[R] Integration with variable bounds

[Ravi Varadhan]

You get 0 because you did not specify lower and upper bounds that define the hyper-rectangle; therefore, the default is used which is (0,1)^4.

Specify the proper lower and upper bounds.  

Ravi.

____________________________________________________________________

Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University

Ph. (410) 502-2619
email: rvaradhan at jhmi.edu


----- Original Message -----
From: Dmlong21 <dmlong at bios.unc.edu>
Date: Tuesday, March 29, 2011 3:11 pm
Subject: Re: [R] Integration with variable bounds
To: r-help at r-project.org


> Thanks for the tip but all I get is 0 for the integral.  Any other
> suggestions?
> 
> int <- function(y){
> u2 = y[1]
> z2 = y[2]
> u1 =y[3]
> z1 = y[4]
> 
> reg.nonzero <- (u2 > z1 & u2 < z2) & (z2 > z1 & z2 < 12) & (u1 > 4 & 
> u1 <
> z1) & (z1 > 4 & z1 < 12)
> 
> 
> ff <- ifelse (reg.nonzero, u1*(z1-u1)*u2*(z2-u2)*exp(-0.027*(12-z2)), 
> 0)
> 
> 
> return(ff)
> 
> 
> }
> 
> 
> cuhre(4,1,int)
> 
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