On Jul 26, 2013, at 8:44 AM, Tiago V. Pereira wrote:
> I am trying to double integrate the following expression:
>
> # expression
> (1/(2*pi))*exp(-y2/2)*sqrt((y1/(y2-y1)))
>
> for y2>y1>0.
>
> I am trying the following approach
>
> # first attempt
>
> library(cubature)
> fun <- function(x) { (1/(2*pi))*exp(-x[2]/2)*sqrt((x[1]/(x[2]-x[1])))}
> adaptIntegrate(fun, lower = c(0,0), upper =c(5, 6), tol=1e-8)
>
> However, I don't know how to constrain the integration so that y2>y1>0.
Generally incorporating boundaries is accomplished by multiplying the integrand
with logical vectors that encapsulate what are effectively two conditions:
Perhaps:
fun <- function(x) { (x[1]<x[2])*(x[1]>0)* (1/(2*pi))*exp(-x[2]/2)*
sqrt((x[1]/(x[2]-x[1])))}
That was taking quite a long time and I interrupted it. There were quite a few
warnings of the sort
1: In sqrt((x[1]/(x[2] - x[1]))) : NaNs produced
2: In sqrt((x[1]/(x[2] - x[1]))) : NaNs produced
Thinking the NaNs might sabotage the integration process, I added a conditional
to the section of that expression that was generating the NaNs. I don't really
know whether NaN's are excluded from the summation process in adaptIntegrate:
fun <- function(x) { (x[1]<x[2])*(x[1]>0)* (1/(2*pi))*exp(-x[2]/2)*
if(x[1]>x[2]){ 0 }else{ sqrt((x[1]/(x[2]-x[1]))
)} }
adaptIntegrate(fun, lower = c(0,0), upper =c(5, 6) )
I still didn't have the patience to wait for an answer, but I did plot the
function:
fun2 <- function(x,y) { (x<y)*(x>0)* (1/(2*pi))*exp(-y/2)* sqrt((x/(y-x)))}
persp(outer(0:5, 0:6, fun2) )
So at least the function is finite over most of its domain.
--
David Winsemius
Alameda, CA, USA
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