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CPS2051 Mentje G. et al.
               (2006)) and the single loss approximation (SLA) method (see e.g. Böcker and
               Klüppelberg (2005)). For a detailed comparison of numerical approximation
               methods, the interested reader is referred to de Jongh et al. (2016).
               2.2 Scenario modelling
                   The  Basel  Accord  (see  BCBS  196,  (2011))  suggests  the  use  of  scenario
               assessments to improve severity distribution estimation. BCBS refers to three
               types  of  scenarios  namely  the  individual  scenario  approach,  the  interval
               approach  and  the  percentile  approach.  In  the  remainder  of  the  paper  we
               discuss a percentile approach suggested by de Jongh et al. (2015), which we
               believe  is  the  most  practical  of  the  existing  approaches  available  in  the
               literature.
                   As discussed in de Jongh et al. (2015), we advocate the use of the so-called
               one-in- year scenario approach. In the one-in- years scenario approach, the
               experts are asked to answer the question: ‘What loss level   is expected to be
                                                                        
               exceeded once every  years?’. Popular choices for  vary between 5 and 100
               and often 3 values for  are used. As an example, one bank used  = 7, 20 and
               100 and motivated the first choice as the number of years of historical data
               available to them. In this case the largest loss in the historical data may serve
               as a guide for choosing   since this loss level has been reached once in 7
                                         7
               years. If the experts judge that the future will be better than the past, they may
               want to provide a lower assessment for   than the largest loss experienced
                                                        7
               so far. If they foresee deterioration they may judge that a higher assessment
               is more appropriate. The other choices of  are selected in order to obtain a
               scenario spread within the range that one can expect reasonable improvement
               in accuracy from the experts’ inputs. Of course the choice of  = 100 may be
               questionable because judgments on a one-inhundred years loss level are likely
               to fall outside many of the experts’ experience. In the banking environment,
               they may take additional guidance from external data of similar banks which
               in effect amplifies the number of years for which historical data are available.
                   If the annual loss frequency is () distributed and the true underlying
               severity distribution is , and if the experts are of oracle quality in the sense of
               actually knowing  and , then the assessments provided should be
                                                        1
                                            =  −1  (1 −  ).                       (1)
                                           
                                                       
                   To see this, let   denote the number of loss events experienced in  years
                                  
               and let   denote the number of these that are actually greater than   . Then
                        
                                                                                   
                 ~() and the conditional distribution of    given    is binomial with
                                                                        
                                                               
                 
               parameters    and  1  −   = P(X  ≥   ) =  1  −  ( )  with  ~  and   =
                                                     
                                                                   
                                         
                             
                                                                                       
                ( ). Therefore  = [(  |  )] =  [  (1  −  )] =  (1  −  (  )) .
                                                                    
                                                 
                                                                                      
                   
                                   
                                                            
                                             
                                                             1
               Requiring that  =1, yields (1) and  = 1  −   .
                                                    
                                 
                                                             
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