library(qrmtools)
library(copula)
library(combinat) # for permn()
library(sfsmisc) # for eaxis()
doPDF <- FALSE
We start by considering the following setup in the homogeneous case, that is, when all marginal distributions are equal.
qF2 <- function(p, th = 2) qPar(p, shape = th) # Par(2) quantile function
pF2 <- function(q, th = 2) pPar(q, shape = th) # Par(2) distribution function
dim <- 8 # variable dimension (we use 8 or 100 here)
Investigate the helper function \(h(s,t)\) (the function for the inner
root-finding to compute \(D(s)\); see
dual_bound()
).
d <- 8 # dimension
s <- c(1, 5, 10, 100, 500, 1000)
t <- sapply(seq_along(s), function(i) {
res <- exp(seq(log(1e-3), log(s[i]/d), length.out = 257))
res[length(res)] <- s[i]/d # to avoid numerical issues (t > s/d)
res
})
f <- sapply(seq_along(s), function(i)
sapply(t[,i], function(t.)
qrmtools:::dual_bound_2(s[i], t = t., d = d, pF = pF2) -
qrmtools:::dual_bound_2_deriv_term(s[i], t = t., d = d, pF = pF2)))
palette <- colorRampPalette(c("maroon3", "darkorange2", "royalblue3"), space = "Lab")
cols <- palette(6)
if(doPDF)
pdf(file = (file <- paste0("fig_worst_VaR_hom_dual_h_Par=2_d=",d,".pdf")),
width = 6, height = 6)
plot(t[,1], f[,1], type = "l", log = "x", xlim = range(t), ylim = range(f), col = cols[1],
xlab = "t", ylab = expression("h(s,t) for d = 8 and F being Par(2)"))
lines(t[,2], f[,2], col = cols[2])
lines(t[,3], f[,3], col = cols[3])
lines(t[,4], f[,4], col = cols[4])
lines(t[,5], f[,5], col = cols[5])
lines(t[,6], f[,6], col = cols[6])
abline(h = 0, lty = 2)
legend("topright", lty = rep(1,6), col = cols,
bty = "n", legend = as.expression(lapply(1:6,
function(i) substitute(s==s., list(s. = s[i])))))
if(doPDF) dev.off()
As we know, \(h(s,s/d) = 0\). We also see that \(s\) has to be sufficiently large in order to find a root \(h(s,t) = 0\) for \(t < s/d\).
Now let’s plot the dual bound \(D(s)\) for various \(\theta\) (this checks the outer root-finding).
theta <- c(0.5, 1, 2, 4) # theta values
s <- seq(48, 2000, length.out = 257) # s values
D <- sapply(theta, function(th)
sapply(s, function(s.)
dual_bound(s., d = d, pF = function(q) pPar(q, shape = th)))) # (s, theta) matrix
if(doPDF)
pdf(file = (file <- paste0("fig_worst_VaR_hom_dual_D_s_Par=",
paste0(theta, collapse="_"),"_d=",d,".pdf")),
width = 6, height = 6)
plot(s, D[,1], type = "l", ylim = range(D), col = "maroon3",
ylab = substitute("Dual bound D(s) for d ="~d.~"Par("*theta*") margins", list(d. = d)))
lines(s, D[,2], col = "darkorange2")
lines(s, D[,3], col = "royalblue3")
lines(s, D[,4], col = "black")
legend("topright", lty = rep(1,4),
col = c("maroon3", "darkorange2", "royalblue3", "black"),
bty = "n", legend = as.expression(lapply(1:4,
function(j) substitute(theta==j, list(j = theta[j])))))
if(doPDF) dev.off()
Check Wang_h_aux()
.
d <- 8 # dimension
alpha <- 0.99 # confidence level
c <- seq(0, (1-alpha)/d, length.out = 129) # domain of h
h.aux <- qrmtools:::Wang_h_aux(c, level = alpha, d = d, qF = qF2)
par(mar = c(5, 4+1, 4, 2) + 0.1) # increase space (for y axis label)
plot(c, h.aux, type = "l", xlab = "c (in initial interval)",
ylab = expression(frac(d-1,d)~{F^{-1}}(a[c])+frac(1,d)~{F^{-1}}(b[c])))
Check the objective function \(h(c)\) (that is, Wang_h()
with
numerical integration) on its domain. Note that \(h\) is used to determine the root in the
open interval \((0,(1-\alpha)/d)\) for
computing worst value-at-risk.
h <- sapply(c, function(c.) qrmtools:::Wang_h(c., level = alpha, d = d, qF = qF2))
if(doPDF)
pdf(file = (file <- paste0("fig_worst_VaR_",alpha,"_hom_Wang_h_Par=2_d=",d,"_num.pdf")),
width = 6, height = 6)
plot(c, h, type = "l", xlab = "c (in initial interval)",
ylab = substitute("h(c) for"~~alpha~"= 0.99 and d ="~d.~"Par(2) margins",
list(d. = d)))
abline(h = 0, lty = 2)
if(doPDF) dev.off()
Check the objective function \(h\) at the endpoints of its domain.
sapply(c(0, (1-alpha)/d), function(c.)
