Reserving – introduction I.
One of the most important function of
actuaries.
Every year-end closure there is a requirement
to calculate reserves for future liabilities and
grant the realistic P&L.
Now there is a lot of type of reserves with
special calculation rules.
From 2016 there was a change in this
viewpoint: it will be new types of reserves.
Insurance mathematics II. lecture
Reserving – introduction II.
Legal regulation: 43/2015
Regulation of Government
Requirement: all reserves have to
calculate per business lines (not
per products)
Reserves are important because of
measure also (total reserves are
more than 2.000 billion HUF at
2015).
Insurance mathematics II. lecture
Unearned premium reserve I.
Example:
we take out a household policy at 01.12.2016
with 24.000 HUF yearly premium. We agree
yearly payment frequency and we pay total
premium in December. What
is the realistic

P&L situation at 31.12.2016?
30.11.2017
01.12.2016 31.12.2016
2.000
HUF
22.000 HUF
Insurance mathematics II. lecture
Unearned premium reserve II.
Suppose that no any claim related to this policy. What is
the real P&L figure in 2016 and 2017?
If we would book whole premium for 2016 then we
would not book any premium for 2017, but we are
in risk during 11 months! But we got really the
whole premium in 2016.
The solution:
1.
We book whole premium for 2016.
2.
We calculate unearned premium reserve for 2016.
It means that this part of premium will be the offset
of risk in 2017. The value of UPR is:
11
 24.000  22.000 HUF
12
Insurance mathematics II. lecture
Unearned premium reserve III.
Generalizing:
We note
d the premium due to payment frequency;
T the duration of the payment which continues
into next year;
K the duration till date of year-closure, then:
T K
UPR  d 
T
Insurance mathematics II. lecture
Unearned premium reserve IV.
Open problems with this definition:
-the lengths of months are not equal (in the example the
real UPR is not 21.962 HUF ?);
- the risk can be not equal in the whole period (example:
fleets);
- based on Hungarian regulation it should not reserve
UPR more then 1 year;
- difference between Hungarian and international
regulation. The base of UPR is the yearly premium in
international standard, but the payment regarding
frequency in Hungarian regulation.
Insurance mathematics II. lecture
Mathematical reserve
Now we are dealing with mathematical
reserve in non-life insurance, i.e. related to
liability insurance and accident insurance.
(Apart from these types there are
mathematical reserve regarding life
insurance and health insurance also.)
Insurance mathematics II. lecture
Mathematical reserve in
liability insurance I.
This type is used generally if the insurer has to pay
annuity based on liability insurance (typically in Hungary
MTPL). The total future annuity payments are estimated
with the next formula:
n 1
l xk
1
MR  

 ( S k  (1  d )  b)
k
(1  i)
k 0 l x
where:
-x is an age of annuitant;
- lx comes from mortality table (number of x ages);
- n is the unexpired years regarding annuity;
- Sk is the yearly annuity in the k-th year (with taxes);
Insurance mathematics II. lecture
Mathematical reserve in
liability insurance II.
- i is the technical interest (the yield which the
insurer will reach till end of annuity with
guarantee; now based on the regulation the
maximum of technical interest is 0 according to
liability insurance);
- d,b are cost factors; typically one of them is 0.
