|
|
QUESTION: What is the effect of Autocorrelation On
Statistical Process Control Charts ANSWER: Statistical
methods can be very useful in summarizing information and very powerful in
testing hypothesis. As in all things, there can be drawbacks. To proceed with
the application of a statistical test one has to be careful about validating
the assumptions under which the test is valid. It is often possible to remedy
the violation and then safer to proceed. One of the most often violated
assumptions is that the observations are independent. Unfortunately, the real
world operates in ignorance of many statistical assumptions, which can lead
to problems in analysis. The good news is that these problems may be easily
overcome, so long as they are recognized and dealt with correctly. The Value of
Information The assumption
of independence of observations implies that the most recent data point
contains no more information or value than any other data point, including
the first one that was measured. In practice this means that the most recent
reading has no special effect on estimating the next reading or measurement.
In summary it provides no information about the next reading. If this
assumption, i.e. independence of all readings, holds this implies that the
overall mean or average is the best estimate of future values. If however
there is some serial or autoprojective (autocorrelation) structure the best
estimate of the next reading will depend on the recently observed values. The
exact form of this prediction is based on the observed correlation structure.
Stock market prices are an example of an autocorrelated data set. Weather
patterns move slowly , thus daily temperatures have been found to be
reasonably described by an AR(2) model which implies that your temperature is
a weighted average of the last two days temperature. Try it and see if
doesn't work ! A Hypothethical |
|
The problem
seemed straightforward and simple to Joe. |
|
|
His boss Jill
called him into his office and said. "We |
|
|
took 25
measurements at a fixed interval of time on line #1. |
|
|
We then made a
change in the process and took 25 more |
|
|
measurements at
the next 25 periods. We have two columns of |
|
|
numbers fifty
readings in all. I want you to test if there |
|
|
is a significant
difference in these two means." |
|
|
|
|
|
BEFORE AFTER |
|
|
|
|
|
1 9.3514475830
9.8966583638 |
|
|
2 9.5681955479
9.9944607413 |
|
|
. 9.5466960924
10.3626602774 |
|
|
. 9.3969162112
10.3679648003 |
|
|
.
9.7971314258 10.3122677744 |
|
|
. 9.5415274115
10.3294613471 |
|
|
. 9.7555390371
10.5880309702 |
|
|
. 10.0000733326
10.4541222131 |
|
|
. 9.6649460022
10.1278926708 |
|
|
. 9.8526793092
10.2218148549 |
|
|
. 10.0348204485
10.0686794649 |
|
|
. 9.7034438999
10.0273011663 |
|
|
. 9.4997914890
10.3240292077 |
|
|
. 10.0091622391
9.8474173752 |
|
|
. 10.3309466494
9.8520087711 |
|
|
. 10.2246080180
9.7593820424 |
|
|
. 10.0623380360
9.9021653502 |
|
|
. 10.2635470167
9.8952197365 |
|
|
. 9.9144741061
9.9344370445 |
|
|
. 9.8057220172
10.1769312079 |
|
|
. 9.6633186971
10.0315961092 |
|
|
. 9.8584971928
10.3399870500 |
|
|
. 9.9664655952
10.3744613997 |
|
|
. 10.1030273833
10.2208646771 |
|
|
25 10.0938308150
10.5819767510 |
|
|
|
|
|
|
|
|
|
|
|
SERIES
STATISTICS IN TERMS OF THE DIFFERENCED AND TRANSFORMED DATA |
|
|
|
|
|
BEFORE AFTER |
|
|
|
|
|
Differencing
Orders Applied to Y NONE NONE |
|
|
Power
Transformation (Lambda) on Y 1.00 1.00 |
|
|
Number of
Effective Observations n 25 25 |
|
|
Mean of the
Series Y 9.840366 10.15967 |
|
|
Variance of the
Series Y .7024076E-01 .5511425E-01 |
|
|
Standard
Deviation of the Series Y .2650297 .2347642 |
|
|
Standard Error
of the Mean / n .5300595E-01 .4695285E-01 |
|
|
Mean divided by
its Standard Error /[ / n] 185.6465 216.3803 |
|
|
|
|
|
DIFFERENCES IN
TWO MEANS 9.84 - 10.16 = -.3193 |
|
|
|
|
|
POOLED STANDARD
DEVIATION (24*.070+24*.055) = .255517 |
|
|
|
|
|
STANDARD ERROR
OF DIFFERENCE .25517/ npooled = .070811 |
|
|
|
|
|
T VALUE DIFF
divided by its Standard Error d/[ / n] = -4.509 |
|
|
|
|
|
Joe went back to
his office and pulled his favorite stat |
|
|
book written by
one of his school professors. As he re-read |
|
|
the material on
testing the hypothesis of the difference |
|
|
between two
means he recalled that if the test for normality |
|
|
was valid within
each group then he could use a parametric |
|
|
approach.