qrmtools:::Wang_h(c., level = alpha, d = d, qF = qF2)) # -Inf, 0
## [1] -Inf 0
\(-\infty\) is not a problem for
root finding (for \(\theta > 1\);
for \(\theta <= 1\) it is
NaN
, see below, and thus a problem!), but the \(0\) at the right endpoint is a problem.
method <- "Wang.Par" # this also holds for (the numerical) method = "Wang"
th <- 0.99
qrmtools:::Wang_h(0, level = alpha, d = d, method = method, shape = th) # NaN => uniroot() fails
## [1] NaN
## Note: Wang_h() is actually already NaN for c <= 1e-17
qrmtools:::Wang_h_aux(0, level = alpha, d = d, method = method, shape = th) # Inf
## [1] Inf
A proper initial interval \([c_l,u_l]\) with \(0<c_l<=c_u<(1-\alpha)/d\) (containing the root) is thus required. We have derived this in Hofert et al. (2017, ``Improved Algorithms for Computing Worst Value-at-Risk’’)).
Check the objective function \(h(c)\) (Wang_h()
without
numerical integration)
d <- dim # dimension
alpha <- 0.99 # confidence level
theta <- c(0.1, 0.5, 1, 5, 10, 50) # theta values
palette <- colorRampPalette(c("darkorange2", "maroon3", "royalblue3", "black"), space = "Lab")
cols <- palette(length(theta))
c <- seq(0, (1-alpha)/d, length.out = 2^13+1)
## => They all go further down to 0 if length.out is increased.
## Smaller theta thus corresponds to a larger derivative in the root
## Root-finding thus requires higher precision for smaller theta
h <- matrix(, nrow = length(c), ncol = length(theta))
for(j in 1:length(theta))
h[,j] <- sapply(c, function(c.)
qrmtools:::Wang_h(c., level = alpha, d = d, method = "Wang.Par", shape = theta[j]))
z <- h
z[z <= 0] <- NA # > 0 => makes log-scale possible
if(doPDF)
pdf(file = (file <- paste0("fig_worst_VaR_",alpha,"_hom_Wang_h_Par_d=",d,".pdf")),
width = 6, height = 6)
plot(NA, xlim = range(c), ylim = range(z, na.rm = TRUE), log = "y", yaxt = "n", xlab = "c",
ylab = substitute("h(c) for"~~alpha~"= 0.99, d ="~d.~"Par("*theta*") margins",
list(d. = d)))
eaxis(2)
for(k in 1:length(theta))
lines(c, z[,k], col = cols[k])
legend("topleft", bty = "n", lty = rep(1,length(theta)), col = cols,
legend = as.expression(lapply(1:length(theta),
function(k) substitute(theta==k, list(k = theta[k])))))
if(doPDF) dev.off()
After dealing with various numerical issues, we can now look at some example calculations of best/worst value-at-risk. We first plot value-at-risk as a function of \(\alpha\).
d <- dim # dimension
alpha <- 1-2^seq(-0.001, -10, length.out = 128) # confidence levels; concentrated near 1
theta <- c(0.1, 0.5, 1, 5, 10, 50) # theta values
VaR <- simplify2array(sapply(alpha, function(a)
sapply(theta, function(th) VaR_bounds_hom(a, d = d, method = "Wang.Par",
shape = th)), simplify = FALSE))
## => (best/worst VaR, theta, alpha)-matrix
if(doPDF)
pdf(file = (file <- paste0("fig_worst_VaR_hom_Wang_Par_d=",d,".pdf")),
width = 7, height = 7)
par(mar = c(5, 4+1, 4, 2) + 0.1) # increase space (for y axis label)
plot(NA, xlim = range(alpha), ylim = range(VaR), log = "xy", yaxt = "n",
xlab = expression(1-alpha),
ylab = as.expression(substitute(underline(VaR)[alpha](L^{"+"})~"(dashed) and"~bar(VaR)[alpha](L^{"+"})~
"(solid) for d ="~d.~"and Par("*theta*") margins",
list(d. = d))))
eaxis(2)
for(k in 1:length(theta)) {
lines(1-alpha, VaR[2,k,], col = cols[k]) # worst VaR
lines(1-alpha, VaR[1,k,], col = cols[k], lty = 2) # best VaR
}
legend("topright", bty = "n", lty = rep(1,length(theta)), col = cols,
legend = as.expression(lapply(1:length(theta),
function(k) substitute(theta==k, list(k = theta[k])))))
if(doPDF) dev.off()
Now consider best/worst value-at-risk as a function of \(d\).
d <- seq(2, 1002, by = 20) # dimensions
alpha <- 0.99 # confidence level
theta <- c(0.1, 0.5, 1, 5, 10, 50) # theta values
VaR <- simplify2array(sapply(d, function(d.)