If there is an L limit in the policy related to
annuity the formula will change as follows:
MR  min( L, MR)
*
Insurance mathematics II. lecture
Mortality table example
Age
lx male
lx female
Age
0
100,000
100,000
51
1
99,647
99,614
99,605
99,593
99,584
99,574
99,565
99,556
99,547
99,537
99,527
99,515
99,501
99,484
99,462
99,435
99,403
99,365
99,323
99,273
99,220
99,164
99,106
99,046
98,985
98,924
98,862
98,801
98,740
98,680
98,620
98,560
98,500
98,439
98,377
98,312
98,245
98,175
98,101
98,019
97,928
97,827
97,715
97,593
97,458
97,307
97,137
96,942
96,719
96,471
99,652
99,621
99,603
99,587
99,574
99,564
99,557
99,550
99,543
99,535
99,527
99,517
99,508
99,497
99,485
99,469
99,453
99,435
99,416
99,397
99,377
99,356
99,336
99,315
99,294
99,273
99,250
99,225
99,199
99,171
99,140
99,106
99,067
99,025
98,979
98,927
98,868
98,802
98,728
98,646
98,562
98,462
98,350
98,231
98,108
97,981
97,854
97,715
97,554
97,384
52
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
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
95
96
97
98
99
100
101
lx male
96,200
95,912
95,609
95,295
94,966
94,613
94,236
93,820
93,364
92,866
92,314
91,711
91,057
90,368
89,644
88,853
87,997
87,057
85,883
84,467
82,718
80,516
77,722
74,173
70,288
66,273
62,164
58,001
53,900
49,845
45,822
41,822
37,840
33,879
29,950
26,072
22,281
18,619
15,145
11,924
9,026
6,513
4,437
2,819
1,646
867
404
161
53
14
-
lx female
97,203
97,006
96,796
96,581
96,353
96,103
95,835
95,556
95,269
94,963
94,633
94,269
93,848
93,392
92,903
92,362
91,788
91,184
90,526
89,737
88,796
87,640
86,185
84,305
81,895
79,202
76,249
73,066
69,722
66,195
62,466
58,517
54,339
49,931
45,303
40,484
35,523
30,492
25,492
20,646
16,094
11,983
8,442
5,566
3,386
1,870
920
393
141
41
-
Mathematical reserve in
liability insurance III.
The formula is uncomplicated, but the estimation of
parameters is not easy:
- in the most cases the insurer knows just the initial
annuity. In the long run it can be a lot of change regarding
the health status of annuitant, inflation, etc.;
-the mortality of annuitant is different comparing the total
mortality rate (but there are no separate mortality table of
annuitant). It can cause unexpected profit or loss.
Because of above reasons the insurer usually tends to
pay lump sum (typically 50-60% of virtual mathematical
reserve).
Insurance mathematics II. lecture
Mathematical reserve in
accidental insurance
The formula is similar as in liability insurance, but
the change of annuity is not so frequent –
because usually the measure of annuity is exact
in the policy.
Insurance mathematics II. lecture
Claims reserve I.
Reasons:
- lag in reporting of claims;
- lag in payment of claims.
There are two types of claims reserve:
-If the insurer has known the claims but the claims are
not totally paid, there are Outstanding Claims Reserve
(OS Reserve);
-If the insurer has not yet known the claims, it can be
used IBNR reserve (incurred but not reported).
Insurance mathematics II. lecture
Claims reserve II.
There are two different possible approach:
- separate assumption for OS and IBNR reserve or
- together estimating with statistical methods.
Now in Hungary it is used the separate approach
generally, but because of Solvency II. in the future the
second approach will come into view also.
The measure of lag is characteristic for products, for
example the CASCO and accident products have usually
higher speed run-off, and MTPL and other liability
products have usually slower run-off.
Insurance mathematics II. lecture
Claims reserve III.
For assumption it can consider inflation and the yield of
reserves also.
It is interesting and generally unanswerable question
how is the most useful splitting of portfolio for the
assumption:
The target is to find homogenous groups of risks. It can
be per products or per business lines or sometimes in
one product there is useful to further splitting (for
example, in MTPL splitting between annuities and nonannuities, or splitting big claims and non-big claims).
Insurance mathematics II. lecture
Outstanding claims reserve
and claims handling reserve
In separate OS reserve assumption methods the actuaries have
no a lot of tasks. The claim experts have experience how much
can be the ‘best estimate’ of the different claim event.
The actuaries have just two tasks: calculate claims handling
reserve with the next formula and calculate so-called ‚initial’ claims
reserve.
where
CHP
CHR 
 CLR
CP
CHR – claims handling reserve;
CHP – claims handling payment in current year;
CP – claims payment in current year;
CLR – claims reserve (OS or IBNR).
Insurance mathematics II. lecture
IBNR reserve I.
There are a lot of different algorithm to evaluate IBNR
reserve. Before the detailed description of these methods
it can be useful to define several basic ideas.