Furthermore if he could accept the hypothesis of |
|
|
constant
variance (similar dispersion within the two groups) |
|
|
he could use the
standard test of the equivalence of two |
|
|
means. He also
noted that if the variances (dispersions) |
|
|
were different
he could use the same test but with an |
|
|
adjusted degrees
of freedom. Continuing his readings he |
|
|
found that if these
readings were correlated he would be |
|
|
better off using
the paired-t test approach. The paired |
|
|
t-test reflected
the presence of a correlative structure |
|
|
between each
pair of readings. For example: |
|
|
|
|
|
TIRE WEAR |
|
|
POS1 POS2 |
|
|
CAR 1 Y 1 1 Y 2
1 |
|
|
CAR 2 Y 1 2 Y 2
2 |
|
|
.. .. .. |
|
|
CAR 25 Y25 1 Y25
2 |
|
|
|
|
|
Joe had 50
readings but they weren't independent as |
|
|
they were taken
chronologically. But they weren't correlated |
|
|
in the way the
textbook pointed out i.e. Y1 AND Y26 Y2 AND |
|
|
Y27. There might
be a relationship between Y1 and Y2 Y2 and |
|
|
Y3 etc. |
|
|
|
|
|
The text was
fastidious about normality checking |
|
|
otherwise you
would have to use a non-parameteric test like |
|
|
Wilcoxin's t and
rigorous about separating the paired t-test |
|
|
from the
standard and even seemingly pre-possessed by |
|
|
testing
constancy of variance before proceeding nothing |
|
|
existed
whatsoever about his particular problem. Thus |
|
|
assured that he
could proceed because no textbook mentioned |
|
|
what to do with
autocorrelated data Joe whipped out his |
|
|
favorite piece
of PC software and went ahead with his test. |
|
|
|
|
|
|
|
|
Sometime that
night in a recurring dream Joe heard a |
|
|
faint whisper
"the residuals/errors have to be N.I.I.D. |
|
|
where D means
distributed". Joe said to himself I verified |
|
|
the n meaning
normality and I verified the i meaning |
|
|
identically by
verifying constant variance but what is that |
|
|
other I in
N.I.I.D.? |
|
|
|
|
|
The other I that
was bothering Joe means independent. |
|
|
If you have data
that is chronological then the degree of |
|
|
dependence
between successive readings is critical as it may |
|
|
cause you to
conclude incorrectly. |
|
|
|
|
|
This example of
textbooks totally ignoring time series |
|
|
data reflects an
error of omission that is slow to cure. The |
|
|
textbook authors
cover themselves by stating the assumptions. |
|
|
Unwashed readers
misuse the textbook solution because they |
|
|
were not told in
clear language that the approach did not |
|
|
apply to their
problem. Caveat emptor or caveat Joe |
|
|
statistician. We
attempt here to expand upon simple but |
|
|
incorrect
