Jay,
If we assume our sample is complete and the data represent the whole
population, then I don't see how your logic applies. But I think you may be
right anyway, if we are predicting the effect from the cause.


Donald,

I think I understand. In the model y=x1+x2, the expected value of y(  when
using the regression equation y=a+ b1x1 +e)  is equal to x1. The expected
value of x2 (error) given x1 is zero, since x1 and x2(error) are
uncorrelated. Thus, in the extremes of x1, greater errors are expected when
predicting y from x1, since the errors are not all likely to equal zero.

The picture changes however when we predict x1 from y.
In the regression model x1=y, the errors (x2) will be smaller in the
extremes of y since x1 and x2 are correlated in the extremes of y. That is,
when x1 and x2 are uniformly distributed. But when x1 and x2 are not uniform
but normal, the smaller errors expected in the extremes of y (predicting x1)
are not much smaller, since x1 and x2 will be nearer correlated at zero.

The explanation of the latter model is based on Bunge's notion of conjoint
and disjoint causes. In conjoint causation (uniform causes), all the levels
of the causes are combined. A crosstabulation of y by x1and x2 will have
data in the corners of the box. Thus the most extreme values of y will be
the combinations of extreme correlated values of x1 and x2.  But with
normally distributed causes, the causation is disjoint, the corners of the
cross tabulations will tend to be empty, because there will be so few
instances where correlated extremes of the causes are paired. Consequently,
the range of y will be less in the disjoint cause than in the conjoint
cause. The extreme value of y will tend to be caused by extremes of either
x1 or x2 (disjoint) but not by both at the same time (conjoint).

So we are probably all correct.  When predicting the effect from a cause,
the error is greatest in the extremes of y, the values of x1 and x2 diverge
with deviations from the intercept.  When predicting a cause from the
effect, the errors will be smallest in the extremes of the effect. This
assuming the causes are uniform/conjoint.

At least in theory.

Best,

Bill



"Donald Burrill" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> On Wed, 25 Sep 2002 [EMAIL PROTECTED] wrote in part:
>
> > When two lines cross, where they cross they will be equal and error
> > will be zero.  As the nonparallel lines extend, they grow further
> > apart.  I see a potential problem with this model... there is only one
> > line, the regression of y on x.
>
> But measured with uncertainty:  y = b_0 + b_1*x + e.
>  The uncertainty in b_0 adds to the uncertainty in e, implying a kind of
> probabilistic sheaf of parallel lines that are most dense at b_0.
>  The uncertainty in b_1 implies a probabilistic fan of lines of slightly
> different slopes centered on (and most dense at) the slope b_1, and all
> intersecting at (x_bar, b_0 + b_1*x_bar).
>  Combining both of these uncertainties leads to a confidence band of
> hyperbolic shape (for a given confidence level) that is narrowest (the
> two arcs of the hyperbola are closest together) at the mean of x.
>
> > "Gus Gassmann" <[EMAIL PROTECTED]> had written in part:
> > >
> > > If you have two lines, y = beta_0 + beta_1 x and y = b_0 + b_1 x,
> > > where b_i is close to beta_i, then the lines diverge and the distance
> > > between them increases as you move away from the center (where they
> > > typically intersect).  So your confidence limits widen as you move
> > > out into the tails.
>
> Gus was describing the difference between the "true" regression line
> (using population values beta_0 and beta_1) and the regression line
> estimated from the sample (using the sample values b_0 and b_1).
>  Same shape of result as I've described for the uncertainty about the
> estimated line, of course.
>   -- Don.
>  -----------------------------------------------------------------------
>  Donald F. Burrill                                            [EMAIL PROTECTED]
>  56 Sebbins Pond Drive, Bedford, NH 03110                 (603) 626-0816
>  [Old address:  184 Nashua Road, Bedford, NH 03110       (603) 471-7128]
>
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