Hi,
I am using the uniroot function in order to carry out a bivariate Monte Carlo simulation using the logistics model. I have defined the function as: BV.FV <- function(x,y,a,A) (((x^(-a^-1)+y^(-a^-1))^(a-1))*(y^(a-1/a))*(exp(-((1^(-a^-1)+y^(-a^-1))^a)+y^-1)))-A and the procedure is as follows: Randomly generate values of A~(0,1), y0 = -(lnA)^-1 Where: A=Pr{X<xi|Y=yi-1} and a is the dependency between X and y (0.703) Use y0 to determine x where x = x(i) and y = y0(i-1) in the above equation. Unfortunately when the randomly defined A gets to approximately 0.46 the intervals I provide no longer have the opposite signs (both -ve). I have tried various different upper limits but I still can't seem to find a root. Does anyone have any suggestions? Thanks, Doug |
<quote author="dpender">
Hi, I am using the uniroot function in order to carry out a bivariate Monte Carlo simulation using the logistics model. I have defined the function as: BV.FV <- function(x,y,a,A) (((x^(-a^-1)+y^(-a^-1))^(a-1))*(y^(a-1/a))*(exp(-((1^(-a^-1)+y^(-a^-1))^a)+y^-1)))-A and the procedure is as follows: Randomly generate values of A~(0,1), y0 = -(lnA)^-1 Where: A=Pr{X<xi|Y=yi-1} and a is the dependency between X and y (0.703) Use y0 to determine x where x = x(i) and y = y0(i-1) in the above equation. Unfortunately when the randomly defined A gets to approximately 0.46 the intervals I provide no longer have the opposite signs (both -ve). </quote> Try curve(BV.FV(x,y=y0,a=.703,A=.7),from=1,to=10) for different values of a, A and y0. I have a feeling that your function or something else is not quite correct. Berend |
In reply to this post by dpender
On Fri, Feb 04, 2011 at 04:35:00AM -0800, dpender wrote:
> > Hi, > > I am using the uniroot function in order to carry out a bivariate Monte > Carlo simulation using the logistics model. > > I have defined the function as: > > BV.FV <- function(x,y,a,A) > (((x^(-a^-1)+y^(-a^-1))^(a-1))*(y^(a-1/a))*(exp(-((1^(-a^-1)+y^(-a^-1))^a)+y^-1)))-A > > and the procedure is as follows: > > Randomly generate values of A~(0,1), y0 = -(lnA)^-1 > > Where: A=Pr{X<xi|Y=yi-1} and a is the dependency between X and y (0.703) > > Use y0 to determine x where x = x(i) and y = y0(i-1) in the above equation. > > Unfortunately when the randomly defined A gets to approximately 0.46 the > intervals I provide no longer have the opposite signs (both -ve). > > I have tried various different upper limits but I still can't seem to find a > root. > > Does anyone have any suggestions? Hi. The expression for BV.FV() contains only one occurrence of x and it may be separated from the equation. The solution for x may be expressed as get.x <- function(y,a,A) ((A/(y^(a-1/a))/(exp(-((1^(-a^-1)+y^(-a^-1))^a)+y^-1)))^(1/(a-1))-y^(-a^-1))^(-a) a <- 0.703 A <- 0.45 y <- - 1/log(A) uniroot(BV.FV, c(1, 20), y=y, a=a, A=A)$root [1] 3.477799 get.x(y, a, A) [1] 3.477794 As pointed out in another reply, the equation does not always have a solution. It seems to tend to infinity in the following for (A in seq(0.45, 0.4677, length=20)) { y <- - 1/log(A) cat(get.x(y, a, A), uniroot(BV.FV, c(1, 1000), y=y, a=a, A=A)$root, "\n") } 3.477794 3.477791 3.637566 3.637565 3.812851 3.812851 4.006213 4.006214 4.220838 4.220834 4.460738 4.460739 4.731048 4.731052 5.038453 5.038473 5.391845 5.391843 5.803329 5.803329 6.289873 6.28988 6.876084 6.876084 7.5992 7.599201 8.518665 8.518683 9.736212 9.736212 11.44329 11.44328 14.05345 14.05345 18.68155 18.68155 29.97659 29.9766 219.3156 219.3155 Hope this helps. Petr Savicky. ______________________________________________ [hidden email] 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. |
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