textbook solutions to cover possible time series |
|
|
problem sets. |
|
|
|
|
|
This replay
studies the impact of non-random error terms |
|
|
on time series
data. We created four time series each of |
|
|
length 50 which
were generated via simulation. Simulation |
|
|
starts with an
error process which is used as the input. |
|
|
This random
input is then used with a known deterministic |
|
|
model to create
a realization of a series. |
|
|
|
|
|
For example the
series WN was based on the following: |
|
|
|
|
|
We generated 300
normally distributed uncorrelated and independent |
|
|
samples A(T) and
then using the model |
|
|
|
|
|
Y(T) - 10.0000 =
A(T) |
|
|
|
|
|
or |
|
|
|
|
|
Y(T) = 10.0000 +
A(T) |
|
|
|
|
|
created the
Y(T). |
|
|
|
|
|
|
|
|
|
|
|
A second series
AR was based on the following: |
|
|
|
|
|
We generated 300
normally distributed uncorrelated and |
|
|
independent
samples A(T) and then using the model: |
|
|
|
|
|
[Y(T) - 10.0000]
[(1- .7B)]**+1 = A(T) |
|
|
|
|
|
created the
Y(T). |
|
|
|
|
|
Specifically |
|
|
|
|
|
Y(T) - 10.0000 =
A(T) |
|
|
|
|
|
[Y(T) - 10.0000]
[(1- .7B)]**+1 = A(T) |
|
|
|
|
|
[Y(T)][(1-
.7B)]**+1 - [10.0000] [(1- .7B)]**+1 = A(T) |
|
|
|
|
|
Y(T) - .7*Y(T-1)
- 10.0000 [(1- .7)] = A(T) |
|
|
|
|
|
Y(T) - .7*Y(T-1)
- 3. = A(T) |
|
|
|
|
|
Restating |
|
|
|
|
|
Y(T) = .7*Y(T-1)
+ 3. + A(T) |
|
|
|
|
|
thus |
|
|
|
|
|
Y(1) = .7*Y(0) +
3. + A(1) |
|
|
|
|
|
and since Y(0)
is unknown we will use as a starting condition the |
|
|
value 0. |
|
|
|
|
|
Y(1) = 0. + 3. +
A(1) |
|
|
|
|
|
and then |
|
|
|
|
|
Y(2) = .7*Y(1) +
3. + A(2) |
|
|
etc. |
|
|
|
|
|
After completion
of these 300 we will discard the first 250 |
|
|
therefore
eliminating the initial conditions bias of setting Y(0)=0. |
|
|
In this way we
generate a series of numbers which is autocorrelated. |
|
|
|
|
|
A third series
WNL was based on the following: |
|
|
|
|
|
Y(T) - 9.7500 =
A(T) + .50 * LINTAT26(T) |
|
|
|
|
|
where LINTAT26 =
0 for T=1 2 ....25 |
|
|
= 1 for T=26 27
50 |
|
|
or |
|
|
|
|
|
Y(T) = 9.7500 +
A(T) + .50 * LINTAT26(T) |
|
|
|
|
|
A fourth series
ARL was based on the following: |
|
|
|
|
|
[Y(T) - 10.0000]
[(1- .7B)]**+1 = A(T) + .25 * LINTAT26(T) |
|
|
|
|
|
Let us look at
some descriptive statistics of these four series: |
|
|
First the two
statistics summarizing central tendency and dispersion. |
|
|
|
|
|
|
|
|
FIELD # 1 FIELD
# 2 N FIELD # 3 |
|
|
NAME MEAN STAND |
|
|
|
|
|
WN 10.00 50 1.00 |
|
|
AR 10.00 50 1.00 |
|
|
WNL 10.00 50 .30 |
|
|
ARL 10.00 50 .30 |
|
|
|
|
|
|
|
|
|
|
|
Now a statistic
which measures the internal relationship within a |
|
