The smartest solution [Bioanalytics]

posted by ElMaestro  – Denmark, 2019-03-16 22:35  – Posting: # 20041
Views: 540

Ladies and gentlemen,

I present to you the Philadelphia variation: The smartest and most empirical solution to a very, very small (and mainly theoretical?) problem.

Let us apply weights 1/C^z in such a fashion that the sum of relative absolute residuals is smallest. Idea borrowed from ISR.
And it may make good sense to look at the relative magnitude of residuals since this is what runs pass criteria are based on.

Therefore, here is something to play around with:
Conc=c(1, 2, 4, 8, 20, 50, 100, 150, 200, 300)
Ratio=c(0.5303, 0.1074, 0.2092, 0.4121, 0.9886, 2.3197, 5.0343, 7.7656, 10.2105, 14.9564)

ObjF=function(z)
{
 w=1/(Conc^z)
 M=lm(Ratio ~Conc, weight=w)
 return(sum(abs(resid(M)/Conc)))
}

##now let us find the value of z which gives the smallest sum of absolute relative residuals
optimize(ObjF,  c(0, 10))


Note: the fit is not very good, r squared is rather low, but that is besides the point. You get my drift, I hope.

z then defines the weighting scheme which can be said to give the smallest overall amount of percent-wise prediction error on the calibration curve. Not a bad place to start.

You can modify the idea as you please, perhaps you want to define ObjF via the Ratio and not via the Conc, or perhaps you want to return another type of objective altogether. Various things that don't work include but aren't limited to abs sum of residuals, sum of residuals, and more.:cool:

Thank me later. :-D

if (3) 4

x=c("Foo", "Bar")
b=data.frame(x)
typeof(b[,1]) ##aha, integer?
b[,1]+1 ##then let me add 1


Best regards,
ElMaestro

“(...) targeted cancer therapies will benefit fewer than 2 percent of the cancer patients they’re aimed at. That reality is often lost on consumers, who are being fed a steady diet of winning anecdotes about miracle cures.” New York Times (ed.), June 9, 2018.

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