Reporting/payment year
…………..
X
X
1, n
1,1
…………..
Accident year
X i, j
X n ,n
Run-off triangles
X i , j means the total amount of
the claims which are
occurred in i-th year and
reported/paid in j-th year
Insurance mathematics II. lecture
IBNR reserve II.
Development year
X 1,1
…………..
Accident year
X n ,1
…………..
X i, j
X 1,n
Lagging triangles
X i , j means the total amount of claims
which are occurred in i-th year and
reported/paid in (i+j-1)-th year
Insurance mathematics II. lecture
IBNR reserve III.
Development year
X 1,1
…………..
Accident year
X n ,1
…………..
X i, j
X 1,n
Cumulated triangles
X i , j means the total amount of claims
which are occurred in i-th year and
reported/paid till (i+j-1)-th year
Insurance mathematics II. lecture
IBNR reserve IV.
The cumulated triangle is complete, if there is
no any reporting/paying after n-th year (difficult to
say).
It is possible to make run-off triangles for number
of claims also.
Hungarian regulation requires for IBNR
calculation just using run-off triangles.
For assumption the cumulated triangle will be the
basic usually.
Insurance mathematics II. lecture
IBNR reserve V.
Denote
X i , n  the claims which are reported/paid after n-th
year regarding claims occurred in i-th year.
Our target is estimating the next formula (for each i):
IBNRi  X i ,n   X i ,n 1i
Our best estimate is as follows:
Xˆ i ,n   E ( X i ,n  X u ,v , u  v  n  1)
Our problem is that in practice usually we do not know covariance
and common distribution of claims. That is why we simplify in the
methods which we are using for calculation of IBNR.
Insurance mathematics II. lecture
Methods of IBNR calculation
Grossing Up method I.
Example:
1
2011
2
3
4
5
Development year
223; 311; 252; 127; 29
Accident year
2012 254; 378; 249; 153
2013 312; 411; 276
2014 359; 435
Lagging triangle
2015 384
Insurance mathematics II. lecture
Methods of IBNR calculation
Grossing Up method II.
Example:
1
2
3
4
5
Development year
2011 223; 534; 786; 913; 942
Accident year
2012 254; 632; 881; 1034
2013 312; 723; 999
2014 359; 794
Cumulated triangle
2015 384
Insurance mathematics II. lecture
Methods of IBNR calculation
Grossing Up method III.
Example:
Is the cumulated triangle complete? No, we have
data from earlier years as follows:
Year
Claims until 5th
year
Total
Ratio (5th
year/Total)
2007
780
830
93,98%
2008
810
890
91,01%
2009
800
860
93,02%
Total
2390
2580
92,64%
Insurance mathematics II. lecture
Methods of IBNR calculation
Grossing Up method IV.
Example:
942
 1017
0,9264
Assumption for 2015:
The base of assumption will be 2015 as next table shows:
1. year
2. year
3. year
4. year
5. year
Total
2015
223
534
786
913
942
1017
Ratio
21,9%
52,5%
77,3%
89,8%
92,64%
100%
We assume that the run-off of next years will be equal as 2015.
Insurance mathematics II. lecture
Methods of IBNR calculation
Grossing Up method V.
Example:
Occurring year
5
Total
IBNR
2011 223; 534; 786; 913; 942
1017
75
2012 254; 632; 881; 1034
1152
118
2013 312; 723; 999
1292
293
2014 359; 794
1512
718
2015 384
1751
1367
Total
2571
1
2
3
4
Insurance mathematics II. lecture
Methods of IBNR calculation
Grossing Up method VI.
Generalizing:
1. Based on earlier year we estimate
dn 
X 1,n
X 1,n 
If we have no any data from earlier year we can use data from similar
products or OS reserves.
2. Further factors:
d n 1 
X 1,n 1
Xˆ
1, n 
d n2 
X 1,n  2
Xˆ
…. d1 
1, n 
3. Ultimate payment estimation:
X 2,n1
ˆ
X 2, n  
d n1
X
X
Xˆ 3,n  3,n2 …. Xˆ n,n   n,1
d n2
d1
Insurance mathematics II. lecture
X 1,1
Xˆ
1, n 
Methods of IBNR calculation
Grossing Up method VII.