|
series: |
|
|
A U T O C O R R
E L A T I O N |
|
|
F U N C T I O N |
|
|
|
|
|
|
|
|
LAG WN WNL AR
ARL |
|
|
|
|
|
|
|
|
1 -.127 .695
.571 .632 |
|
|
2 -.137 .664
.384 .472 |
|
|
3 .240 .681 .345
.422 |
|
|
4 -.084 .548
.140 .228 |
|
|
5 .017 .528 .082
.163 |
|
|
6 -.080 .432
-.017 .061 |
|
|
7 .038 .380
-.017 .045 |
|
|
8 -.016 .352
-.041 .023 |
|
|
|
|
|
|
|
|
|
|
|
Thus we can
readily see that the level shift has |
|
|
contaminated the
ACF so that the identification of the |
|
|
underlying model
is more difficult than simply looking for |
|
|
decays and
cutoffs. Now a second statistic which measures |
|
|
the internal
relationship: |
|
|
|
|
|
P A R T I A L |
|
|
A U T O C O R R
E L A T I O N |
|
|
F U N C T I O N |
|
|
|
|
|
LAG WN WNL AR
ARL |
|
|
|
|
|
|
|
|
1 -.127 .695
.571 .632 |
|
|
2 -.156 .349
.086 .121 |
|
|
3 .209 .305 .144
.141 |
|
|
4 -.052 -.097
-.186 -.187 |
|
|
5 .067 .017 .027
.030 |
|
|
6 -.153 -.152
-.132 -.118 |
|
|
7 .064 -.008
.102 .105 |
|
|
8 -.072 .007
-.074 -.040 |
|
|
|
|
|
|
|
|
The replay
illustrates bringing in these four series and |
|
|
then plotting
them. The next step is to test the hypothesis |
|
|
of the
difference between two means: |
|
|
|
|
|
HO: MEAN OF THE
FIRST 25 = MEAN OF THE SECOND 25 |
|
|
HA: "
" <>" " |
|
|
|
|
|
A basic
statistical test presented in a first course in |
|
|
Business
Statistics is whether or not two means are |
|
|
statistically
significantly different from one another. The |
|
|
student learns
to compute: |
|
|
|
|
|
1. two means and
two standard deviations |
|
|
2. test the
equivalence of the standard deviations |
|
|
and if equal
compute the pooled standard deviation p |
|
|
|
|
|
and then given
the equivalence of the two standard deviations |
|
|
|
|
|
3. test the
statistical significance of the observed |
|
|
difference
between the two means using |
|
|
|
|
|
Standard Error
of the Mean Diff = p/ (1/n1) + (1/n2) |
|
|
|
|
|
This is exactly
equal to the linear model test for a |
|
|
regression where
the model is: |
|
|
|
|
|
Y(T) = + A(T) +
W0 * LINTAT26(T) |
|
|
|
|
|
note that A(t)
must be a white noise normal gaussian and |
|
|
consequently
each of the A's must be independent and |
|
|
identically
distributed. Let us look at the results of the |
|
|
test of the
significance for the dummy variable LINTAT26. |
|
|
|
|
|
|
|
|
In this section
we are assuming that the true state of |
|
|
nature is an
error process that is N.I.I.D. All of the |
|
|
subsequent tests
and tables make that assumption. We know |
|
|
of course what
reality is by virtue of the simulation. |
|
|
|
|
|
WN |
|
|
|
|
|
THE ESTIMATED
