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Unformatted text preview: Res Ec 797A: Forecasting Order Of Topics And Readings For Lecture 9 (October 4, 2011) Source Assignment
5 BM BM BM Enders Pagegs) Topic
Three Pages Due October 12, 2011 (The majority of the three pages
Numbered 13 relates to speciﬁc SAS instructions rather than to assignment
tasks.)
Four Pages SAS/ETS Time Series Forecasting System Notes: FineTuning
Numbered 14 And Additional Explanations (with speciﬁc reference to
Assignment 5) A Clean Break: Difference Equations 42 Summary of BM pages 4363: Background on Difference
Equations
4348 First installment of a set of notes on difference equations and their use in time series. Chapter 1 Difference equations: Pages 114; 1417; 2238
(Start reading this material. Expect it to be tough). University of Massachusetts
Department of Resource Economics RESEC 797A Fall 2011
Assignment 5 HOLTWINTERS EXPONENTIAL SMOOTHING In this exercise you will be estimating and forecasting with exponential smoothing for Variables
1 and 2. Split your data into withinsample and postsample parts using the same time periods as
you did for the previous exercises. 1. Use the appropriate twoparameter or threeparameter exponential smoothing algorithm in
SAS for each series. Produce onestepahead withinsample and outof—sample forecasts.
Compute withinsample and outof—sample RMSEs. Compare the accuracy of these with naive
nochange forecasts; i.e., use Theil’s U. You need to get the data into a form that SAS can read. This can be a major obstacle. If it
consists of a text file and the date is not in a form recognizable by SAS, the code shown below
and the notes that follow will get you started. /* this program reads a text ﬁle of two columns, the variable is called series */
/* date is in the form yyyy.q for quarterly data *x’
/* SAS does not appear to read the date in this form *f
/* the function intnx is used to calculate the date *!
data templ (keep = date empl);
inﬁle 'd:\sas\table51.dat';
input num $ empl;
date=intnx('qtr', '1jan85'd, _n_1);
format date ych. ;
run; Notes:
1. SAS can read an Excel spreadsheet. You can import it into SAS. 2. The above code will read a single column of data also if the input line is changed to:
input empl ; 3. The INTNX function assigns a SASreadable date to each observation. My series started from
January 1985. Insert your starting date in similar way. Use 'qtr', 'month' , or 'year' as appropriate. 4. If your data are monthly change the format command to:
format date yymmc. ; Make sure you include the "dot" after yymmc If yearly, use: format date year. (with the "dot" also). 5. If you PROC PRINT the output, it should look like The SAS System 11:03 Thursday, September 29, 2011 OBS EMPL DATE 1 416.0 1985:1
2 446.8 198512
3 461.9 l985:3
4 465.7 198524 6. Go to the Tool Bar. Click on the commands: SolutionslAnalysisTime Series Forecasting System.
Click on the "Browse ..." box to the right of "Data Set". Highlight the library "WORK". Highlight the SAS data set "TEMPl ".  You should ﬁnd that the "Time 1D" and "Interval" are correctly ﬁlled in, based on the information
contained in the SAS code on the previous page. Click OK. Don't forget to do this!
7. You are now in the "Time Series Forecasting" dialog box. Click "Develop Models". 8. You are now in the "Series Selection" dialog box. You now need to select the variable with which you will be working. Click on "empl/", located in the white box under "Time Series Variables".
Then, click "OK". 9. You are back in the "Develop Models" dialog box. You need to specify the ranges of your within
sample and postsample data sets. Click on the "Set Ranges..." box located to the right of
"Evaluation Range:". 10. You are now in the "Time Ranges Speciﬁcation" dialog box. You will notice that the starting
and ending points of your data are displayed next to "Period of Fit". You will want to change this
range so that you exclude the last 8 or 12 observations (the reserved data). Change the ”Period of
Evaluation" range to the same limits for your postsample. (Click on the horizontal pointers to make
changes.) You can set the periods for the "Forecast Horizon", if you desire. (The Forecast Horizon is beyond
the postsample period, and we have not been using this for anything). Also, specify the correct
