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2 Case study 2: shared liabilities – who pays what?

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7.2 Case study 2: shared liabilities – who pays what?

333

Pareto distribution, and (3) a lognormal distribution. We will also compare our
theoretical results with those of simulations. We will also study the corresponding results for aggregate losses, examining the expected payouts by the three
parties involved.

7.2.1 Case 1 – exponential losses
Suppose, for illustration purposes, that the individual loss variable X has an
exponential distribution with mean 1 (we are again taking the expected loss as
our monetary unit for convenience). We note that in this case Pr(X > k) = e−k ,
for k ≥ 0.
We can find expressions for E[C], E[Y] and E[Z] in turn quite easily by
evaluating the integrals which define these expectations. But it is easier to find
E[Z] first and then deduce the others. We have
E[Z] =

{x − (D + M)}e−x dx =

D+M

= e−(D+M)

ue−(u+D+M) du

0

ue−u = e−(D+M) .

0

Alternatively, we can introduce the reinsurance claim variable Z ∗ , where
Z ∗ ≡ Z | (X > D + M) ≡ X − (D + M) | (X > D + M).
Then in this exponential case we have (by appealing to the lack of memory
property (2.17)) that Z ∗ has the same distribution as X, with E[Z ∗ ] = 1. Using
the relationship
E[Z] = E[Z ∗ ] Pr(X > D + M),
we then have again E[Z] = e−(D+M) .
Now consider the variable Y + Z (≡ X − C), which is the amount attributable
to the direct insurer and reinsurer together and is easily defined as follows:
Y +Z =

0
if X ≤ D
X − D if X > D.

The variable Y + Z has the same structure as Z, with D + M replaced by D.
Since E[Z] = e−(D+M) , it follows immediately that E[Y + Z] = e−D , and hence
that E[Y] = e−D − e−(D+M) .
Finally, since 1 = E[X] = E[C] + E[Y] + E[Z], we find E[C] = 1 − e−D .
To sum up:
E[C] = 1 − e−D , E[Y] = e−D − e−(D+M) and E[Z] = e−(D+M) .

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Case studies

Note The fact that the expressions for E[C], E[Y] and E[Z] are the same as
those for Pr(X ≤ D), Pr(D < X ≤ D + M) and Pr(X > D + M), respectively, is
a feature of this particular case, in which X ∼ Exp(1).
We now find the expected sizes of losses conditional on the loss having a
value less than D, between D and D + M, and greater than D + M. We know
that E[Z ∗ ] = E[X − (D + M) | (X > D + M)] = 1, and so we have
E[X | (X > D + M)] = D + M + 1.
Similarly, we have E[X − D | (X > D)] = 1, and so E[X | (X > D)] = D + 1.
Now, from
E[X] = E[X | (X ≤ D)] Pr(X ≤ D) + E[X | (X > D)] Pr(X > D),
we have 1 = E[X | (X ≤ D)](1 − e−D ) + (D + 1)e−D , and so
E[X | (X ≤ D)] = 1 −

De−D
.
1 − e−D

Finally, from
E[X] = E[X | (X ≤ D)] Pr(X ≤ D)
+ E[X | (D < X ≤ D + M)] Pr(D < X ≤ D + M)
+ E[X | (X > D + M)] Pr(X > D + M),
we have
De−D
(1 − e−D )
1 − e−D
+ E[X | (D < X ≤ D + M)](e−D − e−(D+M) )

1= 1−

+ (D + M + 1)e−(D+M) ,
from which we get
E[X | (D < X ≤ D + M)] = 1 + D −

Me−M
.
1 − e−M

To sum up:
De−D
,
1 − e−D
Me−M
,
E[X | (D < X ≤ D + M)] = 1 + D −
1 − e−M
E[X | (X > D + M)] = D + M + 1.
E[X | (X ≤ D)] = 1 −

Some numerical values are presented in Table 7.7. We note the obvious fact that
Pr(X ≤ D) increases as the deductible D increases. For fixed retention M, both

7.2 Case study 2: shared liabilities – who pays what?

335

Table 7.7. Probabilities for exponential losses

Pr(X ≤ D)
Pr(D < X ≤ D + M)
Pr(X > D + M)

M=3
D = 0.1

M=3
D = 0.3

M=5
D = 0.1

M=5
D = 0.3

0.0952
0.8598
0.0450

0.2592
0.7039
0.0369

0.0952
0.8987
0.0061

0.2592
0.7358
0.0050

Table 7.8. Expected values for exponential losses

E[C]
E[Y]
E[Z]
E[XLO]
E[XMED]
E[XHI]

