n <- 5
lower <- -1
upper <- 3
df <- 4
corr <- diag(5)
corr[lower.tri(corr)] <- 0.5
delta <- rep(0, 5)
prob <- pmvt(lower=lower, upper=upper, delta=delta, df=df, corr=corr)
print(prob)
pmvt(lower=-Inf, upper=3, df = 3, sigma = 1) == pt(3, 3)
# Example from R News paper (original by Edwards and Berry, 1987)
n <- c(26, 24, 20, 33, 32)
V <- diag(1/n)
df <- 130
C <- c(1,1,1,0,0,-1,0,0,1,0,0,-1,0,0,1,0,0,0,-1,-1,0,0,-1,0,0)
C <- matrix(C, ncol=5)
### scale matrix
cv <- C %*% V %*% t(C)
### correlation matrix
dv <- t(1/sqrt(diag(cv)))
cr <- cv * (t(dv) %*% dv)
delta <- rep(0,5)
myfct <- function(q, alpha) {
lower <- rep(-q, ncol(cv))
upper <- rep(q, ncol(cv))
pmvt(lower=lower, upper=upper, delta=delta, df=df,
corr=cr, abseps=0.0001) - alpha
}
### uniroot for this simple problem
round(uniroot(myfct, lower=1, upper=5, alpha=0.95)$root, 3)
# compare pmvt and pmvnorm for large df:
a <- pmvnorm(lower=-Inf, upper=1, mean=rep(0, 5), corr=diag(5))
b <- pmvt(lower=-Inf, upper=1, delta=rep(0, 5), df=300,
corr=diag(5))
a
b
stopifnot(round(a, 2) == round(b, 2))
# correlation and scale matrix
a <- pmvt(lower=-Inf, upper=2, delta=rep(0,5), df=3,
sigma = diag(5)*2)
b <- pmvt(lower=-Inf, upper=2/sqrt(2), delta=rep(0,5),
df=3, corr=diag(5))
attributes(a) <- NULL
attributes(b) <- NULL
a
b
stopifnot(all.equal(round(a,3) , round(b, 3)))
a <- pmvt(0, 1,df=10)
attributes(a) <- NULL
b <- pt(1, df=10) - pt(0, df=10)
stopifnot(all.equal(round(a,10) , round(b, 10)))
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