sapply(theta, function(th) VaR_bounds_hom(alpha, d = d., method = "Wang.Par",
shape = th)), simplify = FALSE))
## => (best/worst VaR, theta, d)-matrix
if(doPDF)
pdf(file = (file <- paste0("fig_worst_VaR_hom_Wang_Par_alpha=",alpha,"_in_d.pdf")),
width = 7, height = 7)
par(mar = c(5, 4+1, 4, 2) + 0.1) # increase space (for y axis label)
plot(NA, xlim = range(d), ylim = range(VaR), log = "xy", yaxt = "n",
xlab = expression(d),
ylab = as.expression(substitute(underline(VaR)[a](L^{"+"})~"(dashed) and"~bar(VaR)[a](L^{"+"})~
"(solid) for Par("*theta*") margins",
list(a = alpha))))
eaxis(2)
for(k in 1:length(theta)) {
lines(d, VaR[2,k,], col = cols[k]) # worst VaR
lines(d, VaR[1,k,], col = cols[k], lty = 2) # best VaR
}
legend("topleft", bty = "n", lty = rep(1,length(theta)), col = cols,
legend = as.expression(lapply(1:length(theta),
function(k) substitute(theta==k, list(k = theta[k])))))
if(doPDF) dev.off()
We can also consider best/worst value-at-risk as a function of \(\theta\). Note that, as before, best value-at-risk is the same for all \(d\) here (depicted in a green dashed line below).
d <- c(2, 10, 20, 100, 200, 1000) # dimensions
theta <- 10^seq(-1, log(50, base = 10), length.out = 50) # theta values
alpha <- 0.99 # confidence level
VaR <- simplify2array(sapply(d, function(d.)
sapply(theta, function(th) VaR_bounds_hom(alpha, d = d., method = "Wang.Par",
shape = th)), simplify = FALSE))
## => (best/worst VaR, theta, d)-matrix
if(doPDF)
pdf(file = (file <- paste0("fig_worst_VaR_hom_Wang_Par_alpha=",alpha,"_in_theta.pdf")),
width = 6, height = 6)
par(mar = c(5, 4+1, 4, 2) + 0.1) # increase space (for y axis label)
plot(NA, xlim = range(theta), ylim = range(VaR), log = "xy", yaxt = "n",
xlab = expression(theta),
ylab = as.expression(substitute(underline(VaR)[a](L^{"+"})~"(dashed) and"~
bar(VaR)[a](L^{"+"})~"(solid) for Par("*theta*") margins",
list(a = alpha))))
eaxis(2)
bestVaR <- VaR[1,,1]
for(k in 1:length(d)) {
lines(theta, VaR[2,,k], col = cols[k]) # worst VaR
if(k >= 2) stopifnot(all.equal(bestVaR, VaR[1,,k]))
}
lines(theta, bestVaR, col = "darkgreen", lty = 2) # best VaR
legend("topright", bty = "n", lty = rep(1,length(d)), col = cols,
legend = as.expression(lapply(1:length(d),
function(k) substitute(d==k, list(k = d[k])))))
if(doPDF) dev.off()
Now consider a graphical comparison between the various methods. To stress the numerical challenges, let us include the following implementation of Wang’s approach (with various switches to highlight the individual numerical challenges) for the Pareto case.
## Initial interval for the root finding in case of worst VaR
init_interval <- function(alpha, d, shape, trafo = FALSE, adjusted = FALSE)
{
if(trafo) {
low <- if(shape == 1) {
d/2
} else {
(d-1)*(1+shape)/(d-1+shape)
}
up <- if(shape > 1) {
r <- (1+d/(shape-1))^shape
if(adjusted) 2*r else r
} else if(shape == 1) {
e <- exp(1)
(d+1)^(e/(e-1))
} else {
d*shape/(1-shape)+1
}
c(low, up)
} else {
low <- if(shape > 1) {
r <- (1-alpha)/((d/(shape-1)+1)^shape + d-1)
if(adjusted) r/2 else r
} else if(shape == 1) {
e <- exp(1)
(1-alpha)/((d+1)^(e/(e-1))+d-1)
} else {
r <- (1-shape)*(1-alpha)/d
if(adjusted) r/2 else r
}
up <- if(shape == 1) (1-alpha)/(3*d/2-1)
else (1-alpha)*(d-1+shape)/((d-1)*(2*shape+d))
c(low, up)
}
}
## Function to compute the best/worst value-at-risk in the homogeneous case with
## Par(theta) margins
VaR_hom_Par <- function(alpha, d, shape, method = c("worst", "best"),
trafo = FALSE, interval = NULL, adjusted = FALSE,
avoid.cancellation = FALSE, ...)
