Res Ec 797A Lecture 10 - Fall 2011

Res Ec 797A Lecture 10 - Fall 2011 - Res Ec 797A:...

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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 1-14; 14—17; 22-38 (Start reading this material. Expect it to be tough). Continuing With Difference-Equation 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 45-48 Examples of solving first—order difference equations by the method of iteration (These pages were distributed as Lecture 9 material). BM 49 Solving two first-order difference equations BM 50 A general solution and a homogeneous solution for a first- order difference equation BM 51-57 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 58-59 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 first-order, 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 1-3‘ = ‘ + b __ y] a YO [1- a :| So, the homogeneous solution for yt = ayl_l +b is yr : alyo A down-to-earth 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 second-order difference equations. 50 Second-Order 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 t-l and t-2 would be a second order difference equation. A typical second-order, linear, nonhomogeneous difference equation: yr + yl-l " 2y1-2 =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 first-difference situation requires another approach due to the presence of the additional 2nd-difference 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 equation--also known as the characteristic equation--which possesses two quadratic roots--or 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 second-order 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 1-14 so that we can pick apart Enders pages 14-17 and pages 22-3 8. Pages 19—21 provide economic examples that will interest some people. Pages 38-41 introduces notation for lag operations that will be useful later in the course. So, for now, focus your efforts on pages 14-17 and 22-38. 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 first-order, nonhomogeneous difference equation: y: = aym + b Recall the general solution to this difference equation: yt = awe-i) + i for a¢1 l-a l-a 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‘(yO-i) is the homogeneous solution. l-a l-a Return to the original difference equation: Yt = ayt-l + b - Suppose your task is to find a valid solution--a 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 k-ak = b k(1-a) = b k = i. 1-a Where have you seen this particular solution before? Summary: general solution FR b b y1 = a‘(yo_ ) + 1-a l-a a—/ W; homogeneous particular solution solution Shortly, we will apply this result to a second-order 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‘_l-0.2yl_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‘/0.81-4(1)(0.2) 0.9i\/0.81-0.80 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 k-O.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 23-24) and where the roots are imaginary (Case 3, pages 25-26). 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 26-29, 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 34-35 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) 2|ai| = |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\/1.69-0.8 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; = 1.3-0.2 = 1.1 st 1—> Necessary condition is not met. We could i=| quit here. (2) i|ai| = |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 : ayt-I + 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(yt-n)(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 Z|ai| <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 ' . Ema-S) : Beat-jail) 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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Res Ec 797A Lecture 10 - Fall 2011 - Res Ec 797A:...

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