[R] Bootstrapping

[Skorp]

hello, 

I am trying to implement the bootstrapping to a set of insurance claim data
in triangular form using the ChainLadder package. I want to obtain the
prediction errors of the reserve estimate using the result from
bootstrapping, here is the output: 

>BootChainLadder(Triangle = incr2cum(data), R = 1000, process.distr =
"gamma")

BootChainLadder(Triangle = incr2cum(data), R = 1000, process.distr =
"gamma")

    Latest        Mean Ultimate  Mean IBNR      SD IBNR     IBNR 75%     
IBNR 95%
2   36,241             36,241                           0                   
0                           0                      0
3   47,380              47,619                         239               
608                      341            1,449
4   45,877             47,062                        1,185             1,072                  
1,701          3,149
5   70,696             74,224                         3,528            
1,850                   4,613          6,683
6   83,797             90,749                         6,952            
2,558                   8,580        11,386
7  105,933           116,061                     10,128            3,173                 
12,212       15,464
8  169,428           186,824                     17,396            4,783                 
20,491       25,602
9  158,634           176,267                     17,633            4,462                 
20,446       25,362
10 123,794         139,182                    15,388            3,955                 
17,783       22,368
11  85,531           111,495                   25,964             4,697                
29,147       33,862

                   Totals
Latest:           927,311
Mean Ultimate:  1,025,723
Mean IBNR:         98,412
SD IBNR:           18,950
Total IBNR 75%:   111,257
Total IBNR 95%:   128,807

Also, would really appreciate if anyone could explain the abbreviations like
SD IBNR, does it stand for Standard Deviation for IBNR? If so, then can i
use this to find the prediction errors?

Thanks.

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