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:
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