MODEL PARAMETERS |
|
|
|
|
|
MODEL COMPONENT
LAG COEFFICIENT STANDARD T-RATIO |
|
|
# (BOP) ERROR |
|
|
|
|
|
Lambda Value
1.000000 |
|
|
1 [ (B)/
(B)]Y(T)=CONSTANT 9.852248 .195795 50.32 |
|
|
|
|
|
INPUT SERIES X1 |
|
|
|
|
|
Lambda Value
1.000000 |
|
|
2 Omega (input)
-Factor # 1 0 .2955076 .276896 1.067 |
|
|
|
|
|
Y(T) = 9.8522 |
|
|
+ X 1(T) [(+
.2955)] |
|
|
+ A(T) |
|
|
|
|
|
|
|
|
WNL |
|
|
|
|
|
THE ESTIMATED
MODEL PARAMETERS |
|
|
|
|
|
MODEL COMPONENT
LAG COEFFICIENT STANDARD T-RATIO |
|
|
# (BOP) ERROR |
|
|
|
|
|
Lambda Value
1.000000 |
|
|
1 [ (B)/
(B)]Y(T)=CONSTANT 9.738417 .280222E-01 347.5 |
|
|
|
|
|
INPUT SERIES X1 |
|
|
|
|
|
Lambda Value
1.000000 |
|
|
2 Omega (input)
-Factor # 1 0 .5237786 .396294E-01 13.22 |
|
|
|
|
|
Y(T) = 9.7384 |
|
|
+ X 1(T) [(+
.5238)] |
|
|
+ A(T) |
|
|
|
|
|
|
|
|
AR |
|
|
|
|
|
THE ESTIMATED
MODEL PARAMETERS |
|
|
|
|
|
MODEL COMPONENT
LAG COEFFICIENT STANDARD T-RATIO |
|
|
# (BOP) ERROR |
|
|
|
|
|
Lambda Value
1.000000 |
|
|
1 [ (B)/
(B)]Y(T)=CONSTANT 9.614634 .182440 52.70 |
|
|
|
|
|
INPUT SERIES X1 |
|
|
|
|
|
Lambda Value
1.000000 |
|
|
2 Omega (input)
-Factor # 1 0 .7707375 .258009 2.987 |
|
|
|
|
|
Y(T) = 9.6146 |
|
|
+ X 1(T) [(+
.7707)] |
|
|
+ A(T) |
|
|
|
|
|
|
|
|
ARL |
|
|
|
|
|
THE ESTIMATED
MODEL PARAMETERS |
|
|
|
|
|
MODEL COMPONENT
LAG COEFFICIENT STANDARD T-RATIO |
|
|
# (BOP) ERROR |
|
|
|
|
|
Lambda Value
1.000000 |
|
|
1 [ (B)/
(B)]Y(T)=CONSTANT 9.840366 .500710E-01 196.5 |
|
|
|
|
|
INPUT SERIES X1 |
|
|
|
|
|
Lambda Value
1.000000 |
|
|
2 Omega (input)
-Factor # 1 0 .3193062 .708110E-01 4.509 |
|
|
|
|
|
Y(T) = 9.8404 |
|
|
+ X 1(T) [(+
.3193)] |
|
|
+ A(T) |
|
|
|
|
|
|
|
|
|
|
|
Notice that the
regression coefficient (.3193062) is the |
|
|
difference
between the two means (second minus first or after |
|
|
minus before )
and the t-ratio (4.509) is identical to the |
|
|
test of the
hypothesis between the two means. |
|
|
|
|
|
S U M M A R Y |
|
|
|
|
|
|
|
|
T WN WNL AR ARL |
|
|
|
|
|
|
|
|
1 1.0677 13.22
2.987 4.509 |
|
|
|
|
|
|
|
|
|
|
|
TRUE STATE OF
NATURE: NO SIGNIFICANT MOVEMENT |
|
|
|
|
|
|
|
|
T WN AR |
|
|
|
|
|
|
|
|
1 1.0677 2.987 |
|
|
|
|
|
|
|
|
TRUE STATE OF
NATURE: SIGNIFICANT MOVEMENT |
|
|
|
|
|
|
|
|
T WNL ARL |
|
|
|
|
|
|
|
|
1 13.22 4.509 |
|
|
|
|
|
|
|
|
|
|
|
Thus the effect
of autocorrelated data in the absence |
|
|
of a significant
level shift is to cause one to reject the |
|
|
null hypothesis
with greater frequency than is warranted. |
|
|
Thus we can
conclude that autocorrelated data leads to false |
|
|
positives
regarding the movement of a mean if no mean |
|
|
movement exists.