4. Reserve assumption:
Vi  Xˆ i ,n  X i ,n1i ; 1  i  n
n
V   Vi
i 1
The above calculation is the basic version, but there are some
modified possibility of this method.
Insurance mathematics II. lecture
Methods of IBNR calculation
Modified Grossing Up methods I.
1. version: If we have data related to earlier year splitting per year
then we can calculate more exact the d factors.
X
Let
and di (1), di (2),..., di (k ); 1  i  n
d i (1)  1,i 1 ; 1  i  n
ˆ
X
1, n 
the other experience d factors. Then the ultimate used factors:
di 
d i (1)  d i (1)  d i (2)  ...  d i (k )
; 1 i  n
k 1
Insurance mathematics II. lecture
Methods of IBNR calculation
Modified Grossing Up methods II.
Example:
1. year
2. year
3. year
4. year
5. year
Total
2007
Ratio (%)
170
20,5%
422
50,8%
660
79,5%
750
90,4%
780
94%
830
100%
2008
Ratio (%)
140
15,7%
435
48,9%
680
76,4%
782
87,9%
810
91%
890
100%
2009
Ratio (%)
132
15,3%
428
49,8%
670
77,9%
780
90,7%
803
93%
860
100%
2015
Ratio (%)
223
21,9%
534
52,5%
786
77,3%
913
89,8%
942
92,7%
1017
100%
50,5%
77,8%
89,7%
92,7%
100%
Average 18,4%
ratio
Insurance mathematics II. lecture
Methods of IBNR calculation
Modified Grossing Up methods III.
Example:
Year
Total (ultimate)
payment
IBNR
2011
1017
75
2012
1153
119
2013
1284
285
2014
1572
778
2015
2090
1706
Total
2964
Insurance mathematics II. lecture
Methods of IBNR calculation
Modified Grossing Up methods IV.
2. version: If we have data related to earlier year splitting per year
then we can calculate more exact the d factors.
X
Let
and di (1), di (2),..., di (k ); 1  i  n
d i (1)  1,i 1 ; 1  i  n
ˆ
X
1, n 
the other experience d factors. Then the ultimate used factors:
di  min( di (1); di (1); di (2);...; di (k )); 1  i  n
Insurance mathematics II. lecture
Methods of IBNR calculation
Modified Grossing Up methods V.
Example:
1. year
2. year
3. year
4. year
5. year
Total
2007
Ratio (%)
170
20,5%
422
50,8%
660
79,5%
750
90,4%
780
94%
830
100%
2008
Ratio (%)
140
15,7%
435
48,9%
680
76,4%
782
87,9%
810
91%
890
100%
2009
Ratio (%)
132
15,3%
428
49,8%
670
77,9%
780
90,7%
803
93%
860
100%
2015
Ratio (%)
223
21,9%
534
52,5%
786
77,3%
913
89,8%
942
92,7%
1017
100%
48,9%
76,4%
87,9%
91%
100%
Minimum 15,3%
ratio
Insurance mathematics II. lecture
Methods of IBNR calculation
Modified Grossing Up methods VI.
Example:
Year
Total (ultimate)
payment
IBNR
2011
1035
93
2012
1177
143
2013
1308
309
2014
1625
831
2015
2502
2118
Total
3493
Insurance mathematics II. lecture
Methods of IBNR calculation
Modified Grossing Up methods VII.
Xˆ 1,n ; di (1); 1  i  n
3. version: We estimate
it we judge ultimate payment for 2.year:
as in the 1. version. After
X
Xˆ 2,n  2,n1
d n1 (1)
With this result we define the d factors and estimate ultimate payment as follows:
d n  2 (2) 
X 2,n  2
X
; ...; d1 (2)  2,1
Xˆ 2,n 
Xˆ 2,n 
d n  2 (1)  d n  2 (2)
2
X 3,n2

d n2
d n2 
Xˆ 3,n
After it we continue this process till each d factors and payments will be calculated.
