Re: Regression Estimate for a in PI LAD Regression
From: Osher Doctorow (mdoctorow_at_comcast.net)
Date: 11/05/04
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Date: Fri, 5 Nov 2004 13:45:22 +0000 (UTC)
On 4 Nov 04 17:11:03 -0500 (EST), Osher Doctorow wrote:
>V. Kargin (Cornerstone Research, N.Y.) and A. Onatski in "Dynamics
>of interest rate curve by functional autoregression," arxiv.org/pdf/
>math.ST/0411047.pdf, use a very roughly analogous idea with
functional
>analysis at least in part to predict dynamics of interest rate
>curves in econometrics. Onatski is in the economics department
>of Columbia University. Their paper is dated October 2004.
>Kargin and Onatski's paper is of interest for several reasons. First
>of all, they don't use the usual least squares regression to estimate
Referring to the second posting back, notice that:
1) y = (1 + y - x)x + a
has 1 + y - x replacing b (sample coefficient) or beta (population
coefficient) where beta is supposed to be constant but 1 + y - x is
not necessarily constant. This means that it is even more surpris-
ing that the sufficient statistics or their squares and products
which turn out to have their sums equal to a or its estimate (also
called a if confusion is not likely) are similar to the multiplica-
tive sufficient statistics from classical simple linear regression
with constant beta and b estimating beta there.
What this means is even simpler: simple linear regression is itself
explained in its local (but notice not usually global) successes
because it is a rough approximation locally to what is really a
nonlinear regression given by (1) above.
Several cases are especially of interest, and they raise the ques-
tion of why classical statisticians have not tried to replace beta
by a variable in y = beta x + alpha (which can also be written y =
bx + a if no confusion occurs). It is probably because they did not
realize that the estimated beta in simple regression, b = rs_y/s_x,
has any fundamental significance in the sense that rho (the popula-
tion parameter of r) should be modelled under different cases.
Here are several interesting cases, with provisional results that
I have obtained but which could be beneficially cross-checked several
more times.
Case 1. 1 + y - x = b (constant b)
Case 2. 1 + y - x = bx (constant b)
Case 3. 1 + y - x = bx^2 (+ a) (constant b and/or a)
Case 4. 1 + y - x = b exp(kx) (+ a_ (a, b, k constants, k > 0)
Case 5. 1 + y - x = b exp(-kx) (+a) (a, b, k constants, k > 0)
Cases 1-3 seem to be logically O.K., while cases 4 and 5 seem to
have anomalies when considering y values at x = 0 and x = 1
simultaneously (remember that 1 + y - x has y < = x and x, y in
[0, 1], which puts some constraints on a, b, k. When checking, I
find it easiest to write y(0) = yo, y(1) = y1, where y = y(x).
Note also that there is another constraint, namely that y =
(1 + y - x)x + a is supposed hold in all 5 cases above. This gives
2 simultaneous equations in addition to range and inequality
constraints for each case.
For example, if 1 + y - x = b (constant b), then y = x + b - 1 but
also y = (1 + y - x)x + a, so x + b - 1 = (1 + y - x)x + a, so
x + b - 1 = x + xy - x^2 + a or b = 1 + xy - x^2 + a, so if x = 0
then b = 1 + a and if x = 1 then b = 1 + y - 1 + a = y + a so that
y(1) = 1. As for y, from 1 + y - x = b we get y = x + b - 1 =
x + 1 + a - 1 = x + a and from y = (1 + y - x)x + a we get y = bx +
a so x + a = bx + a so b = 1. But b = 1 + a so 1 = 1 + a so a = 0.
So finally y = x. As a reminder, these are population equations.
If we tried 1 + y - x = b exp(kx) + a, the corresponding simul-
taneous equations for x = 0 and x = 1 when subtracted seem to
require a + b = 1 if y ever = x on [0, 1]. The case when y does
not equal x on the interval will have to be rechecked probably.
Osher Doctorow
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