However if the mean has significantly |
|
|
moved the
presence of autocorrelated data masks the movement |
|
|
in the mean. |
|
|
|
|
|
In this section
we are assuming that the true state of |
|
|
nature is an
error process that is N.I.I.D. With |
|
|
|
|
|
[Y(t) ] [(1-
.7B)]**+1 = A(t) |
|
|
|
|
|
note that this
model of the errors was found by |
|
|
empirically
studying the first 25 observations of ARL. |
|
|
|
|
|
WN |
|
|
|
|
|
THE ESTIMATED
MODEL PARAMETERS |
|
|
|
|
|
MODEL COMPONENT
LAG COEFFICIENT STANDARD T-RATIO |
|
|
# (BOP) ERROR |
|
|
|
|
|
Lambda Value
1.000000 |
|
|
1 [ (B)/
(B)]Y(T)=CONSTANT 11.48754 1.39198 8.253 |
|
|
2
Autoregressive-Factor # 1 1 -.1590011 .140084 -1.135 |
|
|
|
|
|
INPUT SERIES X1 |
|
|
|
|
|
Lambda Value
1.000000 |
|
|
3 Omega (input)
-Factor # 2 0 .2344056 .232329 1.009 |
|
|
|
|
|
Y(T) = 9.9116 |
|
|
+ X 1(T) [(+
.2344)] |
|
|
+ A(T) [(1+
.1590B)]**-1 |
|
|
|
|
|
|
|
|
WNL |
|
|
|
|
|
THE ESTIMATED
MODEL PARAMETERS |
|
|
|
|
|
MODEL COMPONENT
LAG COEFFICIENT STANDARD T-RATIO |
|
|
# (BOP) ERROR |
|
|
|
|
|
Lambda Value
1.000000 |
|
|
1 [ (B)/
(B)]Y(T)=CONSTANT 11.29325 1.36428 8.278 |
|
|
2
Autoregressive-Factor # 1 1 -.1586483 .140084 -1.133 |
|
|
|
|
|
INPUT SERIES X1 |
|
|
|
|
|
Lambda Value
1.000000 |
|
|
3 Omega (input)
-Factor # 2 0 .5150275 .332608E-01 15.48 |
|
|
|
|
|
Y(T) = 9.7469 |
|
|
+ X 1(T) [(+
.5150)] |
|
|
+ A(T) [(1+
.1586B)]**-1 |
|
|
|
|
|
|
|
|
AR |
|
|
|
|
|
THE ESTIMATED
MODEL PARAMETERS |
|
|
|
|
|
MODEL COMPONENT
LAG COEFFICIENT STANDARD T-RATIO |
|
|
# (BOP) ERROR |
|
|
|
|
|
Lambda Value
1.000000 |
|
|
1 [ (B)/
(B)]Y(T)=CONSTANT 4.307894 1.15223 3.739 |
|
|
2
Autoregressive-Factor # 1 1 .5686867 .119324 4.766 |
|
|
|
|
|
INPUT SERIES X1 |
|
|
|
|
|
Lambda Value
1.000000 |
|
|
3 Omega (input)
-Factor # 2 0 .3103636 .454120 .6834 |
|
|
|
|
|
Y(T) = 9.9879 |
|
|
+ X 1(T) [(+
.3104)] |
|
|
+ A(T) [(1-
.5687B)]**-1 |
|
|
|
|
|
|
|
|
ARL |
|
|
|
|
|
THE ESTIMATED
MODEL PARAMETERS |
|
|
|
|
|
MODEL COMPONENT
LAG COEFFICIENT STANDARD T-RATIO |
|
|
# (BOP) ERROR |
|
|
|
|
|
Lambda Value
1.000000 |
|
|
1 [ (B)/
(B)]Y(T)=CONSTANT 4.323042 1.17370 3.683 |
|
|
2
Autoregressive-Factor # 1 1 .5651501 .119329 4.736 |
|
|
|
|
|
INPUT SERIES X1 |
|
|
|
|
|
Lambda Value
1.000000 |
|
|
3 Omega (input)
