# [R] nproc parameter in efpFunctional

Achim Zeileis Achim.Zeileis at uibk.ac.at
Fri Nov 4 21:03:32 CET 2011

```On Fri, 4 Nov 2011, bonda wrote:

> The 2006 CSDA paper is really very informative, perhaps, I'm trying to
> understand the things lying beyond. If we have e.g. k=3, then taking
> nproc=3 for the functional maxBB we get a critical value (boundary)
>
> maxBB\$computeCritval(0.05,nproc=3)
>  1.544421,
>
> and this for nproc=NULL (Bonferroni approximation) will be
>
> maxBB\$computeCritval(0.05)
>  1.358099.

No. In the latter case no Bonferroni approximation is applied. If you want
to use it, you can do so via the rule of thumb

R> maxBB\$computeCritval(0.05/3, nproc = 1)
 1.547175

which essentially matches the critical value computed for nproc = 3. If
you use the more precise value 1 - (1 - 0.05)^(1/3) instead of 0.05/3, you
get a match (up to some small numerical differences).

Setting nproc=NULL is only possible in efpFunctional():
efpFunctional() sets up the computeCritval() and computePval() functions
via simulation methods (unless closed form solutions are supplied). For
the simulation two strategies are available: Simulate nproc = 1, 2, 3, ...
explicitly. Simulate only nproc = 1 and apply a Bonferroni correction. The
last option is chosen if you set nproc=NULL -- it makes only sense if you
aggregate via the maximum across the components.

The resulting computeCritval() and computePval() function always need to
have the correct nproc supplied (i.e., nproc=NULL makes no sense).

> Aggregating 3 Brownian bridges first over components, we obtain time series
> process. Now, we wonder if maximum value of the process (aggregation over
> time) lies over boundary. Which boundary - 1.544421 or 1.358099 - should one
> take? They look too different and, for instance, lead to "unfair computing"
> of empirical size (as rejection rate of null hypothesis) or empirical power
> (as acception rate of alternative).
>
>
>
>
> --
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