R help - Mar 2012 - Quicker way to apply values to a function

Hi all,
myint=function(mu,sigma){
	integrate(function(x) dnorm(x,mu,sigma)/(1+exp(-x)),-Inf,Inf)$value
}

mymu=seq(-3,3,length(1000))
mysigma=seq(0,1,length(500))[-1]

k=1
v=c()
for (j in 1:length(mymu)) {
	for (i in 1:length(mysigma)) {
		v[k]=myint(mymu[j],mysigma[i])
		k=k+1
	}
}


Basically, I want to investigate for what values of mu and sigma, the
integral is divergent.

Is there another way to do this other than loops?

For now, the 'output' vector v is not so informative. Is there a way to
'show' me for what combinations of mu and sigma, the resulting values
'v'
are from?

Thanks.



-----
###################################################
PhD candidate in Statistics
School of Mathematics, Statistics and Actuarial Science, University of Kent
###################################################

--
View this message in context:
http://r.789695.n4.nabble.com/Quicker-way-to-apply-values-to-a-function-tp4497293p4497293.html
Sent from the R help mailing list archive at Nabble.com.

I'd imagine one could solve this problem analytically
(divergence/convergence *almost certainly* [hint cough!]....closed
form value seems hard) but perhaps you want to loop over a matrix
instead:

v <- matrix(NA, ncol = length(mymu), nrow = length(mysigma))
rownames(v) <- mysigma
colnames(v) <- mymu

then in the loop:

v[i,j] <- myint(...)

Michael

On Thu, Mar 22, 2012 at 7:17 PM, casperyc <casperyc at hotmail.co.uk>
wrote:
> Hi all,
> myint=function(mu,sigma){
> ? ? ? ?integrate(function(x) dnorm(x,mu,sigma)/(1+exp(-x)),-Inf,Inf)$value
> }
>
> mymu=seq(-3,3,length(1000))
> mysigma=seq(0,1,length(500))[-1]
>
> k=1
> v=c()
> for (j in 1:length(mymu)) {
> ? ? ? ?for (i in 1:length(mysigma)) {
> ? ? ? ? ? ? ? ?v[k]=myint(mymu[j],mysigma[i])
> ? ? ? ? ? ? ? ?k=k+1
> ? ? ? ?}
> }
>
>
> Basically, I want to investigate for what values of mu and sigma, the
> integral is divergent.
>
> Is there another way to do this other than loops?
>
> For now, the 'output' vector v is not so informative. Is there a
way to
> 'show' me for what combinations of mu and sigma, the resulting
values 'v'
> are from?
>
> Thanks.
>
>
>
> -----
> ###################################################
> PhD candidate in Statistics
> School of Mathematics, Statistics and Actuarial Science, University of Kent
> ###################################################
>
> --
> View this message in context:
http://r.789695.n4.nabble.com/Quicker-way-to-apply-values-to-a-function-tp4497293p4497293.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

On Thu, Mar 22, 2012 at 04:17:20PM -0700, casperyc
wrote:
> Hi all,
> myint=function(mu,sigma){
> 	integrate(function(x) dnorm(x,mu,sigma)/(1+exp(-x)),-Inf,Inf)$value
> }
> 
> mymu=seq(-3,3,length(1000))
> mysigma=seq(0,1,length(500))[-1]
> 
> k=1
> v=c()
> for (j in 1:length(mymu)) {
> 	for (i in 1:length(mysigma)) {
> 		v[k]=myint(mymu[j],mysigma[i])
> 		k=k+1
> 	}
> }
> 
> 
> Basically, I want to investigate for what values of mu and sigma, the
> integral is divergent.
Hi.

The function dnorm(x,mu,sigma)/(1+exp(-x)) has a finite integral
over (-Inf, Inf) for every mu, sigma. The reason is that

  dnorm(x,mu,sigma)

is nonnegative and

  0 < 1/(1+exp(-x)) < 1

So, the integral of dnorm(x,mu,sigma)/(1+exp(-x)) is upper bounded
by the integral of dnorm(x,mu,sigma), which is 1.

Hope this helps.

Petr Savicky.