-Factor # 2 0 .1946314 .123809 1.572 |
|
|
|
|
|
Y(T) = 9.9415 |
|
|
+ X 1(T) [(+
.1946)] |
|
|
+ A(T) [(1-
.5652B)]**-1 |
|
|
|
|
|
|
|
|
S U M M A R Y |
|
|
|
|
|
|
|
|
T WN WNL AR ARL |
|
|
|
|
|
|
|
|
1 1.009 15.48
.6834 1.572 |
|
|
|
|
|
|
|
|
|
|
|
TRUE STATE OF
NATURE: NO SIGNIFICANT MOVEMENT |
|
|
|
|
|
|
|
|
T WN AR |
|
|
|
|
|
|
|
|
1 1.009 .6834 |
|
|
|
|
|
|
|
|
TRUE STATE OF
NATURE: SIGNIFICANT MOVEMENT |
|
|
|
|
|
|
|
|
T WNL ARL |
|
|
|
|
|
|
|
|
1 15.48 1.572 |
|
|
|
|
|
|
|
|
|
|
|
G R A N D S U M
M A R Y |
|
|
|
|
|
|
|
|
TRUE STATE OF
NATURE INDEPENDENT ERRORS CORRELATED ERRORS |
|
|
-------------------
----------------- |
|
|
|
|
|
TRUE STATE OF
NATURE NO SHIFT SHIFT NO SHIFT SHIFT |
|
|
-------- -----
-------- ----- |
|
|
ASSUME
INDEPENDENT 1.068 13.22 2.987 4.509 |
|
|
USE OBSERVED
STRUCTURE 1.009 15.48 .6834 1.572 |
|
|
|
|
|
Recall that in
the case of ARL we commented that the AR |
|
|
structure in the
case where modeling was not done after each |
|
|
reading could
cause a shift in the mean that would go |
|
|
undetected due
to the responsiveness of the one period out |
|
|
forecasts to the
new reading. The replay estimates the AR1 |
|
|
model for the
first 25 readings of ARL and then proceeds to |
|
|
compute updated
forecasts which illustrate the point. We |
|
|
illustrate this
last point by using Autobox to estimate the |
|
|
model for ARL
using the first 25 observations |
|
|
|
|
|
Y(T) = 9.8950 +
A(T) [(1- .5244B)]**-1 |
|
|
|
|
|
|
|
|
and then
forecasting the 26th point. We then |
|
|
incorporate the
actual value for the 26th point and forecast |
|
|
the 27th point. |
|
|
|
|
|
OUT OF SAMPLE
FORECAST VALUES ARL |
|
|
|
|
|
TIME PERIOD |
|
|
26 27 28 29 30
31 32 33 34 35 |
|
|
A C T U A L S |
|
|
9.90 9.99 10.4
10.4 10.3 10.3 10.6 10.5 10.1 10.2 |
|
|
|
|
|
FORECAST |
|
|
ORIGIN |
|
|
25 10.0 |
|
|
26 9.90 |
|
|
27 9.95 |
|
|
28 10.1 |
|
|
29 10.1 |
|
|
30 10.1 |
|
|
31 10.1 |
|
|
32 |
|
|
10.3 |
|
|
33 |
|
|
10.2 |
|
|
34 |
|
|
10.0 |
|
|
|
|
|
|
|
|
Notice how the
forecasts adapt to the level change and |
|
|
thus mask the
special effect of the level shift at time period |
|
|
26. This flaw
was pointed out by Thomas P. Ryan and is |
|
|
easily remedied
by remodeling after each and every new |
|
|
observation has
been recorded. |
|
|
|