{
## Pareto quantile function
qF <- function(p) (1 - p)^(-1/shape) - 1
## Compute \bar{I}
Ibar <- function(a, b, alpha, d, shape)
{
if(shape == 1) log((1-a)/(1-b))/(b-a) - 1
else (shape/(1-shape))*((1-b)^(1-1/shape)-(1-a)^(1-1/shape))/(b-a) - 1
}
## Main
method <- match.arg(method)
switch(method,
"worst" = {
## Distinguish according to whether we optimize the auxiliary function
## on a transformed scale
h <- if(trafo) {
## Auxiliary function to find the root of on (1, Inf)
if(shape == 1) {
function(x) x^2 + x*(-d*log(x)+d-2)-(d-1)
} else {
function(x)
(d/(1-shape)-1)*x^(-1/shape + 1) - (d-1)*x^(-1/shape) + x - (d*shape/(1-shape) + 1)
}
} else {
## Auxiliary function to find the root of on (0, (1-alpha)/d)
function(c) {
a <- alpha+(d-1)*c
b <- 1-c
Ib <- if(c == (1-alpha)/d) { # Properly deal with limit c = (1-alpha)/d
((1-alpha)/d)^(-1/shape) - 1
} else {
Ibar(a = a, b = b, alpha = alpha, d = d, shape = shape)
}
Ib - (qF(a)*(d-1)/d + qF(b)/d)
}
}
## Do the optimization
if(is.null(interval)) interval <- init_interval(alpha, d, shape,
trafo = trafo, adjusted = adjusted)
c <- uniroot(h, interval = interval, ...)$root
if(trafo) # convert value back to the right scale (c-scale)
c <- (1-alpha)/(c+d-1)
if(avoid.cancellation) {
t1 <- (1-alpha)/c-(d-1)
d * ((c^(-1/shape)/d) * ((d-1)*t1^(-1/shape) + 1) - 1) # = qF(a)*(d-1) + qF(b)
} else {
a <- alpha+(d-1)*c
b <- 1-c
qF(a)*(d-1) + qF(b)
}
},
"best" = {
max((d-1)*0 + (1-alpha)^(-1/shape)-1, # Note: Typo in Wang, Peng, Yang (2013)
d*Ibar(a = 0, b = alpha, alpha = alpha, d = d, shape))
},
stop("Wrong 'method'"))
}
For the comparison, consider the following setup.
alpha <- 0.99 # confidence level
d <- dim # dimension
n.th <- 32 # number of thetas
th <- seq(0.2, 5, length.out = n.th) # thetas
qFs <- lapply(th, function(th.) {th.; function(p) qPar(p, shape = th.)}) # n.th-vector of Pareto quantile functions
pFs <- lapply(th, function(th.) {th.; function(q) pPar(q, shape = th.)}) # n.th-vector of Pareto dfs
N <- 1e4 # number of discretization points for RA(); N = 1e5 does not improve the situation
Now compute the values and plot them.
res <- matrix(, nrow = n.th, ncol = 7)
colnames(res) <- c("Wang", "straightforward", "transformed", "Wang.Par",
"dual", "RA.low", "RA.up")
pb <- txtProgressBar(max = n.th, style = if(isatty(stdout())) 3 else 1) # setup progress bar
for(i in seq_len(n.th)) {
## "Wang" (numerical integration with smaller uniroot() tolerance; still
## numerically critical -- we catch "the integral is probably divergent"-errors here)
Wang.num.res <- tryCatch(VaR_bounds_hom(alpha, d = d, qF = qFs[[i]])[2], error = function(e) e)
res[i,"Wang"] <- if(is(Wang.num.res, "simpleError")) NA else Wang.num.res
## Our straightforward implementation
res[i,"straightforward"] <- VaR_hom_Par(alpha, d = d, shape = th[i])
## Our straightforward implementation based on the transformed auxiliary function
res[i,"transformed"] <- VaR_hom_Par(alpha, d = d, shape = th[i], trafo = TRUE)
## "Wang.Par" (using a smaller uniroot() tolerance and adjusted initial interval)
res[i,"Wang.Par"] <- VaR_bounds_hom(alpha, d = d, method = "Wang.Par", shape = th[i])[2]
## "dual" (with uniroot()'s default tolerance)
res[i,"dual"] <- VaR_bounds_hom(alpha, d = d, method = "dual",
interval = crude_VaR_bounds(alpha, qF = qFs[[i]], d = d),
pF = pFs[[i]])[2]
## Rearrangement Algorithm
set.seed(271) # use the same random permutation for each theta
RA. <- RA(alpha, qF = rep(qFs[i], d), N = N)
res[i,"RA.low"] <- RA.$bounds[1]
res[i,"RA.up"] <- RA.$bounds[2]
## Progress
setTxtProgressBar(pb, i) # update progress bar
}
close(pb) # close progress bar
res. <- res/res[,"dual"] # standardize (by dual method)
ylim <- range(res., na.rm = TRUE)
if(doPDF)
pdf(file = (file <- paste0("fig_worst_VaR_",alpha,"_hom_comparison_d=",
d,"_N=",N,".pdf")), width = 7, height = 7)
par(mar = c(5, 4+1, 4, 2) + 0.1) # increase space (for y axis label)
plot(th, res.[,"Wang"], type = "l", ylim = ylim,
xlab = expression(theta), ylab = substitute("Standardized (by dual method)"~
bar(VaR)[0.99]~"for d ="~d.~"Par("*theta*") margins", list(d. = d)),
col = "gray60", lty = 2, lwd = 5.5) # Wang (with numerical integration)
lines(th, res.[,"straightforward"], col = "maroon3", lty = 1, lwd = 1) # still bad (although we have an initial interval)
lines(th, res.[,"transformed"], col = "black", lty = 1, lwd = 1) # okay
lines(th, res.[,"Wang.Par"], col = "royalblue3", lty = 2, lwd = 2.5) # Wang Pareto (wo num. integration)
lines(th, res.[,"RA.low"], col = "black", lty = 3, lwd = 1) # lower RA bound
lines(th, res.[,"RA.up"], col = "black", lty = 2, lwd = 1) # upper RA bound
legend("topright", bty = "n",
col = c("gray60", "maroon3", "black", "royalblue3", "black", "black"),
lty = c(2,1,1,2,3,2), lwd = c(5.5,1,1,2.5,1,1),
legend = c("Wang (num. int.)", "Wang Pareto (straightforward)",
"Wang Pareto (transformed)", "Wang Pareto (wo num. int.)",
"Lower RA bound", "Upper RA bound"))
if(doPDF) dev.off()
So what goes wrong with our straightforward implementation? Let’s
start with the obvious. As we can infer from the x-axis scale in the
plots of \(h\) above,
uniroot()
’s default tolerance
tol = .Machine$double.eps^0.25
(\(\approx 0.0001220703\) here) is too large
to determine the root accurately. In fact, if we choose a smaller
tol
, we have no problem.
