Constant multiplication

[manowar]

I am trying to run several regressions to determine the changes in the dependent variable when I alter the constant and independent variables. The regression in its unaltered original form is:

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p c l k

Where p is output, c is the constant or intercept, l is labor and k is capital. I have no problem altering the independent variables by multiplying them (say by 500), but when I try to multiply c I get:

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illegal lag or index specification for coefficient C

.

Is there any way to do this?

[EViews Glenn]

You have to create a new variable

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series cc = 500 equation eq1.ls p cc l k

[manowar]

All you've done there is create a new variable whose value for each observation is 500, you didn't multiply the existing constant by anything. I want to multiply each of the observations of my constant by 500.

[startz]

All you've done there is create a new variable whose value for each observation is 500, you didn't multiply the existing constant by anything. I want to multiply each of the observations of my constant by 500.

What do you think the observation is for a constant?

[manowar]

What do you think the observation is for a constant?

Not sure I understand your question, but you do know that the constant is made up of more than one value?

The observations of my constant already have values, I don't need a new constant whose value for each observation is 500, I want to modify the values/observations of my existing constant by multiplying them.

[startz]

What do you think the observation is for a constant?

Not sure I understand your question, but you do know that the constant is made up of more than one value?

The observations of my constant already have values, I don't need a new constant whose value for each observation is 500, I want to modify the values/observations of my existing constant by multiplying them.

Using "c" in a regression is exactly the same as entering a variable that equals 1.0 for every observation. For that reason, Glenn's answer is correct. But then, Glenn's answers are almost always correct.

[EViews Glenn]

I don't know about that. Given the last couple of posts, it's clear that I didn't understand the original question.

[manowar]

Got it, thanks.