Insurance mathematics II. lecture
Methods of IBNR calculation
Modified Grossing Up methods VIII.
Example:
1. year
2. year
3. year
4. year
5. year
Total
2011
Ratio (%)
223
21,9%
534
52,5%
786
77,3%
913
89,8%
942
92,6%
1017
100%
2012
Ratio (%)
254
22,1%
632
54,9%
881
76,5%
1034
89,8%
2013
Ratio (%)
312
24%
723
55,6%
999
76,9%
2014
Ratio (%)
359
24,6%
794
54,3%
2015
Ratio (%)
384
23,1%
1152
100%
1299
100%
1461
100%
1659
100%
Insurance mathematics II. lecture
Methods of IBNR calculation
Modified Grossing Up methods IX.
Example:
Year
Total (ultimate)
payment
IBNR
2011
1017
75
2012
1152
118
2013
1299
300
2014
1461
667
2015
1659
1275
Total
2434
Insurance mathematics II. lecture
Methods of IBNR calculation
Modified Grossing Up methods X.
Xˆ 1,n ; di (1); 1  i  n
4. version: We estimate
it we judge ultimate payment for 2.year:
as in 1. version. After
X
Xˆ 2,n  2,n1
d n1 (1)
With this result we define the d factors and estimate ultimate payment as follows:
d n  2 (2) 
X 2,n  2
X
; ...; d1 (2)  2,1
Xˆ 2,n 
Xˆ 2,n 
d n2  min( d n2 (1); d n2 (2))
X 3,n2
ˆ
X 3,n 
d n2
After it we continue this process till each d factors and payments will be calculated.
Insurance mathematics II. lecture
Methods of IBNR calculation
Modified Grossing Up methods XI.
Example:
1. year
2. year
3. year
4. year
5. year
Total
2011
Ratio (%)
223
21,9%
534
52,5%
786
77,3%
913
89,8%
942
92,6%
1017
100%
2012
Ratio (%)
254
22,1%
632
54,9%
881
76,5%
1034
89,8%
2013
Ratio (%)
312
24%
723
55,6%
999
76,5%
2014
Ratio (%)
359
24,6%
794
52,5%
2015
Ratio (%)
384
21,9%
1152
100%
1306
100%
1512
100%
1753
100%
Insurance mathematics II. lecture
Methods of IBNR calculation
Modified Grossing Up methods XII.
Example:
Year
Total (ultimate)
payment
IBNR
2011
1017
75
2012
1152
118
2013
1306
307
2014
1512
718
2015
1753
1369
Total
2587
Insurance mathematics II. lecture
Methods of IBNR calculation
Link ratio methods I.
We suppose that
c j (i ) 
X i , j 1
X i, j
X
1. Determining cn  1  i ,n 
dn
X i ,n
ratio does not depend on significantly for i
is similar then in the Grossing Up method.
(with experience of earlier years or OS reserve). Other
cj
factors will be
defined as function of actual c j (i ) . With choosing different function will
be defined the different version of link ratio method.
2. Ultimate payment and IBNR reserve estimation:
Xˆ 1,n  cn  X 1,n
V1  Xˆ 1,n  X 1,n  (cn  1)  X 1,n
Insurance mathematics II. lecture
Methods of IBNR calculation
Link ratio methods II.
2. Ultimate payment and IBNR reserve estimation continued:
Xˆ 2,n  cn  cn1  X 2,n1
V2  Xˆ 2,n  X 2,n1  (cn  cn1  1)  X 2,n1
…
Xˆ n,n  cn  cn1  ...  c1  X n,1
…
Basic version:
1. modification:
Vn  Xˆ n,n  X n,1  (cn  cn1  ...  c1  1)  X n,1
c j  c j (1); 1  j  n  1
cj 
c j (1)  c j (2)  ...  c j (n  j )
n j
Insurance mathematics II. lecture
; 1  j  n 1
Methods of IBNR calculation
Link ratio methods III.