tol <- 2.2204e-16
wVaR.tol <- sapply(th, function(th.)
VaR_hom_Par(alpha = alpha, d = d, shape = th., tol = tol))
plot(th, wVaR.tol/res[,"dual"], type = "l", ylim = ylim,
xlab = expression(theta), ylab = "Wang's approach (straightforward) but with smaller tol")
So is this the solution to all the problems? No. Consider the setup as before, where \(d\) is running (we catch the appearing errors here).
alpha <- 0.99 # confidence level
d <- seq(2, 1002, by = 20) # dimensions
theta <- c(0.1, 0.5, 1, 5, 10, 50) # theta values
VaR <- simplify2array(sapply(d, function(d.)
sapply(theta, function(th) {
res <- tryCatch(VaR_hom_Par(alpha, d = d., shape = th, tol = tol),
error = function(e) e)
if(is(res, "simpleError")) { warning(conditionMessage(res)); NA }
else res
}), simplify = FALSE))
By using warnings()
we see that we cannot even determine
the root in all cases as the function values are numerically
not of opposite sign at the endpoints of the theoretically
correct initial interval containing the root. We can solve this
(non-elegantly) by simply adjusting the lower bound.
VaR <- simplify2array(sapply(d, function(d.)
sapply(theta, function(th) VaR_hom_Par(alpha, d = d., shape = th, tol = tol, adjusted = TRUE)),
simplify = FALSE))
So we see that we need a smaller tolerance and to extend the initial interval in order to get reasonable results.
What about using the transformed auxiliary function directly? Is this a numerically more stable approach?
## [1] Inf
As we can see, there is a problem as well. To understand it, consider
the following minimized version of VaR_hom_Par()
(for
computing worst value-at-risk for \(\theta\neq
1\) only).
h <- function(x)
(d/(1-th)-1)*x^(-1/th + 1) - (d-1)*x^(-1/th) + x - (d*th/(1-th) + 1)
interval <- init_interval(alpha, d, th, trafo = TRUE)
x <- uniroot(h, interval = interval)$root
(c <- (1-alpha)/(x+d-1)) # convert back to c-scale
## [1] 1.869497e-31
## [1] Inf
As we can see, \(c>0\) is so
small that \(b = 1-c\) is numerically
equal to 1 and thus the Pareto quantile evaluated as \(\infty\). This cancellation can be avoided
by simplifying the terms (note that
qPar(1-c, shape = th) = c^(-1/th)-1
).
## [1] 162.5923
We can apply this to the variable combinations as before. Note that due to the partly extreme values of \(d\) and \(\theta\), we also need to adjust the range here in order for a root to be found.
d <- seq(2, 1002, by = 20) # dimensions
theta <- c(0.1, 0.5, 1, 5, 10, 50) # theta values
VaR. <- simplify2array(sapply(d, function(d.)
sapply(theta, function(th) VaR_hom_Par(alpha, d = d., shape = th,
trafo = TRUE, adjusted = TRUE,
avoid.cancellation = TRUE)),
simplify = FALSE))
We can now compare the values (based on the non-transformed/transformed auxiliary function) in a graph.
ylim <- range(VaR, VaR.)
if(doPDF)
pdf(file = (file <- paste0("fig_worst_VaR_",alpha,"_hom_comparison_num_problems.pdf")),
width = 7, height = 7)
par(mar = c(5, 4+1, 4, 2) + 0.1) # increase space (for y axis label)
plot(d, VaR[1,], type = "l", ylim = ylim, log = "y", yaxt = "n", xlab = "d",
ylab = as.expression(substitute(bar(VaR)[a](L^{"+"})~
"directly (solid) or with transformed h (dashed) for Par("*theta*") margins",
list(a = alpha))), col = cols[1])
eaxis(2)
for(k in 2:length(theta)) lines(d, VaR [k,], col = cols[k])
for(k in 1:length(theta)) lines(d, VaR.[k,], col = cols[k], lty = 2, lwd = 2)
legend("right", bty = "n",
col = cols, lty = rep(1, length(theta)),
legend = as.expression( lapply(1:length(theta), function(k)
substitute(theta==th, list(th = theta[k]))) ))
if(doPDF) dev.off()
We see that the results differ slightly if both \(d\) and \(\theta\) are large. It remains numerically challenging to appropriately compute worst value-at-risk in this case. However, both of the above approaches seem to work for dimensions even as large as \(10^5\).
