To forecast HEPI values out of sample –
We will assume for now that the deviations of
the HEPI from its linear trend are i.i.d. N(0,2).
Point Forecasts –
Under these assumptions, the point
forecasts of the HEPI are the same as the
point forecasts of the trend values:
hepiforeT+h,T = -7.06 + 4.96*(T+h),
where, in this example, T = 44.
So my point forecasts of the HEPI for
2005-2010 (under the assumption of a
linear trend and i.i.d. deviations from the
trend) are:
Year
2004
2005
2006
2007
2008
2009
2010
Forecast
231.5 (Actual)
216.1
221.1
226.1
231.0
236.0
240.9
The implied forecasted HEPI inflation
rates are:
Year
2005
2006
2007
2008
2009
2010
Inflation Rate Forecast
-6.9%
2.27
2.22
2.17
2.12
2.08
So, if the Regents had been presented with
this forecast of the HEPI inflation rate for
2005 and used it to determine your current
tuition, your tuition would be nearly 7% less
than it was a year ago. (However, since this
model has been underestimating the HEPI
over the last part of the sample, my bet is
that this forecast will severely underestimate
the actual 2005 HEPI inflation rate.)
Interval Forecasts –
Recall that under our current assumptions, a
95-percent confidence interval for the h-step
ahead forecast of y is
yˆ T  h ,T  1.96ˆ
The y-hats are the point forecasts we
computed above. -hat comes from the
EViews regression output:
̂
(
= standard error of the regression
= 11.38
̂ = sqrt(sum of squared residuals/(T-2))
= sqrt(5435.15/42)
= 11.38)
So our point forecasts along with the lower and
upper bounds for the 95% confidence interval are:
Year
2005
2006
2007
2008
2009
2010
Lower Bound
Point Forecast Upper Bound
193.8
198.8
203.7
208.7
213.6
216.1
221.1
226.1
231.0
236.0
238.5
243.4
248.4
253.4
258.3
218.6
240.9
263.3
To contruct these intervals in EViews –
Call hepiup95 the upper bound of the 95% interval
and call hepilow95 the lower bound of the 95%
interval. Then using the Genr command construct the
two series:
hepiup95 = hepifore+1.96*11.4
hepilow95 = hepifore-1.96*11.4.)
Forecasting the HEPI with other trend models –
These forecasts were constructed using the linear
trend model, but we saw from the graph of the HEPI
and the graph of the deviations of the HEPI from the
estimated linear trend that a quadratic or log linear
trend model may be more appropriate for this series.
Consider, for example, the quadratic trend model
hepit = 0 + 1t + 2t2 + t
where t denotes the deviation of hepit from its trend.
To construct the forecasts of hepi2005,… for this
model, we will continue to assume that the ’s are
i.i.d. N(0,2). In this case:
hepiforeT  h,T  ˆ0  ˆ1 (T  h)  ˆ2 (T  h) 2
where the -hats are the OLS estimates of the ’s
obtained from the regression of hepi on t, t2 and a
constant.
This regression can be run from the EViews
workfile in two ways:
A.
1. Create a new variable, tsquare, using the
Genr command and entering
tsquare = t^2
2. Click on hepi, then simultaneosly click
on “control” and “t” then “control” and
“tsquare”. The names hepi, t, and t-square
should now be highlighted.
3. Right-click on any of the highlighted
areas, then left-click on “open” and “as
equation”. This should open the Equation
Specification window, with hepi as the
first variable (the dependent variable),
followed by t, tsquare, and c. Make sure
the method and sample are correctly set.
Then click “OK” to run the regression.
B.
1. Click on hepi, then simultaneosly click
on “control” and “t”. The names hepi
and t should now be highlighted.
2. Right-click on any of the highlighted
areas, the left-click on “open” and “as
equation”. This should open the
Equation Specification window, with
hepi as the first variable, followed by t
and c. In the Equation Specification box,
add t^2 to the list of variables anywhere
after hepi, e.g.,
hepi t t^2 c
Check the method and sample, then
click OK to run the regression.
