In article <[EMAIL PROTECTED]>,
[EMAIL PROTECTED] (Donald F. Burrill) wrote:
> Why would one wish to avoid multiple linear regression, which is so
much
> more convenient than the sequence of simple linear regressions
alluded to
> below? Two possible reasons occur to mind:
> (1) for the challenge of finding out how to do it (in which case I
> wouldn't want to spoil your fun!);
> (2) this is an assigned homework problem (in which case it would be
a
> shame to spoil your instructor's intent).
Third possible reason:
I am a lowly computer programmer who has only one recipe for making
multiple linear regressions in SQL: via a sequence of simple ones. This
is my aforementioned schedule is so important: it is an algorithm. Aha!
I'm found out!
>
> On Sat, 12 Feb 2000 [EMAIL PROTECTED] wrote:
>
> > I am told that I can solve for these three unknowns (B1, B2 and B3)
by
> > doing simple linear regression to obtain "residuals"; from the
> > residuals come the unknowns. For example, I know that with just two
> > unknowns (B1 and B2) in:
> >
> > y = B0 + B1 * x1 + B2 * x2 + e [1]
> >
> > I can obtain B1 after the following schedule of calculations:
> >
> > regress y on x2 which yields R1 (residual one)
> > regress x1 on x2 which yields R2
> > regress R1 on R2 which leads to a slope value that is B1
> >
> > Now the pitiful questions: how do I get B2?
>
> Same way you got B1.
>
Yes, but what happens when you have three independent variables: x1, x2
and x3? I am afraid that I will have to regress in all possible
combinations of x1, x2 and x3. So my little schedule would be quite
large right?
I hope this is not the task at hand:
y on x1 ---> Ryx1
x2 on x1 ---> Rx2x1
Ryx1 on Rx2x1 ---> (?)
y on x1 ---> Ryx1
x3 on x1 ---> Rx3x1
Ryx1 on Rx3x1 ---> (?)
y on x2 ---> Ryx2
x1 on x2 ---> Rx1x2
Ryx2 on Rx1x2 ---> (?)
y on x2 ---> Ryx2
x3 on x2 ---> Rx3x2
Ryx2 on Rx3x2 ---> (?)
y on x3 ---> Ryx3
x1 on x3 ---> Rx1x3
Ryx3 on Rx1x3 ---> (?)
y on x3 ---> Ryx3
x1 on x2 ---> Rx1x2
Ryx3 on Rx1x2 ---> (?)
If this is absurd please tell me WHY it is.
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