This example implements the Swiss Alps copulas of Hofert, Vrins (2013, "Sibuya copulas").
## lower resolution - less size (default dpi = 72): knitr::opts_chunk$set(dpi = 48)
Lambda <- function(t) pmax(log(t), 0) LambdaInv <- function(t) (t != 0)*exp(t) ## <=> ifelse(t == 0, 0, exp(t))
M1 <- function(t) 0.01*t M1Inv <- function(t) t/0.01 M2 <- function(t) { f <- function(t.) { ct <- ceiling(t.) if(ct %% 2) t. - (ct-1)/2 else ct/2 } unlist(lapply(t, f)) } M2Inv <- function(t) { f <- function(t.) { if(t. == 0) return(0) t. + ceiling(t) - 1 } unlist(lapply(t, f)) }
S1 <- function(t, H) exp(-(M1(t) + Lambda(t)*(1-exp(-H)))) S1Inv <- function(u, H, upper=1e6) unlist(lapply(u, function(u.) uniroot(function(x) S1(x,H=H)-u., interval=c(0, upper))$root)) S2 <- function(t, H) exp(-(M2(t) + Lambda(t)*(1-exp(-H)))) S2Inv <- function(u, H, upper=1e6) unlist(lapply(u, function(u.) uniroot(function(x) S2(x,H=H)-u., interval=c(0, upper))$root))
p and its inverse (for $p_i(t_k-)$ and $p_i(t_k), i = 1,2$):
p1 <- function(t, k, H) exp(-M1(t)-H*k) p1Inv <- function(u, k, H, upper=1e6) unlist(lapply(u, function(u.) uniroot(function(x) p1(x,k=k,H=H)-u., interval=c(0, upper))$root)) p2 <- function(t, k, H) exp(-M2(t)-H*k) p2Inv <- function(u, k, H, upper=1e6) unlist(lapply(u, function(u.) uniroot(function(x) p2(x,k=k,H=H)-u., interval=c(0, upper))$root))
and the wrappers, which work with p1(), p2(), p1Inv(), p2Inv()
as arguments:
p <- function(t, k, H, I, p1, p2){ if((lI <- length(I)) == 0){ stop("error in p") }else if(lI==1){ if(I==1) p1(t, k=k, H=H) else p2(t, k=k, H=H) }else{ # lI == 2 c(p1(t, k=k, H=H), p2(t, k=k, H=H)) } } pInv <- function(u, k, H, I, p1Inv, p2Inv){ if((lI <- length(I)) == 0){ stop("error in pInv") }else if(lI==1){ if(I==1) p1Inv(u, k=k, H=H) else p2Inv(u, k=k, H=H) }else{ # lI == 2 c(p1Inv(u[1], k=k, H=H), p2Inv(u[2], k=k, H=H)) } }
C <- function(u, H, Lambda, S1Inv, S2Inv) { if(all(u == 0)) 0 else u[1]*u[2] * exp(expm1(-H)^2 * Lambda(min(S1Inv(u[1],H), S2Inv(u[2],H)))) }
Compute the singular component (given $u_1=$ u1
, find $u_2=$ u2
on the singular component)
$u_2 = S2(S1^{-1}(u_1))$:
s.comp <- function(u1, H, S1Inv, S2){ if(u1 == 0) 0 else S2 (S1Inv(u1,H), H) }
Generate one bivariate random vector from C
:
rC1 <- function(H, LambdaInv, S1, S2, p1, p2, p1Inv, p2Inv) { d <- 2 # dim = 2 ## (1) U <- runif(d) # for determining the default times of all components ## (2) -- t_{h,0} := initial value for the occurrence of the homogeneous ## Poisson process with unit intensity t_h <- 0 t <- 0 # t_0; initial value for the occurrence of the jump process k <- 1 # indices for the sets I I_ <- list(1:d) # I_k; indices for which tau has to be determined tau <- rep(Inf,d) # in the beginning, set all default times to Inf (= "no default") ## (3) repeat{ ## (4) ## k-th occurrence of a homogeneous Poisson process with unit intensity: t_h[k] <- rexp(1) + if(k == 1) 0 else t_h[k-1] ## k-th occ. of non-homogeneous Poisson proc. with integrated rate function Lambda: t[k] <- LambdaInv(t_h[k]) ## (5) pvec <- p(t[k], k=k, H=H, I=c(1,2), p1=p1, p2=p2) # c(p_1(t_k), p_2(t_k)) I. <- (1:d)[U >= pvec] # determine all i in I I <- intersect(I., I_[[k]]) # determine I ## (6)--(10) if(length(I) > 0){ default.at.t <- U[I] <= p(t[k], k=k-1, H=H, I=I, p1=p1, p2=p2) tau[I[default.at.t]] <- t[k] Ic <- I[!default.at.t] # I complement if(length(Ic) > 0) tau[Ic] <- pInv(U[Ic], k=k-1, H=H, I=Ic, p1Inv=p1Inv, p2Inv=p2Inv) } ## (11) -- define I_{k+1} := I_k \ I I_[[k+1]] <- setdiff(I_[[k]], I) ## (12) if(length(I_[[k+1]]) == 0) break else k <- k+1 } ## (14) c(S1(tau[1], H=H), S2(tau[2], H=H)) }
