R help - Apr 2003 - AW: numericDeriv and ecdf

> On only ten points, what did you expect ?  Even with 1000
> observations, estimating a density is difficult, and has
> been the subject of a century of research.  Kernel density
> estimates are among the most successful.  For your immediate
> application, try  plot(density(rnorm(10)), type="l"), etc.
wait, you misunderstood me!

I'd like to see 10 or 9 points with estimated values of 
*numerical* derivatives according to ecdf output. 
And that's it.

Now look into output of numericDerivative in my example:

[1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
attr(,"gradient")
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]    0    0    0    0    0    0    0    0    0     0
 [2,]    0    0    0    0    0    0    0    0    0     0
 [3,]    0    0    0    0    0    0    0    0    0     0
 [4,]    0    0    0    0    0    0    0    0    0     0
 [5,]    0    0    0    0    0    0    0    0    0     0
 [6,]    0    0    0    0    0    0    0    0    0     0
 [7,]    0    0    0    0    0    0    0    0    0     0
 [8,]    0    0    0    0    0    0    0    0    0     0
 [9,]    0    0    0    0    0    0    0    0    0     0
[10,]    0    0    0    0    0    0    0    0    0     0

What could you say now?

With kind regards,
Valery A.Khamenya
---------------------------------------------------------------------------
Bioinformatics Department
BioVisioN AG, Hannover

An empirical CDF is a step function: it does not have a derivative at the 
jump points, and has a zero derivative everywhere else.

What is this function `numericDerivative': do you mean `numericDeriv'?
If so, it seems to be intended for differentiable functions, and
calculates one-sided derivatives.  In your example the one-sided 
derivatives are all zero.

On Fri, 25 Apr 2003, Khamenia, Valery wrote:
> > On only ten points, what did you expect ?  Even with 1000
> > observations, estimating a density is difficult, and has
> > been the subject of a century of research.  Kernel density
> > estimates are among the most successful.  For your immediate
> > application, try  plot(density(rnorm(10)), type="l"), etc.
> 
> wait, you misunderstood me!
> 
> I'd like to see 10 or 9 points with estimated values of 
> *numerical* derivatives according to ecdf output. 
> And that's it.
> 
> Now look into output of numericDerivative in my example:
> 
> [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
> attr(,"gradient")
>       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
>  [1,]    0    0    0    0    0    0    0    0    0     0
>  [2,]    0    0    0    0    0    0    0    0    0     0
>  [3,]    0    0    0    0    0    0    0    0    0     0
>  [4,]    0    0    0    0    0    0    0    0    0     0
>  [5,]    0    0    0    0    0    0    0    0    0     0
>  [6,]    0    0    0    0    0    0    0    0    0     0
>  [7,]    0    0    0    0    0    0    0    0    0     0
>  [8,]    0    0    0    0    0    0    0    0    0     0
>  [9,]    0    0    0    0    0    0    0    0    0     0
> [10,]    0    0    0    0    0    0    0    0    0     0
> 
> What could you say now?
> 
> With kind regards,
> Valery A.Khamenya
> ---------------------------------------------------------------------------
> Bioinformatics Department
> BioVisioN AG, Hannover
> 
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://www.stat.math.ethz.ch/mailman/listinfo/r-help
> 
-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

On Fri, 25 Apr 2003, Khamenia, Valery wrote:
> > On only ten points, what did you expect ?  Even with 1000
> > observations, estimating a density is difficult, and has
> > been the subject of a century of research.  Kernel density
> > estimates are among the most successful.  For your immediate
> > application, try  plot(density(rnorm(10)), type="l"), etc.
>
> wait, you misunderstood me!
>
> I'd like to see 10 or 9 points with estimated values of
> *numerical* derivatives according to ecdf output.
The output of ecdf() is a step function, so mathematically its derivative
is zero except where it is undefined.

Everywhere except the jumps numericDeriv() gives the right answer
Exactly at the jumps the derivative is undefined, so it isn't surprising
that numericDeriv gives the wrong answer.

What result did you want?

	-thomas