When the explanatory ( or independent ) variable is zero then the value taken by explained ( dependent ) variable is the value of the constant . But if you see any book on statistics the fitted regression is Y= a+ bX. ( It is also written as (Y= mX+c) .    So it is a value. . Sven Schreiber is correct . Mine is a simple way . If you plot that actual and fitted value it lies at a point  where X is zero. Incidentally it may be positive or negative also .
Feel free to communicate 


On Sun, May 18, 2014 at 6:43 PM, Huffelpuff <huffelpuff420@gmail.com> wrote:
Thanks, this seems to work.
One last question: I was not able to get these values if I only had my time variable as regressor. I had to add "const" as a regressor too, in order to get the values (225 and 6.25). What exactly is const and what values does it contain (I'm unable to edit this variable)?

Thanks
Peter



On 2014/05/18 01:07 PM, Narandra Dashora wrote:
The estimated equation is 
Model 1: OLS, using observations 1950-1957 (T = 8)
Dependent variable: Stock

             coefficient   std. error   t-ratio   p-value 
  --------------------------------------------------------
  const       225.000       6.21177     36.22     2.95e-08 ***
  Time         -6.25000     0.976086    -6.403    0.0007   ***

Mean dependent var   190.6250   S.D. dependent var   22.90313
Sum squared resid    468.7500   S.E. of regression   8.838835
R-squared            0.872340   Adjusted R-squared   0.851064
F(1, 6)              41.00000   P-value(F)           0.000684
Log-likelihood      -27.63402   Akaike criterion     59.26804
Schwarz criterion    59.42692   Hannan-Quinn         58.19644
rho                 -0.010000   Durbin-Watson        1.650000

Now  substitute time values in the estimated equation 

225- 6.25 X
 Give X values like 1 , 2 3   up to 10
You will get predicted values like this ( by manual calculation)
when X is 5 ( value for 1954 ) the estimated value is 193.75. When X is 6 ( value for 1955) the predicted value is 187.5. By OLS this is the answer.





On Sat, May 17, 2014 at 7:19 PM, Huffelpuff <huffelpuff420@gmail.com> wrote:
Hi,

I'm new to gretl, so forgive my ignorance. I'm aware that gretl provides
forecasting functionality, but I'm interested in using any of the
time-series models in gretl (AR, ARIMA, GARCH, etc) to interpolate a
section of unknown data. If I have something like this:

Year    Stock value
1950    215
1951    220
1952    200
1953    195

Then it is easy in gretl to predict the following years (1954, 1955, and
so forth). But if I have something like this:

Year    Stock value
1950    215
1951    220
1952    200
1953    195
1956    190
1957    185
1958    170
1959    150

I then want to "predict" (or technically interpolate) the values for the
years 1954 and 1955 (the stock value will probably be something between
195 and 190). Is this possible with gretl? If so, how?

Peter
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