2. modification:
c j  max( c j (1); c j (2);...c j (n  j )); 1  j  n  1
3. modification:
cj 
1, j  c j (1)   2, j  c j (2)  ...   n j , j  c j (n  j )
n j
; 1  j  n 1
In 3. modification with special α factors we will get the most popular
‘chain-ladder’ method.
Insurance mathematics II. lecture
Methods of IBNR calculation
Chain ladder method I.
This is the most popular process for IBNR estimation.
cj 

X 1, j  c j (1)  X 2, j  c j (2)  ...  X n  j , j  c j (n  j )
X 1, j  X 2, j  ...  X n  j , j
X 1, j 1  X 2, j 1  ...  X n  j , j 1
X 1, j  X 2, j  ...  X n  j , j
; 1  j  n 1
Insurance mathematics II. lecture

Methods of IBNR calculation
Chain ladder method II.
Example:
Year
Total (ultimate)
payment
IBNR
2011
1017
75
2012
1152
118
2013
1300
301
2014
1458
664
2015
1648
1264
Total
2420
Insurance mathematics II. lecture
Methods of IBNR calculation
Naive loss ratio method
In the next methods we are using premium data also (not just claim
data).
We suppose that the ultimate loss of i-th year will be the 1  i
-th part of the premium.
Then the reserve can be calculated as follows:
Vi  Pi  (1  i )  X i ,n 1i
n
V   Vi
, where
Pi signs the earned premium of i-th year.
i 1
The disadvantage of this method is that IBNR reserve is independent
of actual claim data. Starting company without any own claim data
can use this method.
Insurance mathematics II. lecture
Methods of IBNR calculation
Bornhuetter-Ferguson method I.
This method combines the naive loss ratio and grossing up (or link
ratio) methods.
1. We calculate the ultimate loss payment with naive claim ratio
methods:
Xˆ i ,n  Pi  (1  i )
2. For calculating development factors we are using grossing
up (or link ratio method):
X i ,n
X i ,n 1
X i ,1
1
1
1
dn  
; d n1 

;... d1 

cn X i ,n
cn  cn1 X i ,n
cn  cn1  ...c1 X i ,n
Insurance mathematics II. lecture
Methods of IBNR calculation
Bornhuetter-Ferguson method II.
3. The reserves will be estimated as follows:
1 ˆ
ˆ
V1  (1  d n )  X 1,n   (1  )  X 1,n 
cn
1
V2  (1  d n 1 )  Xˆ 2,n   (1 
)  Xˆ 2,n 
cn  cn 1
….
Vn  (1  d1 )  Xˆ n ,n   (1 
n
1
cn  cn 1  ...  c1
)  Xˆ n,n 
V   Vi
i 1
Insurance mathematics II. lecture
Methods of IBNR calculation
Separation method I.
In this method we do not use cumulated triangles, but we are using
lagging triangles. The used triangles have to be complete.
We suppose that
where
i j
X i , j  c  ni  rj  i  j ; 1  i, j  n
ni signs the number of claims in the i-th year (known),
signs an inflation, c is the average claim amount.
There are two types of this method:
-
arithmetic;
-
geometric.
Now we consider detailed the arithmetic version.
Insurance mathematics II. lecture
Methods of IBNR calculation
Separation method II.
We suppose that
n
r
j 1
j
1
It means that r j signs the expected proportion of claims
development year j without inflation effect. In this case
k
1
is the inflation factor according to first year.
Then we define the next formula:
Bi , j 
X i, j
ni
 c  rj  i  j
And we use the lagging triangle for these new elements as follows
Insurance mathematics II. lecture
Methods of IBNR calculation
Separation method III.
Development year
...
c  r2  2
c  r1  2
c  r2  3 ... c  rn1  n
c  rn  n
………..
Accident year
c  r1  1
c  r1  n
After we define diagonal sums as follows:
Insurance mathematics II. lecture
Methods of IBNR calculation
Separation method IV.
d1  c  r1  1
d2  c  r1  2  c  r2  2  c  (r1  r2 )  2
….
d n1  c  r1  n1  ...  c  rn1  n1  c  (r1  ...  rn1 )  n1  c  (1  rn )  n1
d n  c  r1  n  ...  c  rn  n  c  (r1  ...  rn )  n  c  n
Insurance mathematics II. lecture
Methods of IBNR calculation
Separation method V.