rC <- function(n, H, LambdaInv, S1, S2, p1, p2, p1Inv, p2Inv){ mat <- t(sapply(rep(H, n), rC1, LambdaInv=LambdaInv, S1=S1, S2=S2, p1=p1, p2=p2, p1Inv=p1Inv, p2Inv=p2Inv)) row.names(mat) <- NULL mat } ## Generate copula data n <- 2e4 # <<< use for niceness n <- 4000 # (rather use to decrease *.html and final package size) H <- 10 set.seed(271) U <- rC(n, H=H, LambdaInv=LambdaInv, S1=S1, S2=S2, p1=p1, p2=p2, p1Inv=p1Inv, p2Inv=p2Inv) ## Check margins of U par(pty="s") hist(U[,1], probability=TRUE, main="Histogram of the first component", xlab=expression(italic(U[1]))) hist(U[,2], probability=TRUE, main="Histogram of the second component", xlab=expression(italic(U[2]))) ## Plot U (copula sample) plot(U, pch=".", xlab=expression(italic(U[1])%~%~"U[0,1]"), , ylab=expression(italic(U[2])%~%~"U[0,1]"))
Wireframe plot to incorporate singular component :
require(lattice) wf.plot <- function(grid, val.grid, s.comp, val.s.comp, Lambda, S1Inv, S2Inv){ wireframe(val.grid ~ grid[,1]*grid[,2], xlim=c(0,1), ylim=c(0,1), zlim=c(0,1), aspect=1, scales = list(arrows=FALSE, col=1), # remove arrows par.settings= list(axis.line = list(col="transparent"), # remove global box clip = list(panel="off")), pts = cbind(s.comp, val.s.comp), # <- add singular component panel.3d.wireframe = function(x, y, z, xlim, ylim, zlim, xlim.scaled, ylim.scaled, zlim.scaled, pts, ...) { panel.3dwire(x=x, y=y, z=z, xlim=xlim, ylim=ylim, zlim=zlim, xlim.scaled=xlim.scaled, ylim.scaled=ylim.scaled, zlim.scaled=zlim.scaled, ...) xx <- xlim.scaled[1]+diff(xlim.scaled)*(pts[,1]-xlim[1])/diff(xlim) yy <- ylim.scaled[1]+diff(ylim.scaled)*(pts[,2]-ylim[1])/diff(ylim) zz <- zlim.scaled[1]+diff(zlim.scaled)*(pts[,3]-zlim[1])/diff(zlim) panel.3dscatter(x=xx, y=yy, z=zz, xlim=xlim, ylim=ylim, zlim=zlim, xlim.scaled=xlim.scaled, ylim.scaled=ylim.scaled, zlim.scaled=zlim.scaled, type="l", col=1, ...) }, xlab = expression(italic(u[1])), ylab = expression(italic(u[2])), zlab = list(expression(italic(C(u[1],u[2]))), rot=90)) } ## Copula plot with singular component u <- seq(0, 1, length.out=20) # grid points per dimension grid <- expand.grid(u1=u, u2=u) # grid val.grid <- apply(grid, 1, C, H=H, Lambda=Lambda, S1Inv=S1Inv, S2Inv=S2Inv) # copula values on grid s.comp <- cbind(u, sapply(u, s.comp, H=H, S1Inv=S1Inv, S2=S2)) # pairs (u1, u2) on singular component val.s.comp <- apply(s.comp, 1, C, H=H, Lambda=Lambda, S1Inv=S1Inv, S2Inv=S2Inv) # corresponding z-values wf.plot(grid=grid, val.grid=val.grid, s.comp=s.comp, val.s.comp=val.s.comp, Lambda=Lambda, S1Inv=S1Inv, S2Inv=S2Inv)
For more details, see Trutschnig, Fernandez Sanchez (2014) "Copulas with continuous, strictly increasing singular conditional distribution functions"
Roughly, one defines an Iterated Function System whose attractor is the word "Copula" and starts the chaos game.