The basic idea goes back to Iman and Conover (1982, “A
distribution-free approach to inducing rank correlation among input
variables”). Suppose we have 500 observations of two Pareto
distributions. This is the joint sample under independence (as we
independently draw the uniforms when constructing X
):
## Generate samples
n <- 1000 # sample size
set.seed(271) # set a seed (for reproducibility)
X <- cbind(qPar(runif(n), shape = 2), # Par(2) sample
qPar(runif(n), shape = 2.5)) # Par(2.5) sample
## Plot
if(doPDF)
pdf(file = (file <- paste0("fig_RA_motivation_indep.pdf")),
width = 6, height = 6)
plot(X, xlab = expression(X[1]), ylab = expression(X[2]))
if(doPDF) dev.off()
If we order the realizations in each column to be in increasing order
(so the smallest (largest) entry of the first column lies next to the
smallest (largest) entry of the second column of X
), we
obtain the following (comonotone) sample.
## Generate samples
X. <- apply(X, 2, sort)
## Plot
if(doPDF)
pdf(file = (file <- paste0("fig_RA_motivation_comonotone.pdf")),
width = 6, height = 6)
plot(X., xlab = expression(X[1]), ylab = expression(X[2]))
if(doPDF) dev.off()
We take away that column rearrangements don’t change the marginal distributions but the dependence between the two columns and thus the distributions of their sums. To see the latter (in terms of boxplots and kernel density estimates), consider:
## Samples of the sum of the components
## Note: Their mean is the same (unaffected by the dependence)
Xsum <- rowSums(X)
Xsum. <- rowSums(X.)
## Quick plot
boxplot(Xsum, Xsum., log = "y", names = c("Independence", "Comonotonicity"))
## Kernel density estimates
ran <- range(X, X.)
dXsum <- density(Xsum, from = ran[1], to = ran[2])
dXsum. <- density(Xsum., from = ran[1], to = ran[2])
## Plot
if(doPDF)
pdf(file = (file <- paste0("fig_RA_motivation_kernel_density_estimates.pdf")),
width = 6, height = 6)
plot(dXsum, xlim = c(0, 5), ylim = c(0, max(dXsum$y, dXsum.$y)),
xlab = "x", ylab = "Kernel density estimate at x", main = "", type = "l")
lines(dXsum., col = "royalblue3")
legend("topright", bty = "n", lty = 1,
col = c("black", "royalblue3"), legend = c("Independence", "Comonotonicity"))
if(doPDF) dev.off()
A straightforward (but inefficient) implementation of a basic
rearrange(, sample = FALSE, is.sorted = TRUE)
is the
following. Note that our implementation in qrmtools
allows
for much greater functionality and is faster.
basic_rearrange_worst_VaR <- function(X, tol = 0)
{
N <- nrow(X)
d <- ncol(X)
m.rs.old <- min(rowSums(X))
Y <- X
while (TRUE)
{
for(j in 1:d)
Y[,j] <- sort(Y[,j], decreasing = TRUE)[rank(rowSums(Y[,-j, drop = FALSE]),
ties.method = "first")]
Y.rs <- rowSums(Y)
m.rs.new <- min(Y.rs)
tol. <- abs((m.rs.new - m.rs.old)/m.rs.old)
tol.reached <- if(is.null(tol)) {
identical(Y, X)
} else { tol. <= tol }
if(tol.reached) {
break
} else {
m.rs.old <- m.rs.new
X <- Y
}
}
min(rowSums(Y))
}
We now compare this to the actual implementation
(rearrange(, sample = FALSE, is.sorted = TRUE)
). To this
end, we consider the following setup.
## Build the input matrix (for worst VaR)
alpha <- 0.99 # confidence level
N <- 2^9 # number of discretization points; 512 here
p <- alpha + (1-alpha)*0:(N-1)/N # probabilities for evaluating marginal quantile functions
if(FALSE)
d <- 2^(4:10) # 16, ..., 1024
d <- 2^(4:8) # dimensions 16, ..., 256 (to save time here)
res <- matrix(, nrow = length(d), ncol = 2) # matrix containing run time results
colnames(res) <- c("basic", "sophisticated")
## For each d, measure the run time (some randomness involved here...)
for(i in seq_along(d)) {
mar <- rep(list(qF2), d[i]) # marginal quantile functions
X <- sapply(mar, function(qF) qF(p)) # evaluate marginal quantile functions at p
res[i,] <- c(system.time(basic_rearrange_worst_VaR(X))[["elapsed"]], # basic version
system.time(rearrange(X, sample = FALSE, is.sorted = TRUE))[["elapsed"]])
}
Here’s a plot which shows the improvement in run time (in %). For larger dimensions \(d\), the improvement is very close to 100%.
rearrange()
acts on specific matricesTo see how rearrange()
actually proceeds, consider the
following example. Due to trace = TRUE
, the matrix is
printed after each column rearrangement. A “|” and “=” indicate whether
the respective column has changed or not, respectively, and the last two
printed columns provide the row sums over all other columns but the
current one, as well as the new updated row sums (over all columns)
after the rearrangement. We see that for tol = NULL
the
algorithm stops after the first time \(d\) (here: 3) consecutive rearrangements
left the matrix unchanged.
