, pages 2526). Stability Recall, with the simple ﬁrst difference model, that once we determined the time path of y, we
wanted to determine whether the time path would lead to an equilibrium state or whether it
would take off wildly. We have the same objective with the second difference model. Enders provides a rigorous approach for determining stability. On pages 2629, he shows (for all
three cases mentioned above) that stability conditions are met if the characteristic roots lie within the unit circle. This takes a while to digest, but it is interesting. A very straightforward way of assessing convergence of the time path is see if the linear
combination of homogeneous solutions, i.e.: h h
AI y“ + A2 Y2!
tends toward zero as t—mo. In our previous example recall that fora=0.5—>y:" = (0.5)‘,y" = (0.5)‘+',...,y" = (0.5)“ 1.1+] 1. [+11 (0.4)‘,y" = (0.4)‘+',...,y" = (0.4)“ 2, {+1 2,1+n for a = 0.4 —> y'z“ Notice, then, that the linear combination of these homogeneous solutions converges toward zero as t—>00. Now, what does this mean about the time path of y1 as presented by the general
solution? It means that the time path moves toward some equilibrium value. Other results (regarding Case 1)  If am” <1 and asolz <1, —) convergence as t—mo o If asoll >1 and 3,012 >1, —) explosive as t—mo  If am” >1 and as“ <1, —) the second term dies down while the ﬁrst term tends to deviate
farther and farther. The ﬁrst term eventually dominates the
scene; the time path becomes divergent.  If a501, <1 and as,” >1, —) similar to that directly above 57 Useful rule: 0 Let the root with the higher absolute value be called the dominant root. 0 A time path will be convergent if and only if the dominant root is less than 1
in absolute value. (Note: The nondominant root will exert a deﬁnite inﬂuence on the time path 
at least in the beginning periods) Enders provides an alternative summary on pages 3435 in terms of the equation's
coefﬁcients rather than its characteristic roots. Also, see page 61 of these notes. 1. (1) If all of the coefﬁcients associated with the homogeneous portion of the original difference equation sum to <1, we have the necessary condition for stability. Enders says: 2 ai < l is the necessary condition for stability. i=
These coefﬁcients can be positive or negative. A sufﬁcient condition for stability is that the sum of the absolute value of the coefﬁcients must be <1. Enders says: 11
2 la!
l'l l is a sufﬁcient condition for stability. n If 2 ai = 1, at least one characteristic root equals unity. This is called a unit root
i = process. Examples: Applying The Conditions For Determining Stability Our original example again is:
yt+2 — 0.9 y‘+1 + 0.2 yt = 0, written as sz = 0.9ym  0.2 y‘ Coefﬁcients are: a1=0.9 a2=  0.2 . We showed: 350“ = 0.5; asolz = 0.4 and: yrt = (0.5)‘; y:l = (0.4)‘. All of this means convergence as t—>oo. 58 (2) Apply the previous rules relating to the coeﬂicients of the equation: (1) 2a. : 0.9  0.2 = 0.7 < 1—> Necessary condition for stability is met.
i=l Z
(2) 2ai = 0.9 + 0.2 =1.1 «it laSufﬁcient condition for stability is notmet. ix] 2
(3) We don't have a unit root process because 2 ai i 1.
i—I y‘  1.3yH + .2yl_2 = 0, written as yt = 1.3yH — .2y‘_2 _ 1.3i\/1.690.8 1.3i\/0.89 +
= = l.12,0.18 One root is outside the unit circle. This creates a problem for convergence. If we
applied the conditions that relate to the equation's coefﬁcients, we'd have: (1) 2—121; = 1.30.2 = 1.1 st 1—> Necessary condition is not met. We could
i= quit here. (2) iai = 1.3 + 0.2 = 1.5 st 1—> Sufﬁcient condition is notmet.
i= 2
(3) We don't have a unit root process because 2 ai 9t 1. i= Introducing A Stochastic Component To The Difference Equation Return to our 011' ginal difference equation:
yl : aytI + b! What if at is deﬁned to be a random disturbance and it replaces b above? Wehave
yt : aym + 8! where 81 is an i.i.d. disturbance and E(e_t)=0. Also assume E(8t at)= 62 59 and E(et 85)=0. This is called a random walk model if a=1. The behavior of y: is stochastic. It now pays to address the expected value of y, its variance, its covariance, and the magnitude
of "a". Speciﬁcally, we now pay attention to: E(yt) for all t
E(yt  pf = var(yl) for all t E(ytn)(yl_S—n) = cov(yt,y‘_s) for all t,s 6O Attempting To Determine The Stability Of A Difference Equation Using:
0 Rules Pertaining to the Equation's Coefﬁcients: The Easy Way
 The Equation's Characteristics Roots: The Ultimate Way If the If the Conclusions about Need we go further
necessary sufﬁcient stability using rules by calculating the
condition: condition: pertaining to the coefﬁcients: characteristic roots? l l l l
Holds Stability No No Conclusion Yes
Doesn't
Hold No need to address
the sufﬁcient condition No Stability No If 2;“ < l, we have the necessary condition for stability. Stability requires Ea. <1. However, even if Zai < 1, it is premature to conclude
i= i=l anything until we look at the sufﬁcient condition. But, if the necessary condition reveals that £3. 2 l, we can conclude that there is no stability. A sufﬁcient condition for stability is thatzlai < 1. If this occurs, we have stability. So, if Zai <1 we can conclude that there is stability. i=l If, however, Elail 21 (and the necessary condition has been satisﬁed), we can't conclude anything. We must look at the characteristic roots to reach a conclusion. 61 Stability and Stationarity (D Everything we've done so far had to do with the series becoming stable or converging. (2) From equation 3 on page 42: @
l Suppose that b is G) _
y‘ = a y“ + b1 _, replaced with 81, a —> y‘ = a ybl + at
stochastic disturb
y.=a y0+b1 ance (white noise), Y1=a Yo+81
y, = a yl + b2 assumed to have y2 = a yl + 32
nice properties: I . . ' E(€i)=E(sil)=w=02 Throu 11 re eated substituti n'
Through repeated substitutlon: g p 0 ' . EmaS) : Beatjail) yl = a‘y0 + ia‘J‘Sk
= O for all s k=l
l Enders, Chapter 2, page 50 i
This is the general solution. This is the general 501mm”
1 i
This is similar to Enders, page
54, equation 2.10. I
yl=a‘yo+2a"kbk
k 1 ® The seiies becoming stable
or converging depends upon
the magnitude of “a” l
relative to the value 1_ (6) Note: Introducing explicitly
the stochastic disturbance at
puts y in the position of being
a random variable. So, ex
plore the behavior of y through expected value operations. The y sequence (3
E(yt)¢E(y1+s) <— cannot be stationary. Additional, new concept.
(Enders, page 54) y is weakly stationary or
However, if t is large, E(yt) approaches (— A <— Coval‘iance stationary if
l E(yt+s). Similar reasoning applies to problem E(yl) = E(yl_s) = u (< 00)
the variance and autocovariance. Em“ _ “)2] = Em,” _ “)2] = a;
(Enders’ page 55) Bay.  u) (y... — L0] = E[(y..,  u) (y...  #0] (9 “Thus, if a sample is generated by a
process that has recently begun, the = Y; for an S realizations may not be stationary.” l
Enders, page 53 Enders, p. 52: The stability condition is a
necessary condition for the series to be stationary. 62 ...
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