IFS <- local({ ## Using `local`, so `n` is part of IFS n <- 23 list(function(x) c(3*x[1]/n, x[2]/4), function(x) c(-(x[2]-1)/n, x[1]/2+1/4), function(x) c(3*x[1]/n, x[2]/4+3/4), function(x) c((3*x[1]+4)/n, x[2]/4), function(x) c(-(x[2]-5)/n, x[1]/2+1/4), function(x) c((3*x[1]+4)/n, x[2]/4+3/4), function(x) c(-(x[2]-7)/n, x[1]/2+1/4), function(x) c(-x[2]/n+9/n, 3*x[1]/4), function(x) c((3*x[1]+8)/n, x[2]/4+3/4), function(x) c(x[1]/n+10/n, x[2]/8+1/2+1/8), function(x) c(2*x[1]/n+9/n, x[2]/4+1/4+1/8), function(x) c(-x[2]/n+13/n, (3*x[1]+1)/4), function(x) c((3*x[1]+12)/n, x[2]/4), function(x) c((3*x[1]+12)/n, x[2]/4), function(x) c(-x[2]/n+15/n, (3*x[1]+1)/4), function(x) c((3*x[1]+16)/n, x[2]/4), function(x) c(-(x[2]-21)/n, 3*x[1]/4), function(x) c((3*x[1]+20)/n, x[2]/4+3/4), function(x) c((x[1]+21)/n, x[2]/4+1/4+1/8), function(x) c(-(x[2]-23)/n, 3*x[1]/4)) })
B <- 20 # replications n.steps <- 20000 # number of steps AA <- vector("list", length=B) set.seed(271) for(i in 1:B) { ind <- sample(length(IFS), size=n.steps, replace=TRUE) # (randomly) 'bootstrap' functions res <- matrix(0, nrow=n.steps+1, ncol=2) # result matrix (for each i) pt <- c(0, 0) # initial point for(r in seq_len(n.steps)) { res[r+1,] <- IFS[[ind[r]]](pt) # evaluate randomly chosen functions at pt pt <- res[r+1,] # redefine point } AA[[i]] <- res # keep this matrix } A <- do.call(rbind, AA) # rbind (n.steps+1, 2)-matrices n <- nrow(A) stopifnot(ncol(A) == 2, n == B*(n.steps+1)) # sanity check
$X :=$ Rotate $A$ by $-45^{o} = -\pi/4$ :
phi <- -pi/4 X <- cbind(cos(phi)*A[,1] - sin(phi)*A[,2]/3, sin(phi)*A[,1] + cos(phi)*A[,2]/3) stopifnot(identical(dim(X), dim(A)))
Now transform the margings by their marginal ECDF's so we get uniform margins.
Note that, it is equivalent but faster to use rank(*, ties.method="max")
:
U <- apply(X, 2, function(x) ecdf(x)(x)) ## Prove equivalence: stopifnot(all.equal(U, apply(X, 2, rank, ties.method="max") / n, tolerance = 1e-14))
Now, visually check the margins of U
; they are perfectly uniform:
par(pty="s") sfsmisc::mult.fig(mfcol = c(1,2), main = "Margins are uniform") hist(U[,1], probability=TRUE, main="Histogram of U[,1]", xlab=quote(italic(U[1]))) hist(U[,2], probability=TRUE, main="Histogram of U[,2]", xlab=quote(italic(U[2])))
whereas U
, the copula sample, indeed is peculiar and contains the word
"COPULA" many times if you look closely (well, the "L" is defect ...):
par(pty="s") plot(U, pch=".", xlab = quote(italic(U[1]) %~% ~ "U[0,1]"), asp = 1, ylab = quote(italic(U[2]) %~% ~ "U[0,1]"))
This is an implementation of Example 2.3 in https://arxiv.org/pdf/0906.4853
library(abind) # for merging arrays via abind() library(lattice) # for cloud() library(sfsmisc) # for polyn.eval()
Implement the random number generator:
##' @title Generate samples from the Sierpinski tetrahedron ##' @param n sample size ##' @param N digits in the base-2 expansion ##' @return (n, 3)-matrix ##' @author Marius Hofert rSierpinskyTetrahedron <- function(n, N) { stopifnot(n >= 1, N >= 1) ## Build coefficients in the base-2 expansion U12coeff <- array(sample(0:1, size = 2*n*N, replace = TRUE), dim = c(2, n, N), dimnames = list(U12 = c("U1", "U2"), sample = 1:n, base2digit = 1:N)) # (2, n, N)-array U3coeff <- apply(U12coeff, 2:3, function(x) sum(x) %% 2) # (n, N)-matrix Ucoeff <- abind(U12coeff, U3 = U3coeff, along = 1) ## Convert to U's t(apply(Ucoeff, 1:2, function(x) polyn.eval(coef = rev(x), x = 2))/2^N) # see sfsmisc::bi2int }
Draw vectors of random numbers following a "Sierpinski tetrahedron copula":
set.seed(271) U <- rSierpinskyTetrahedron(1e4, N = 6)
Use a scatterplot matrix to check all bivariate margins:
pairs(U, gap = 0, cex = 0.25, col = "black", labels = as.expression( sapply(1:3, function(j) bquote(U[.(j)])) ))
All pairs "look" independent but, of course, they aren't:
cloud(U[,3] ~ U[,1] * U[,2], cex = 0.25, col = "black", zoom = 1, scales = list(arrows = FALSE, col = "black"), # ticks instead of arrows par.settings = list(axis.line = list(col = "transparent"), # to remove box clip = list(panel = "off"), standard.theme(color = FALSE)), xlab = expression(U[1]), ylab = expression(U[2]), zlab = expression(U[3]))
print(sessionInfo(), locale=FALSE)
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