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## Copyright (C) 2009 Esteban Cervetto <estebancster@gmail.com>
##
## Octave is free software; you can redistribute it and/or modify it
## under the terms of the GNU General Public License as published by
## the Free Software Foundation; either version 3 of the License, or (at
## your option) any later version.
##
## Octave is distributed in the hope that it will be useful, but
## WITHOUT ANY WARRANTY; without even the implied warranty of
## MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
## General Public License for more details.
##
## You should have received a copy of the GNU General Public License
## along with Octave; see the file COPYING. If not, see
## <http://www.gnu.org/licenses/>.
## -*- texinfo -*-
## @deftypefn {Function File} {@var{ultimate} =} ultimateld (@var{s}, @var{quotas})
## Calculate the ultimate values by the Loss Development (Chainladder) method.
##
## @var{s} is a mxn matrix that contains the run-off triangle, where m is the number of accident-years
## and n is the number of periods to final development. @var{s} may contain u = m-n complete years.
## The value @var{s}(i,k), 1<=i<=m, 0<=k<=n-1 represents the cumulative losses from accident-period i
## settled with a delay of at most k years.
## The values @var{s}(i,k) with i + k > m must be zero because is future time.
## The 1xn vector @var{quotas} is a set of cumulative quotas calculated by some method.
##
## The LD method asumes that exists a development pattern on the individual factors.
## This means that the identity
##
## @verbatim
## E[S(i,k) ]
## LDI(k) = -------------
## E[S(i,k-1) ]
##
## @end verbatim
## holds for all k = @{0, @dots{}, n-1@} and for all i = @{1, @dots{}, m@}.
##
## This follows to
##
## @verbatim
## l=n-1 1
## quotas(k) = II -------
## l=k+1 LDI(l)
## @end verbatim
##
## and the @var{ultimate} value is
##
## @verbatim
## ULTIMATE(i) = S(i,n-i-1) / QUOTAS(n-i-1)
## @end verbatim
##
## @seealso {bferguson, quotaad, quotapanning}
## @end deftypefn
function ultimate = ultimateld (S, quotas)
[m,n] = size (S); #triangle with m years (i=1,2,u,...u+1,u+2,....m) and n periods (k=0,1,2,...n-1)
u = m - n; #rows of the upper square
S = fliplr(triu(fliplr(S),-u)); #ensure S is triangular
if (size(quotas) ~= [1,n])
error(strcat("quotas must be of size [1,",num2str(n),"]" ));
end
#calculate the ultimate value
if (u==0)
ultimate = flipud(diag(fliplr(S))) ./ quotas';
else
ultimate = [(flipud(diag(fliplr(S),-u)) ./ quotas')', S(1:u,n)]';
end
ultimate = flipud(ultimate);
end