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Re: [Phys-L] LSF Slope Versus Average of Single Point Slopes

Is the LSF method developed based on the distribution of errors being a Gaussian distribution?

If so, wouldn't a better test of DP's question involve generating a non-Gaussian distribution of errors? (Of course, if LSF is a completely general, distribution-free operation, never mind).

-----Original Message-----
From: phys-l-bounces@mail.phys-l.org [mailto:phys-l-bounces@mail.phys-
l.org] On Behalf Of Donald Polvani
Sent: Saturday, May 05, 2012 2:44 PM
To: Phys-L@Phys-L.org
Subject: Re: [Phys-L] LSF Slope Versus Average of Single Point Slopes

Thanks to the list and, in particular, John Denker and John Mallinckrodt, for
your comments on this issue.

Despite the fact that weighted averaging, or maximum a posteriori
techniques, may be more optimal approaches, I was still bothered that taking
a simple average of the (unweighted) point slopes (APS) seemed to produce
an "equivalent" result to the LSF (maximum likelihood) slope. So, I took John
Mallinckrodt's lead and computed the slopes and slope standard deviations
each approach should yield assuming only errors in y and with y errors not a
function of x.

As he pointed out for y = mx (no y-intercept), the APS and LSF solutions for
slope are:

m_APS = (1/N)*S_n(y_n/x_n) (1)
m_LSF = S_n(x_n*y_n)/S_n(x_n^2) (2)

Where S_n(x_n) means the sum over n of x_n and n = 1,2,3 ...N.

I worked out the standard deviation for the slopes of each approach
(s_m_APS and s_m_LSF) and obtained:

s_m_APS = (s_y/N)*sqrt(S_n(1/x_n^2)) (3)
s_m_LSF = s_y*(1/sqrt(S_n(x_n^2))) (4)

Where s_y is the standard deviation of the individual deviations dy_n = y_n -
m*x_n

In my original post, I was judging the APS and LSF slopes solely on their
average values (as obtained from a Monte Carlo simulation). "By eye" there
wasn't much to choose between the two results for m. However, I should
have looked at the standard deviations s_m_APS and s_m_LSF as a better
discriminator.

I made a new (Excel) Monte Carlo simulation for the following problem:

y = mx with m = 1, x = 1,2,3,4,5 (i.e. no errors in x)

The y values were generated from zero mean Gaussian random distribution
with a standard deviation of 0.1. I ran 500 trials and averaged the individual
results for m_APS and m_LSF. I also computed the standard deviations of
the 500 individual results.

Here are typical results for one set of 500 trials:

Average of m_APS = 0.9982
Average of m_LSF = 0.9989
s_m_APS = 0.0238
s_m_LSF = 0.0134

Note that the Excel ratio of s_m_APS/s_m_LSF = 1.78. From Equations (3)
and (4) above, and assuming approximately equal values for s_y from the
two approaches, we get:

s_m_APS/s_m_LSF = (1/N)*sqrt(S_n(1/x_n^2))*(sqrt(S_n(x_n^2))) = 1.79
for this example (using the x_n above) in good agreement with the Excel
result.

So, I feel better. For this simple example, with equal weighting for the point
slopes and the LSF maximum likelihood approach, the LSF approach does
have the smaller standard deviation and, in that sense, is more "optimum"
than the APS result. While the average results for m_APS and m_LSF change
value for each new set of 500 Monte Carlo trials (with m_APS sometimes
better than m_LSF), the ratio of s_m_APS/s_m_LSF stays pretty much
around 1.79, indicating the superiority of the LSF approach (and also giving
me confidence in Excel's LSF zero y-intercept routine).

Don

Dr. Donald G. Polvani
Adjunct Faculty, Physics
Anne Arundel Community College

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