From the last equation we will get the first assumptions:
n
c  ̂n   B j ,n1 j
j 1
rˆn 
B1,n
̂n
c  ̂n
After that we could calculate recursively the next factors as follows:
d
c  ˆn 1  n 1
1  rˆn
c  ˆn  2 
d n2
1  rˆn  rˆn 1
B1,n 1  B2,n 1
rˆn 1 
c  ˆ  c  ˆ
n
rˆn2 
n 1
B1,n2  B2,n2  B3,n2
c  ˆ  c  ˆ  c  ˆ
n
n 1
n2
…
Insurance mathematics II. lecture
Methods of IBNR calculation
Separation method VI.
Then we will assume the future inflations:
ˆn1; ˆn2 ;...; ˆ2n
After we fill the remaining part of the lagging triangle:
Xˆ i , j  ni  rˆj  cˆi  j ; n  2  i  j  2n
The assumption for the IBNR reserve will be the next:
Vi  ni 
n
 rˆ  cˆ
j  n 1i
j
i j
n
V   Vi
i 1
Insurance mathematics II. lecture
Methods of IBNR calculation
Separation method VII.
Example:
1
2
3
Occurring year
2013 375; 250; 188
2014
273; 341
2015
324
Insurance mathematics II. lecture
Methods of IBNR calculation
Separation method VIII.
We have information about the number of claims also:
n1  8; n2  14,67; n3  17
Insurance mathematics II. lecture
Methods of IBNR calculation
Separation method IX.
Than we can calculate the missing part of the triangle:
The IBNR reserve will be the sum of these elements:
Insurance mathematics II. lecture
Methods of IBNR calculation
Separation method X.
The geometric separation method will be calculated similar, just the
beginning assumption is different:
n
r 1
i
i 1
In the estimator formulas we use the products of diagonal instead of sums of
diagonal.
Insurance mathematics II. lecture
Methods of IBNR calculation
Which method do will use in practice?
There are no one ‘true’ method which is the most useful in each case
(products, business lines). What can the actuary do in practice?
The most useful possibility is calculating IBNR reserve with methods as
much as possible, back-testing the results, and year by year fining the
method. There is important to consider result of earlier IBNR as follows:
RIBNR  OpIBNR  ClaimPayment  ClOS  ClIBNR
Insurance mathematics II. lecture
Bonus reserve I.
Insurance mathematics II. lecture
Bonus reserve II.
Insurance mathematics II. lecture
Bonus reserve III.
Then we will get for bonus reserve:
If the premium refund is affected when the policy is claim-free then the
formula will be easier as follows:
Insurance mathematics II. lecture
Other reserves I.
Bonus reserve for life insurance
This reserve is affected just in life insurance (it has to be calculated in case
of plus yield return to policyholder).
Equalization reserve
This reserve can be calculated for those business line which are profitable
(it has to be calculated, but the measure of reserve can be 0).
If the result of business line is negative then the reserve has to be reduced
with the measure of negative result.
Large claim reserve
This reserve has to be calculated in case of huge risks (for example:
nuclear power station).
Insurance mathematics II. lecture
Other reserves II.
Cancellation reserve
Based on own experience it has to be reserved the expected cancelled part
of premium in the future. It is important to take into consideration the
unearned premium reserve (it is prohibited to reserve for unearned part of
premium) and in life insurance the saving part of the premium (it is
prohibited to reserve for the saving part of the premium).
Example:
If we have 10.000 HUF written premium for a policy which has 1 month
delay at year-end, and based on our experience 5% the probability that
premiums delaying 1 month won’t be paid then the cancellation reserve will
be 10.000·5%=500 HUF. But if there is 2.500 UPR related to this policy
then we have just (10.000-2.500)·5%=375 HUF cancellation reserve. (If the
UPR more than 10.000 HUF then it is prohibited to reserve.)
Insurance mathematics II. lecture
Other reserves III.
Insurance mathematics II. lecture