On Jul 17, 2011, at 8:13 PM, Jochen1980 wrote:

Hi David,

my first attempt to work through your code was successful, my predicted line is pretty close to the ecdf-function. I have no idea why you inverted the
gumbel-function and what the advantage of this strategy is?

I was advised by a person who knows way more statistics than I do that the strategy of minimizing the error around a regular step function was statistically suspect (presumably even if the x value were a random process). So I inverted the solution and put the random values on the y-axis, leaving the stair-step values on the x-axis.. I think claiming an "advantage" might be premature (especially since both methods yield very similar values) and would probably require some testing before acceptance. There is of course the fitdistr function in MASS and there is a nice vignette on "Fitting Distribution in R" some place around (yep, first google hit):

http://cran.r-project.org/doc/contrib/Ricci-distributions-en.pdf



I interpret your (1:100/100)-trick like this: you build a sequence of 100 Tics. In addition to this 100 is identical to the number of created points.

The ecdf function object (at least as I understood from its plotted result) sets up its y-values (in the function's environment) to be a regular sequence determined by the number of original points and allows you to recover the x-values with the knots function.

--
David


In my real world situation, ecdf() will produce ties, same values, and this results that I do not know what number knot() will return, then this will
result in a non-symmetric-data.frame()-error.

R-Code:
############### Tutorial 3, Gamma-Distribution-Points to Gumbel
################
# --- create Points --- #
http://r.789695.n4.nabble.com/ecdf-to-nls-how-to-transform-data-td3671754.html
print( "Tutorial 3" )
p <- rgamma( 100, 2)
print( p )
ep <- ecdf( p )
x <- knots(ep)
hist( p, main="Tutorial 3" )
ypre <- seq( from=1, to=100, by=1 )
y <- ypre / 100
df <- data.frame( x=x, y=y )
plot( df )
res <- nls( y ~ exp( - exp( - ( x - mue ) / beta )), start=list(mue=2,
beta=2), data=df, trace=TRUE )
print( summary(res) )
#print( str(res))
lines( x, predict( res, list( x=x )), lty = 1, col = "blue" ) # predicted
vals

Console-Output:
[1] "Tutorial 3"
 [1] 4.75748951 5.36642043 2.03025702 1.80216440 1.40745277 0.91393251
 [7] 1.30575505 2.07451301 1.18500815 0.24972503 1.36604865 2.36786796
[13] 1.37386798 1.72390843 1.93700139 0.78722468 2.92828385 1.51165415
[19] 4.17000960 2.98356424 1.20195996 1.85899533 0.94863757 1.50438053
[25] 0.50854274 3.32760075 2.21334316 0.58463817 2.26985676 2.88033389
[31] 2.00718903 1.19043470 1.62945752 0.32010478 0.54837755 1.32965131
[37] 3.32024573 0.53825226 3.30077557 0.46426513 1.00681720 0.59276030
[43] 1.31431609 2.27822419 2.56404713 0.52459218 1.05228996 2.23799673
[49] 1.14438962 2.41311612 1.40254244 1.20379073 0.44195457 0.95880408
[55] 1.45254027 4.49645001 2.18826796 1.07161515 1.62617544 3.15506003
[61] 2.15611491 2.43261017 1.40293461 2.79977886 3.44503201 3.25282551
[67] 0.91570531 0.70243512 5.67971774 3.55532192 1.71515046 2.97718949
[73] 0.47145131 0.32879089 0.60735231 0.92388638 4.29277857 1.73681839
[79] 0.09387383 2.81281295 1.84419058 2.63070353 1.52298124 2.89761263
[85] 1.05251987 1.24258703 3.09130229 2.66738574 3.17035060 2.15287327
[91] 2.63042751 1.69736779 3.21126033 1.56920999 2.68985477 0.82952068
[97] 3.62230865 1.65286592 2.76798783 2.08091935

Formula: y ~ exp(-exp(-(x - mue)/beta))

Parameters:
    Estimate Std. Error t value Pr(>|t|)
mue  1.382744   0.005926   233.3   <2e-16 ***
beta 0.984836   0.008816   111.7   <2e-16 ***




--
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David Winsemius, MD
West Hartford, CT

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