number of periods in your "Holdout Sample". (This is the postsample.) Finally, click "OK". 11. You are back in the "Develop Models" dialog box. Click "No Models" located in the big, white
box. Click on "Fit Smoothing Model". Choose an appropriate exponential smoothing model. This will most likely be either Holt's or Winters' multiplicative depending on your data (annual would
only need Holt's). Click "OK". 12. Go to "View" on the Tool Bar and check out what you've done! Play with it! For example,
click on "Forecast Table" to get your series of onestep ahead forecasts, or click on "Parameter Estimates" or beta hat. You will see the estimates of the smoothing parameters and the
level, trend, and seasonal factors. Click on sigmasquared and get summary statistics including
RMSE. University of Massachusetts
Department of Resource Economics RESEC 797A SAS/ETS Time Series Forecasting System Notes
FineTuning and Additional Explanations Return to the SAS program presented at the beginning of Assignment 5. It was a program that read in quarterly data
for use in the SAS Econometric TimeSeries (ETS) environment. Recall that we did nothing with this program or with the data set used in the program except to illustrate how to get the data into a form that SAS ETS could use
immediately. We now use the program presented at the beginning of Assignment 5 and its data set to show how to derive relevant
withinsample and postsample information. Reading in data by way of the original SAS program:
The data in Table 51 are employment data. So, I will refer to them as “empl” .  data temp1 (keep = date empl);
infile 'a:\tab1e51.dat‘;
input num $ empl;
date=intnx(’qtr',‘1jan85'd,_n_1);
format date ych.; ' proo print data=temp1 ;
run; Output: The SAS System 083 DATE EMPL
 1 1985.1 416.0
2 1985.2 446.8
3 1985.3 461.9
4 1985.4 465.7
5 1986.1 445.9
6 1986.2 471.3
7 1986.3 486.6
8 1986.4 484.2
9 1987.1 449.2
10 1987.2 483.2
11 1987.3 489.6
12 1987.4 484.3
13 1988.1 476.5
14 1988.2 507.0 15 1988.3 516.3
16 1988.4 510.8 The steps presented in Assignment 5 provide a very detailed explanation pertaining to estimating an exponential
smoothing model. After going through these steps several times. reinforcement begins to take hold and things begin
to be automatic. It's time to ﬁnetune things. That is, think of the information We need to generate so that we can
put Assignment 5 in the same format as all other assignments. Speciﬁcally, our plan is to obtain ouestep—ahead forecasts for the withinsample and postsample periods for Variable l and Variable 2. Implied here is obtaining the
RMSE for both the withinsample and postsample periods. To get withinsample RMSE Complete everything through Step 8 on page 2 of the Assignment 5 handout. After clicking “OK”, you are in the
“Develop Models" dialog box. The very first statistic you desire to generate is the withinsample RMSE. Click on
the “Set Ranges. . ." box. You are now in the “Time Ranges Specification” dialog box. Let the end points of both
the “Period of Fit" range and the “Period of Evaluation" range be those that define your withinsample period. In
my simple example, this is 198511 to 1986z4. Click “OK". You are back in the “Develop Models" dialog box. Click “No Models” located in the big, white box. Click on “Fit
Smoothing Model”. Choose an appropriate exponential smoothing model. Try Winters‘ multiplicative on these
quarterly data. Click “OK”. Notice that I am back to the “Develop Models” dialog box, “Winters Method —
Multiplicative” is highlighted in black, and the Root Mean Square Error value 3.11602 is reported. Looking at output from estimation Go to the Tool Bar and click “View". Then, click “Parameter Estimates". You will see the estimates of the
smoothing parameters and the level, trend, and seasonal factors. My results are shown below. Paramet er Estimates EMPL
Winters Method  Multiplicative PARM VALUE STDERR T P
LEVEL Smoothing Weight 0.00100 0.2098 0.004766 0.9964
TREND Smoothing Weight 0.00100 35.1800 0.0000284 1.0000
SEASONAL Smoothing Weight 0.54940 0.6589 0.8338 0.4424
Residual Variance (sigma squared) 15.53528 . . .
Smoothed Level 480.54870 . . .
Smoothed Trend 6.13097 . . .
Smoothed Seasonal Factor 1 0.95875 . . Smoothed Seasonal Factor 2 1.00621 . . Smoothed Seasonal Factor 3 1.02571 . . .