A <- matrix(c(1:4, 2*(1:4)-1, 2^(0:3)), ncol = 3)
rearrange(A, tol = NULL, sample = FALSE, is.sorted = TRUE, trace = TRUE)
##
## [1,] 1 1 1
## [2,] 2 3 2
## [3,] 3 5 4
## [4,] 4 7 8
## | -col sum
## [1,] 4 1 1 2 6
## [2,] 3 3 2 5 8
## [3,] 2 5 4 9 11
## [4,] 1 7 8 15 16
## | -col sum
## [1,] 4 5 1 5 10
## [2,] 3 7 2 5 12
## [3,] 2 3 4 6 9
## [4,] 1 1 8 9 10
## | -col sum
## [1,] 4 5 2 9 11
## [2,] 3 7 1 10 11
## [3,] 2 3 4 5 9
## [4,] 1 1 8 2 10
## | -col sum
## [1,] 3 5 2 7 10
## [2,] 2 7 1 8 10
## [3,] 4 3 4 7 11
## [4,] 1 1 8 9 10
## = -col sum
## [1,] 3 5 2 5 10
## [2,] 2 7 1 3 10
## [3,] 4 3 4 8 11
## [4,] 1 1 8 9 10
## = -col sum
## [1,] 3 5 2 8 10
## [2,] 2 7 1 9 10
## [3,] 4 3 4 7 11
## [4,] 1 1 8 2 10
## = -col sum
## [1,] 3 5 2 7 10
## [2,] 2 7 1 8 10
## [3,] 4 3 4 7 11
## [4,] 1 1 8 9 10
## $bound
## [1] 10
##
## $tol
## [1] 0
##
## $converged
## [1] TRUE
##
## $opt.row.sums
## [1] 6 9 9 10 10 10 10
##
## $X.rearranged
## [,1] [,2] [,3]
## [1,] 3 5 2
## [2,] 2 7 1
## [3,] 4 3 4
## [4,] 1 1 8
##
## $X.rearranged.opt.row
## [1] 2.000000 4.333333 3.666667
This is the highest possible minimal row sum and thus the optimally rearranged matrix. Let’s consider another example.
B <- matrix(rep(1:3, 3), ncol = 3)
rearrange(B, tol = NULL, sample = FALSE, is.sorted = TRUE, trace = TRUE)
##
## [1,] 1 1 1
## [2,] 2 2 2
## [3,] 3 3 3
## | -col sum
## [1,] 3 1 1 2 5
## [2,] 2 2 2 4 6
## [3,] 1 3 3 6 7
## = -col sum
## [1,] 3 1 1 4 5
## [2,] 2 2 2 4 6
## [3,] 1 3 3 4 7
## = -col sum
## [1,] 3 1 1 4 5
## [2,] 2 2 2 4 6
## [3,] 1 3 3 4 7
## = -col sum
## [1,] 3 1 1 2 5
## [2,] 2 2 2 4 6
## [3,] 1 3 3 6 7
## $bound
## [1] 5
##
## $tol
## [1] 0
##
## $converged
## [1] TRUE
##
## $opt.row.sums
## [1] 5 5 5 5
##
## $X.rearranged
## [,1] [,2] [,3]
## [1,] 3 1 1
## [2,] 2 2 2
## [3,] 1 3 3
##
## $X.rearranged.opt.row
## [1] 3 1 1
Here we do not reach an optimal rearrangement (6 would be a better
minimal row sum, attainable via the rearrangement
matrix(c(1,2,3, 2,3,1, 3,1,2), ncol = 3)
). It remains an
open problem how an efficient algorithm can avoid such a
case.
Whether rearrange()
converges or not depends to a large
degree on the underlying sorting algorithm. We found that
unstable sorting algorithms such as C’s qsort()
lead to rearrange()
not terminating on certain input
matrices.
One such example is the above matrix B
, for which the
2nd/3rd column are constantly swapped after the first column being
rearranged since the row sum of all others is (4,4,4)
and
termination then depends on how ranks are assigned to these three equal
numbers.
The current implementation is still not based on a stable
sorting algorithm (as this would largely compromise run time) but a) is
comparably fast and b) terminates on all input matrices. To see the
latter, consider the following example in which we investigate the
possible output matrices of rearrange()
on all input
matrices (obviously only for \(d = 3\)
due to run time; larger \(N\) can be
chosen but note that the effort is \((N!)^{d-1}\) here!). Also note that we do
not actually run the code here (to save run time)
