View source: R/lsConstrain.fit.R
lsConstrain.fit | R Documentation |
Find the vector z that solves:
min{ (x - z)'inv(S)(x - z); Az <= b },
where x is an input vector, S its covariance matrix, A is a matrix of known contrasts, and b is a vector of known constraint constants.
lsConstrain.fit(x, b, s, a, iflag, itmax=4000, eps=1e-06, eps2=1e-06)
x |
vector of length n |
b |
vector of length k, containing constraint constants |
s |
matrix of dim n x n, the covariance matrix for vector x |
a |
matrix of dim k x n, for the contraints |
iflag |
vector of length k; an item = 0 if inequality constraint, 1 if equality constraint |
itmax |
scalar for number of max interations |
eps |
scalar of accuracy for convergence |
eps2 |
scalar to determine close to zero |
List with the following components:
itmax: (defined above)
eps: (defined above)
eps2: (defined above)
iflag: (defined above)
xkt: vector of length k, for the Kuhn-Tucker coefficients.
iter: number of completed iterations.
supdif: greatest difference between estimates across a full cycle
ifault: error indicator: 0 = no error 1 = itmax exceeded 3 = invalid constraint function for some row ASA'=0.
a: (defined above)
call: function call
x.init: input vector x.
x.final: the vector "z" that solves the equation (see z in description).
s: (defind above)
min.dist: the minimum value of the function (see description).
Wollan PC, Dykstra RL. Minimizing inequality constrained mahalanobis distances. Applied Statistics Algorithm AS 225 (1987).
# An simulation example with linear regression with 3 beta's,
# where we have the contraints:
#
# b[1] > 0
# b[2] - b[1] < 0
# b[3] < 0
set.seed(111)
n <- 100
x <- rep(1:3,rep(n,3))
x <- 1*outer(x,1:3,"==")
beta <- c(2,1,1)
y <- x%*%beta + rnorm(nrow(x))
fit <- lm(y ~-1 + x)
s <- solve( t(x) %*% x )
bhat <- fit$coef
a <- rbind(c(-1, 0, 0),
c(-1, 1, 0),
c( 0, 0, 1))
# View expected constraints (3rd not met):
a %*% bhat
# [,1]
# [1,] -1.8506811
# [2,] -0.9543320
# [3,] 0.8590827
b <- rep(0, nrow(a))
iflag <- rep(0,length(b))
save <- lsConstrain.fit(x=bhat,b=b, s=s, a=a, iflag=iflag, itmax=500,
eps=1e-6, eps2=1e-6)
save
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