Hi, David.

accurately reflect how closely the model (GAM) fits the data. I was told
>>
>
This was my presumption; I could be mistaken.


>  that the accuracy of the correlation can be improved using a root mean
>> square deviation (RMSD) calculation on binned data.
>>
>
> By whom? ...  and with what theoretical basis?


I talked with Christian Schunn. He mentioned that using RMSD would produce a
better result for goodness-of-fit (if that term is not synonymous with
correlation, I apologise -- I'm still rather new to this level of
statistics):

http://www.lrdc.pitt.edu/schunn/gof/index.html

It was regarding a chart similar to:

http://i.imgur.com/X0gxV.png

In the chart, the calculation for Pearson's, Spearman's, and Kendall's Tau
provide, in my opinion, an incorrect indicator as to the strength of GAM's
fit to the data. I could be wrong here, too.

His suggestion was to use bin the means (in groups of 5 or so) to reduce the
noise.

I doubt that your strategy offers any statistical advantage, but if you want
> to play around with it then consider:
>
> binned.x <- round( (x + 2.5)/5)
>

> d <-
c(1,3,5,4,3,6,3,1,5,7,8,9,4,3,2,7,3,6,8,9,5,3,1,4,5,8,9,3,3,2,5,7,8,8,5,4,3,2,6,4,3,1,4,5,6,8,9,0,7,7,5,4,3,3,2,1,3,4,5,6,7,9,0,2,4,3,3)
> binned.d <- round( (d + 2.5)/5)
> print(binned.d)
 [1] 1 1 2 1 1 2 1 1 2 2 2 2 1 1 1 2 1 2 2 2 2 1 1 1 2 2 2 1 1 1 2 2 2 2 2 1
1 1
[39] 2 1 1 1 1 2 2 2 2 0 2 2 2 1 1 1 1 1 1 1 2 2 2 2 0 1 1 1 1

That doesn't make sense to me. My impression was that I should try to put
every 5 values in a bin, average that bin, then calculate the RMSD between
the observed values and the values from GAM. In other words (o is observed
and m is model):

  bins <- 5

  while( length(o) %% bins != 0 ) {
    o <- o[-length(o)]
  }
  omean <- apply( matrix(o, bins), 2, mean )

  while( length(m) %% bins!= 0 ) {
    m <- m[-length(m)]
  }
  mmean <- apply( matrix(m, bins), 2, mean )

  sqrt( mean( omean - mmean ) ^ 2 )

But that feels sloppy, error prone, and fragile.

Joris mentioned that I could try using tapply with
cut(d,round(length(d)/5)). I couldn't figure out how to get the means back
from the factors.

Dave

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