The results –
Model Fit with Quadratic Trend
Dependent Variable: HEPI
Method: Least Squares
Date: 09/07/05 Time: 14:33
Sample: 1961 2004
Included observations: 44
Variable
Coefficient
Std. Error
t-Statistic
Prob.
T
C
T^2
1.770474
17.42289
0.070968
0.218784
2.134468
0.004714
8.092322
8.162635
15.05384
0.0000
0.0000
0.0000
R-squared
Adjusted R-squared
S.E. of regression
Sum squared resid
Log likelihood
Durbin-Watson stat
0.995381
0.995156
4.506589
832.6832
-127.1235
0.108927
Mean dependent var
S.D. dependent var
Akaike info criterion
Schwarz criterion
F-statistic
Prob(F-statistic)
104.6295
64.74760
5.914704
6.036354
4417.526
0.000000
Model Fit with Linear Trend
Dependent Variable: HEPI
Method: Least Squares
Date: 09/07/05 Time: 14:35
Sample: 1961 2004
Included observations: 44
Variable
Coefficient
Std. Error
t-Statistic
Prob.
T
C
4.964024
-7.060994
0.135053
3.489237
36.75606
-2.023650
0.0000
0.0494
R-squared
Adjusted R-squared
S.E. of regression
Sum squared resid
Log likelihood
Durbin-Watson stat
0.969849
0.969132
11.37578
5435.149
-168.3953
0.047114
Mean dependent var
S.D. dependent var
Akaike info criterion
Schwarz criterion
F-statistic
Prob(F-statistic)
104.6295
64.74760
7.745239
7.826338
1351.008
0.000000
Note that all of the coefficients in the
estimated quadratic trend model appear to be
statistically significant and the R2 of the
regression is approximately 99.5%.
What does the estimated quadratic trend
look like? What do the deviations from the
trend look like?
How do the HEPI forecasts compare across
the two models, continuing to assume that
the deviations from trend are i.i.d. N(0,2)?
Point forecasts:
1. Assuming a linear trend,
hepiforeT+h,T = -7.06 + 4.96*(T+h)
2. Assuming a quadratic trend,
hepifore T+h,T = 17.42 + 1.77*(T+h)
+ 0.07(T+h)2
HEPI Level:
Year
Forecast (Linear)
2004
2005
2006
2007
2008
2009
2010
231.5 (Actual)
211.3
238.8
216.3
247.0
221.3
255.2
226.2
263.7
231.2
272.2
236.2
Forecast (Quadratic)
280.9
HEPI Inflation Rate:
Year
2005
2006
2007
2008
2009
2010
Forecast (Linear)
-6.9%
2.27
2.22
2.17
2.12
2.08
Forecast (Quadratic)
3.41%
3.35
3.30
3.25
3.20
3.15
So, if the Regents used the forecast generated from
the quadratic trend model to set this year’s tuition,
your tuition would have increased by about 3.4%
from last year’s tuition.
Which of these two forecast models is likely to
perform better?
Your instincts are probably telling you that the
forecast model using the quadratic trend is likely
to perform better. Why? If we just look at the
time series plot of the HEPI data it seems clear
that a quadratic trend is more reasonable than a
linear trend.
What other criteria might we use to help back up
this feeling? In particular, what criteria might we
use when it is not so apparent from the graph
which model to select?
Which estimated model fits the data sample
better? That is, which estimated model has
the
 smaller sum of squared residuals?
 larger R2?
 smaller mean squared error (MSE)
= SSR/T?
The answer to these three questions will always
be the same. (This is clear from the formulas
used to construct SSR, R2, MSE. See p. 85.)
If we apply these to compare the quadratic trend
model to the linear trend model –
SSRq = 833,
SSRL = 5435
R2q = 0.995,
R2L = 0.970
MSEq = 18.93, MSEL = 123.5
This criteria, which is obvious when we look at
the graphs of the fitted trend lines to the actual
data, indicate that the quadratic trend provides a
better fit than the linear trend. However …