Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
that was proposed by Kumar (2010) as reproduced above. But the method seriously fails in that the range of (U,V) resulting from the simulation does not even cover the (0,1)² square! Unless I made an R coding mistake (which is always a possibility).
sim=function(T=1e3,h=.5){ o=matrix(runif(2*T),T,2) a=1-o[,1];b=1-h*(1+2*a*o[,2])+2*h^2*a^2*o[,2] d=1+h*(2-4*a+4*a*o[,2])+h^2*(1-4*a*o[,2] +4*a^2*o[,2]) o[,2]=1-2*o[,2]*(a*h-1)^2/(b+sqrt(d)) return(o)}
which leads to a second degree polynomial equation (in v) when solving the equation F(v|U=u) = w.
sim=function(T=1e3,h=.5){ o=matrix(runif(2*T),T,2) v1=o[,1];w=o[,2] d=(2*h*v1*w-1-h)^2-4*(1-w)*h*(1-h*w*v1^2) o[,2]=-2*h*v1*w+1+h-sqrt(d))/(2*h*(1-h*w*v1^2) return(1-o)}
And with a more likely outcome (Xed checked by comparing F(u,v) with its empirical version for several pairs (u,v)):
R-bloggers.com offers daily e-mail updates about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.