betahat.fun {emulator} | R Documentation |
Determines the MLE (least squares) regression coeffients for the specified regression basis.
The “.A
” form needs only A
(and not Ainv
),
thus removing the need to calculate a matrix inverse. Note that this
form is slower than the other if Ainv
is known in
advance, as solve(.,.)
is slow.
If Ainv
is not known in advance, the two forms seem to
perform similarly in the cases considered here and in the
goldstein
package.
betahat.fun(xold, Ainv, d, give.variance=FALSE, func) betahat.fun.A(xold, A, d, give.variance=FALSE, func)
xold |
Data frame, each line being the parameters of one run |
A |
Correlation matrix, typically provided by
corr.matrix() |
Ainv |
Inverse of the correlation matrix A |
d |
Vector of results at the points specified in xold |
give.variance |
Boolean, with TRUE meaning to return
information on the variance of betahat and
default FALSE meaning to return just the least squares estimator |
func |
Function to generate regression basis; defaults to regressor.basis |
Here, the strategy of using two separate functions, eg foo()
and foo.A()
, one of which inverts A
and one of which
uses notionally more efficient means. Compare the other
strategy in which a Boolean flag, use.Ainv
, has the same
effect. An example would be scales.likelihood()
.
Robin K. S. Hankin
data(toy) val <- toy H <- regressor.multi(val) d <- apply(H,1,function(x){sum((0:6)*x)}) fish <- rep(2,6) A <- corr.matrix(val,scales=fish, power=2) Ainv <- solve(A) # now add suitably correlated Gaussian noise: d <- as.vector(rmvnorm(n=1,mean=d, 0.1*A)) betahat.fun(val , Ainv , d) # should be close to c(0,1:6) # Now look at the variances: betahat.fun(val,Ainv,give.variance=TRUE, d) # now find the value of the a priori expectation (ie the regression # plane) at an unknown point: x.unknown <- rep(0.5 , 6) regressor.basis(x.unknown) %*% betahat.fun(val, Ainv, d) # compare the a-priori with the a-postiori: interpolant(x.unknown, d, val, Ainv,scales=fish) # Heh, it's the same! (of course it is, there is no error here!) # OK, put some error on the old observations: d.noisy <- as.vector(rmvnorm(n=1,mean=d,0.1*A)) # now compute the regression point: regressor.basis(x.unknown) %*% betahat.fun(val, Ainv, d.noisy) # and compare with the output of interpolant(): interpolant(x.unknown, d.noisy, val, Ainv, scales=fish) # there is a difference! # now try a basis function that has superfluous degrees of freedom. # we need a bigger dataset. Try 100: val <- latin.hypercube(100,6) colnames(val) <- letters[1:6] d <- apply(val,1,function(x){sum((1:6)*x)}) A <- corr.matrix(val,scales=rep(1,6), power=1.5) Ainv <- solve(A) betahat.fun(val, Ainv, d, func=function(x){c(1,x,x^2)}) # should be c(0:6 ,rep(0,6). The zeroes should be zero exactly # because the original function didn't include any squares. ## And finally a sanity check: betahat.fun(val, Ainv, d, func=function(x){c(1,x,x^2)})-betahat.fun.A(val, A, d, func=function(x){c(1,x,x^2)})