## Create a (N!)^{d-1}-list of all possible input matrices (with integer
## elements) and 1:N as first column.
N <- 5 # chosen N (<= 6 due to extensive run time)
x_perm <- permn(N) # N!-long list containing all possible permutations of 1:N
N. <- factorial(N)
mat <- vector("list", length = N.^2)
cnt <- 0
for (i in 1:N.) { # double 'for' not time critical here
for (j in 1:N.){
cnt <- cnt+1
mat[[cnt]] <- cbind(1:N, x_perm[[i]], x_perm[[j]]) # matrix with 1:N in first column
}
}
## Rearrange all these (unique) matrices
N. <- factorial(N)^2
system.time(matRAlst <- lapply(1:N., function(i) {
## if(i %% (N./20) == 0) cat(round(100*i/N.),"% done!\n", sep = "");
rearrange(mat[[i]], tol = NULL, sample = FALSE)$X.rearranged
}))
## Go through all rearranged matrices, unique()ify them and count
uniqueLst <- list(matRAlst[[1]])
freq <- c(1)
pb <- txtProgressBar(max = length(matRAlst), style = if(isatty(stdout())) 3 else 1) # setup progress bar
for(i in 2:length(matRAlst))
{
bool <- sapply(1:length(uniqueLst),
function(k) { identical(matRAlst[[i]], uniqueLst[[k]]) }) # is uniqueLst[[k]] in matRAlst?
if(any(bool)) { # matRAlst[[i]] has been observed before
ind <- min(which(bool))
freq[ind] <- freq[ind] + 1 # increase frequency
} else { # matRAlst[[i]] has not been observed before
uniqueLst <- c(uniqueLst, matRAlst[i]) # append it to uniqueLst
freq <- c(freq, 1) # append its frequency
}
## Progress
setTxtProgressBar(pb, i) # update progress bar
}
close(pb) # close progress bar
## Distribution of (final) rearranged matrices
(length(uniqueLst)/N.) * 100 # % of unique rearranged matrices
p.mat <- freq/N. # their corresponding probabilities
plot(p.mat, type = "l", xlab = "Index of unique matrices", ylab = "Probability")
## => Matrices earlier on tend to appear more often (by construction!)
## Distribution of the minimal row sums (= worst VaR estimates)
VaR.est <- sapply(matRAlst, function(x) min(rowSums(x)))
plot(VaR.est)
table(VaR.est)/length(matRAlst) # probabilities
As we can see, rearrange()
terminated on all inputs.
Although it generated quite a different number of rearranged matrices,
the implied minimal row sums (so value-at-risk estimates) were mostly
the same (8 or 9, with about 0.125% also 7).
We now consider the task of computing best and worst value-at-risk (VaR) at confidence level 99% for a portfolio consisting of shares of each of the 20 constituents of the SMI from 2011-09-12 to 2012-03-28. For simplicity, the number of shares of each stock is assumed to be one over the numeric value of the stock price, so that the linearized loss is simply the sum of negative log-returns. The negative log-return of each stock is thus viewed as one loss and value-of-risk of the sum of these 20 losses is of interest. This fits precisely in the framework of the (Adaptive) Rearrangement Algorithm.
First, let’s compute the negative log-returns.
Now let’s fit GPDs to each margin, that is, each SMI constituent. As threshold, we simply choose the 80% quantile for each of margins (more sophisticated methods are available).
res <- vector("list", length = d)
names(res) <- colnames(L)
for(k in seq_len(d)) { # iterate over constituents
## Determine the threshold for constituent k
L. <- L[,k]
thres <- quantile(L., probs = 0.8, names = FALSE) # threshold
## Fit a GPD to the excesses for constituent k
is.exceed <- L. > thres
p.exceed <- mean(is.exceed) # estimated threshold exceedance probability
excess <- L.[is.exceed] - thres # excesses
fit <- fit_GPD_MLE(excess) # fit a GPD to the excesses
stopifnot(fit$converged == 0) # convergence check
shape <- fit$par[["shape"]] # fitted shape xi
scale <- fit$par[["scale"]] # fitted scale beta
stopifnot(is.numeric(shape), is.numeric(scale), scale > 0) # sanity check
## Update res
res[[k]] <- list(loss = L., excess = excess, p.exceed = p.exceed,
threshold = thres, shape = shape, scale = scale)
}
We should also quickly assess the goodness-of-fit of the fitted GPDs
if(doPDF)
pdf(file = (file <- paste0("fig_worst_VaR_bounds_application_GPD_fit_check.pdf")),
width = 6, height = 7)
## Graphical goodness-of-fit assessment
opar <- par(omd = c(0.02, 1, 0.02, 1), mar = rep(1, 4))
layout(matrix(1:d, ncol = 4, byrow = TRUE)) # (5, 4)-layout
for(k in 1:d) {
x <- sort(res[[k]]$excess)
qq_plot(x, FUN = function(p)
qGPD(p, shape = res[[k]]$shape, scale = res[[k]]$scale),
axes = FALSE, frame.plot = TRUE)
mtext("Fitted quantiles", side = 1, line = 0.5, cex = 0.6)
mtext("Empirical quantiles", side = 2, line = 0.5, cex = 0.6)
op <- par(usr = c(0, 1, 0, 1))
text(1/10, 9/10, names(res)[k], cex = 0.7, adj = 0)
y <- qGPD(ppoints(length(x)), shape = res[[k]]$shape, scale = res[[k]]$scale)
R2 <- sprintf("%.4f", cor(x, y)^2) # coefficient of determination R^2
text(9/10, 1/10, substitute(R^2 == r2, list(r2 = R2)), cex = 0.7, adj = 1)
par(op)
}
layout(1) # restore layout
par(opar) # restore plot parameters
if(doPDF) dev.off()