Smoothed Seasonal Factor 4 1.01312 . . . The values for the level and trend smoothed series apply to the last withinsample value. The seasonal factors are the
last four calculated. If your withinsample series had ended on 1987.1, the seasonal factors would be listed in the
order 1987.], 1986.2, 1986.3, 1986.4. (The values would be different from the ones shOWn here since the additional
observation will cause the optimal smoothing parameters to change as well as the values of the level, trend and
seasonal factors series.) The denominator of the residual variance is the number of observations withinsample (here, 8) minus the number of
smoothing parameters (here, 3). Go back to the Tool Bar, click “View” and then “Statistics of Fit”. You will see something similar to the table
below. Notice that the denominator of MSE (presented below) is the number of observations within—sample. The
RMSE is the same value as the one shown previously with the blackened model title. Note that this is the within
sample RMSE. Statistics of Fit EMPL
Winters Method  Multiplicative Statistic Description Value
MSE Mean Square Error 9.70955
RMSE Root Mean Square Error 3.11602
MAPE Mean Absolute Percent Error 0.44325
MAE Mean Absolute Error 2.04143
RSQUARE RSquare 0.979 To get expost RMSE and onestepahead ex post forecasts Click “Options" on the Tool Bar. Then, click “Time Ranges Speciﬁcation". Change the Period of Evaluation to
accommodate the post—sample observations. These include 1987:]  19884. Click “OK”. You will get new
“Statistics of Fit" output. This is postsample information. It appears below. Statistics of Fit EMPL
Winters Method  Multiplicative Statistic Description Value
MSE Mean Square Error 238.77133
RMSE Root Mean Square Error 15.45223
MAPE Mean Absolute Percent Error 2.75793
MAE Mean Absolute Error 13.36496
RSQUARE RSquare 0.428 To the right of the “Statistics of Fit" table in the Model Viewer dialog box is a column of icons. Click the bottom
one. This is the “Forecast Table" that provides within sample and postsample forecasts. I highlighted the postsample results and “printed” that output to my output log. Date Actual Predict Upper Lower Error _Leve_ _Trend_ _Sfactor
1985:1 416.0 416.759 424.484 409.033 0.759 437.628 6.131 0.951
19852 446.8 445.922 453.647 438.197 0.878 443.760 6.131 1.006
1985:3 461.9 460.969 468.694 453.243 0.932 449.892 6.131 1.026
1985:4 465.7 464.318 472.044 456.593 1.382 456.024 6.131 1.020
198621 445.9 439.676 447.401 431.950 6.225 462.161 6.131 0.959
19862 471.3 471.084 478.809 463.358 0.217 468.293 6.131 1.006
198623 486.6 486.645 494.370 478.919 0.045 474.424 6.131 1.026
198614 484.2 490.096 497.821 482.371 5.896 480.549 6.131 1.013
1987:1 449.2 466.605 474.330 458.880 17.405 486.662 6.131 0.939
1987:2 483.2 495.855 503.580 488.130 12.655 492.780 6.131 0.992
1987:3 489.6 511.737 519.462 504.012 22.137 498.889 6.131 1.001
1987:4 484.3 511 .647 519.372 503.921 27.347 504.993 6.131 0.983
1988:1 476.5 480.008 487.733 472.283 3.508 511.120 6.131 0.935
1988:2 507.0 513.175 520.900 505.450 6.175 517.245 6.131 0.986
198823 516.3 524.085 531.810 516.360 7.785 523.368 6.131 0.993
1988:4 510.8 520.709 528.434 512.984 9.909 529.489 6.131 0.973
1989:1 500.995 508.720 493.270 6.131 0.935 535.620 Again, the withinsample data, used for estimation, are from 1985:1 to l986:4. The bold ﬁgures are for the ex post
period plus the one ex ante forecast. The prediction intervals (called, regretably, conﬁdence intervals) are titled U95
and L95 in the model viewer, being the predicted value 1 square root of residual variance x 1.96. When you “print”
the output, you lose the information that they are 95% intervals. Unfortunately you do not get RMSE for naive no
change nor do you get Theil’s U. Summary of BM pages 4363: Background on Difference Equations Start with a difference equation —> This is supposed to characterize the relationship among
observations over time. Find the time path.
Solve for the general solution, which is the time path.
Find the homogeneous solution and then the particular solution. Use the homogeneous solution to ﬁnd the characteristic equation and the characteristic
roots. Determine whether the difference equation leads to stability by comparing the characteristic
roots to the value 1. Show a simpler set of rules (using the coefﬁcients of the difference equation) to look at the
necessary and sufﬁcient conditions for stability. Resort to looking at the magnitude of the characteristic roots if the sufﬁcient conditions for
stability using the coefﬁcients fail. Introduce a stochastic structure to the difference equation as the way to begin addressing
stationarity. 42 Background On Difference Equations Focus of a time series model:  To decompose a series into its trend, seasonal, cyclical, and irregular components. 0 Each of these components is subject to stochastic behavior. 0 The trick is to discover the process that generates each component. 0 Discovery begins with an expression that shows the relationship among observations
through time. This is a difference equation. We manipulate (or solve) the difference
equation to discover the DGP. The solution expresses the value of a variable in terms of its
own lagged values, time, and other variables. 0 By deﬁnition, this ﬁnal expression is a time path. 0 Conclusion: Start with a difference equation; solve it; end up with a time path. Linear difference equations and their solutions play an important role in practically all areas of time
series analysis. If time is regarded as a continuous variable and the equations involve unknown functions and their
derivatives, we ﬁnd ourselves considering differential equations. If time is taken to be a discrete (or integer) valued variable and the equations relate the values of
such variables at different points in time, then we are confronted with difference equations or
recurrence relations.
