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| Syntax |
relative(function y, function F, function dF, ...M, xin, xlow, xup, eps, maxstp, mitr, ...xout, vout, Rout, gout, lamout, Cout) |
| See Also | nlsq , dnlsq |
M 2
----- | y(i) - F(i, x) |
minimize > N log(v ) + ------------------ with respect to (v, x)
----- i i v
i = 1 i
such that xlow < x < xup
j j j
(v, x)
if there is a true, but unknown, value for (v, x)
such that for each i
the elements of the data vector y(i) are independent
normally distributed with mean F(i, x) and variance
v(i).
This estimator and the algorithm used by relative are
discussed in the paper
A relative weighting method for estimating parameters and
variances in multiple data sets, (1997) volume 22, pages 119-135 of
Computational Statistics and Data Analysis.
relative is true if convergence is achieved
and false otherwise.
(i)
Returns the value y(i) as a column vector with the
same type as xin,
where i is an integer scalar between
1 and M.
(i, x)
Returns the value F(i, x) as a column vector with the
same type and dimension as y(i),
where i is an integer scalar between
1 and M
and x is a vector with the same type and
dimension as xin.
(i, x)
The derivatives of F(i, x) with respect to x
as a column vector with the same type and row dimension as y(i).
The column dimension of dF(i, x) is equal to the row dimension of
xin. The (k,j)-th element of the return value dF(i, x)
is the partial derivative of the k-th element of F(i, x)
with respect to the j-th element of x.
xlow(j) < x(j) < xup(j)
must hold for each j.
The column vector xlow
has the same type and dimension as xin and specifies
the lower limits for x
in the extended least squares problem.
The column vector xup
has the same type and dimension as xin and specifies
the upper limits for x
in the extended least squares problem.
The column vector eps
has the same type and dimension as xin and
convergence is accepted when the absolute change in the j-th
element of x is less than or equal eps(j)
for all j.
All of the elements of eps must be greater than 0.
The vector maxstp
has the same type and dimension as xin
and the value maxstp(j)
is the maximum absolute change allowed in the j-th
element if x between iterations of the algorithm.
All of the elements of maxstp must be strictly greater than 0.
The integer scalar mitr specifies the maximum number of
iterations of the relative weighting algorithm to execute before giving
up on convergence.
x
at the k-th iteration of the algorithm.
The input value of vout
has no effect.
The output value of vout has the same type as xin,
its row dimension is M, and its column dimension is equal to
the number of iterations.
The (i,k)-th element of the output is the variance estimate
that corresponds to the i-th measurement set and the k-th iteration.
The input value of Rout has no effect.
The output value of Rout is a row vector with the same
type as xin and column dimension
equal to the number of iterations.
The k-th element of Rout is the reduced
objective function value that
corresponds to the k-th iteration.
(The reduced objective is the sum of the logarithm
terms normalized by the number of data points.
Up to a constant, this is equal to the original extended least squares
objective at the value of v that is optimal for the current
x.)
The input value of gout has no effect.
The output value of gout is a matrix with
row dimension equal to the row dimension of xin
and column dimension equal to the number of iterations.
The (j,k)-th element of gout
is the partial derivative of the reduced
objective function with respect to the
j-th element of x at the k-th iteration.
The input value of lamout has no effect.
The output value of lamout is a row vector with the same type
as xin and lamout(k) is the line search parameter
that corresponds to the difference between the k-th column of
xout and the k+1-th column of xout.
The input value of Cout has no effect.
The output value of Cout is a
square matrix
with the same number of rows as xin.
The matrix Cout is an estimate of the covariance of the
optimal value for x as an estimate of the true parameter vector.
2
x + x t + x t
1 2 3
x which are all 1.
The true value of x and v are
{1, 1, 1} and {.01, .0001} respectively.
These values are approximated by
the values of x and v that solve the
minimization problem.
clear
#
# simulate data set 1
t1 = seq(100) / 100.
y1 = 1. + t1 + t1^2 + .1 * snormal(100, 1)
#
# simulate data set 2
t2 = seq(10) / 10.
y2 = 1. + t2 + t2^2 + .01 * snormal(10, 1)
#
function y(i) begin
global y1, y2
if i == 1 then ...
return y1
else return y2
end
#
function F(i, x) begin
global t1, t2
if i == 1 then ...
return x(1) + x(2) * t1 + x(3) * t1^2
else return x(1) + x(2) * t2 + x(3) * t2^2
end
#
function dF(i, x) begin
global t1, t2
if i == 1 then ...
return [ones(rowdim(t1), 1), t1, t1^2]
else return [ones(rowdim(t2), 1), t2, t2^2]
end
#
M = 2
xin = {.5, .5, .5}
xlow = {0., 0., 0.}
xup = {10., 10., 10.}
eps = {1e-3, 1e-3, 1e-3}
maxstp = {1., 1., 1.}
mitr = 10
xout = novalue
vout = novalue
Rout = novalue
gout = novalue
lamout = novalue
Cout = novalue
relative(function y, function F, function dF, ...
M, xin, xlow, xup, eps, maxstp, mitr, ...
xout, vout, Rout, gout, lamout, Cout )
#
print "minimizing x =", xout.col(coldim(xout))
print "minimizing v =", vout.col(coldim(vout))