[5b02c4]: inst / ultimatecc.m Maximize Restore History

Download this file

ultimatecc.m    95 lines (86 with data), 3.2 kB

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
## 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} =} ultimatecc (@var{s}, @var{v}, @var{quotas})
## Calculate the ultimate values by the Cape Cod 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.
## @var{v} is an mx1 vector of known volume measures (like premiums or the number of contracts).
## @var{quotas} is an 1xn vector of cumulatives quotas.
##
## The Cape Cod method asumes that exists a development pattern on the cumulative quotas (Q).
## This means that the identity
##
## @verbatim
## E[S(i,k) ]
## Q(k) = -------------
## E[S(i,n) ]
## @end verbatim
##
## holds for all k = @{0, @dots{}, n-1@} and for all i = @{1, @dots{}, m@}.
##
## Also, the Cape Cod Method asumes the existence of a value "H" in a way that satisfy
##
## @verbatim
## S(i,n)
## H = E [------]
## V(i)
## @end verbatim
##
## holds for all i = @{1, @dots{}, m@}.
## H is called the Cape Cod loss ratio and it can be prove this value is
##
## @verbatim
## j=n-1
## E S(j,n-j)
## j=0
## quotas(k) = -----------------
## j=n-1
## E Q(n-j)V(j)
## j=0
## @end verbatim
##
## @var{ultimate} returns a row column with the ultimate values. Their values are:
##
## @verbatim
## ULTIMATE(i) = H * V(i)
## @end verbatim
##
## @seealso {bferguson}
## @end deftypefn
function ultimate = ultimatecc (S,V,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(V) ~= [m,1])
usage(strcat("volume V must be of size [",num2str(m),",1]" ));
end
if (size(quotas) ~= [1,n])
usage("quotas must be of dimension [1,n]");
end
# CapeCods K K = S(i+k = n)/quotas*V
if (u==0)
K = sum(diag(fliplr(S))')/ (fliplr(quotas)*V);
else
K = sum([diag(fliplr(S),-u)' S(1:u,n)])/ (fliplr([quotas ones(u)])*V);
end
#ultimate value
ultimate = K * V;
end