Characteristics (regarding both difference equations and differential equations) Example: ym  yt = 2 (1) 0 linear or nonlinear: The example above is linear since no y term (for any period) is raised to a power higher than
1. ‘  homogeneous or nonhomogeneous: The example above is nonhomogeneous since the righthand side (where there is no y term)
is nonzero. 0 ﬁrst or second (or higher) order: The example above is ﬁrst order since there exists only a ﬁrst difference, involving a one
period time lag only. Thus, the equation above is a linear, nonhomogeneous, ﬁrst order difference equation. What does it mean to solve a difference equation? 43 To solve a difference equation (for example, (1) above) means to ﬁnd the time path of y. Call this
time path y(t). This time path is:  a function of time;
 totally free from any differential expressions;
0 consistent with the original difference equation and its initial conditions. (The counterpartsolving a differential equationis a matter of integration). Perhaps the crudest way to solve a difference equation: method of iteration  The original equation gives the pattern of change of y between two consecutive periods 0 Once this pattern is speciﬁed and once we are given an initial value of y (call it yo), we can
ﬁnd y], then y;, and then y3, ..., then yn through repeated application of the pattern speciﬁed
in the difference equation. Example 1: Solve (1) using the method of iteration. Assume an initial value of yo = 15.
Repeat (1): ym  y, = 2
Application through time: y]  yo = 2 3’1 = Yo +2
y2 = y] +2 = (y0+2) + 2 = yo +2(2)
Y3 = Y2 +2 = [Yo+2(2)l + 2 = Yo +3 (2) In general, for any period: yt = yo + t(2) = 15+2t Note: This last expression is consistent with " ﬁnding the time path of y " or " solving the difference equation. "
Example 2: Find the solution to y, = ayH + b. (2)
Notice here that a is constant, b is not.
Start with a given yo. Solve, as in the previous example, using the method of iteration.
Y: = aYtl + bt
3’1 = ayO +b1
y2 = ay1 +b2 = a(ayo+b1) + b2 = azyo + ab] + b2
y3 = ayz +b3 = a(a2yo+ab1+b2) + b3 = a3yo + a2b1+ abz + b3 If we carried out the substitutions for y4 and y5, we would get: 44 Y4 = a4yo + a3b1 + azbz + ab3 + b4
y5 = asyo + a4b1 + 33b; + a2b3 + ab4 + b5 A pattern emerges! In general, we have
I
yt = a‘yo + Z a"kbk (3)
k=l This last line represents the general solution to the difference equation. You can check that the last expression above is the solution to y = ayH + b by substituting the last
expression (for y“) into the righthand side of y = ayH + b. Example 3:
Consider the special case of the previous example where bk = b. In terms of (2), a and now b are constants. Thus, the summation expression in (3) becomes I
a“kb = b22a"k = b(a"l+a"2+...+a+1) k=l t—k M The quantity that is being summed in the last expression is a ﬁnite geometric series.
It abides by the summation formula for a ﬁnite geometric series: _ (
1+a+...+a"2+a"‘ =1a forage 1.
1—a
So, the general solution for y = ayH + b is:
1_ t
Y.=a‘yo+b:—a]
la
The last term can be manipulated to rewrite this equation as:
bbat bal b
= I + = l _— + .—
y! ayo la a 0 la la
=a'(y0i) + —b—,fora¢1.
l—a la 45 Example 4: Now, let's change gears and consider manipulating a rather than b. Speciﬁcally, consider the
situation where a = 1. Is this cause for alarm regarding the previous equation, which again is: y = t[ _i] + _b_
t a yo 1a 1a
b because 1— now appears to be undeﬁned?
 a To see that there is no cause for alarm, retrace the origin of the solution presented above. It is: _ I
Yr = atyo + b:1—a:
la = a‘yo + b[a“‘ +81"2 + + a +1] 7 Noticethat,ifa=l, [21"] + at" + + a + l] = t. This means that:
yt = atyO + bt' Q: What do you notice about the time path of y: when a=1? A: It takes off in the direction and pitch dictated by b. Furthermore, this direction and pitch are
the same, i.e., deterministic, for all values of t. Let ' s summarize on the basis of what we have thus far: 0 An almost trivial issue: For each choice of yo, there is a different corresponding unique
solution to the difference equation.  How do changes in the parameters entering into the difference equation affect the solution
of the difference equation? That is, what is the time path of the variable in question?  Does it settle down and reach some equilibrium state?
 Does it take off? 46 Example 5: Let ' s manipulate "a" further and refresh our memory regarding the original difference being used
in Examples 24: Y. = aym + b Recall its solution: a‘(yoi) + —b— fora¢1 1a la yr Q: Now, if a ¢ 1, what happens to the time path of y if y0 = 1—b— ?
 a b . . . . .
A: y1 = — ; y has ach1eved a statlonary state. That 15, 1t 18 the same value for any y. We also 1  a
refer to this stationary state as an equilibrium state. Putting everything together and starting at the top of Example 5, this value of yr, which is L , 1  a
(and call it y*), is an equilibrium or stationary state for the difference equation y1 = Ely“l + bwhena¢1. We've deﬁned what...
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