M=3
D = 0.1

M=3
D = 0.3

M=5
D = 0.1

M=5
D = 0.3

0.0952
0.8598
0.0450
0.0492
0.9428
4.1

0.2592
0.7039
0.0369
0.1425
1.1428
4.3

0.0952
0.8987
0.0061
0.0492
1.0661
6.1

0.2592
0.7358
0.0050
0.1425
1.2661
6.3

Table 7.9. Expected values (£) when one monetary unit is £5000

E[C]
E[Y]
E[Z]
E[XLO]
E[XMED]
E[XHI]

M = 15 000
D = 500

M = 15 000
D = 1500

M = 25 000
D = 500

M = 25 000
D = 1500

476
4299
225
246
4714
20 500

1296
3520
184
713
5714
21 500

476
4494
30
246
5330
30 500

1296
3679
25
713
6330
31 500

Pr(D < X ≤ D + M) and Pr(X > D + M) decrease as the deductible increases,
while, for fixed deductible, Pr(D < X ≤ D + M) increases and Pr(X > D + M)
decreases as the retention increases.
Let E[XLO], E[XMED] and E[XHI] represent E[X | (X ≤ D)], E[X | (D <
X ≤ D + M)] and E[X | X > D + M], respectively; see Table 7.8. In the case
that our unit (E[X]) represents £5000, we have Table 7.9 (entries in £).
We note that, as we would expect, both E[C] and E[X | (X ≤ D)] increase as
the deductible D increases. For fixed retention M, both E[Y] and E[Z] decrease
as the deductible increases, while for fixed deductible E[Y] increases and E[Z]
decreases as the retention increases.

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Case studies
Table 7.10. Theoretical results for exponential losses

X
C
Y
Z
Z∗
X : 0 < X ≤ 0.3
X : 0.3 < X ≤ 3.3
X : X > 3.3

Number

Expected number

Mean

10 000
10 000
10 000
10 000

369
2592
7039
369

1
0.2592
0.7039
0.0369
1
0.1425
1.1428
4.3

Table 7.11. Simulation results for exponential losses

x
c
y
z
zstar
x: 0 < x <= 0.3
x: 0.3 < x <= 3.3
x: x > 3.3

Number

Min.

Median

Mean

Max.

10 000
10 000
10 000
10 000
394
2606
7000
394

0.000
0.000
0.000
0.000
0.004
0.000
0.300
3.304

0.699
0.300
0.399
0.000
0.662
0.142
0.948
3.962

1.013
0.259
0.713
0.040
1.023
0.144
1.150
4.323

9.336
0.300
3.000
6.036
6.036
0.300
3.299
9.336

To illustrate these results, a simulation of 10 000 losses was carried out
(using R), using X ∼ Exp(1), D = 0.3 and M = 3. The loss vector x was
manipulated to produce vectors c, y, z and zstar containing the values indicated by these vectors’ names, and the data were then summarised. Vectors
containing the losses less than 0.3, between 0.3 and 3.3, and greater than 3.3
were also constructed and summarised.
The corresponding theoretical results are given in Table 7.10. The simulation
results are given in Table 7.11.
Note The simple R code used to produce the required vectors of data was as
follows (xlo, xmed and xhi are the parts of the x-vector such that 0 < x≤ 0.3,
0.3 < x ≤ 3.3 and x> 3.3, respectively):
d
m
x
c
z

=
=
=
=
=

0.3
3
rexp(10000,1)
pmin(x,d)
#’parallel min’ c[i]=min(x[i],d[i])
pmax(0,x-d-m) #’parallel max’

7.2 Case study 2: shared liabilities – who pays what?

337

y = x - c - z
xlo = x[x <= d]
x2 = x[x > d]
xmed = x2[x2 <= d+m]
xhi = x[x > d+m]
zstar = xhi - (d+m)

We extend the situation now to one in which we study an aggregate loss
variable S , where S has a compound Poisson distribution with claim rate λ and
individual claim size variable X ∼ Exp(1). Each loss arises on a policy with
deductible D in force. In addition, an excess of loss reinsurance arrangement
is in place whereby the direct insurer pays a maximum of M on each loss.
Let S C , S I and S R represent the aggregate amounts of the losses paid by the
policyholder (the customer), the direct insurer and the reinsurer, respectively.
Then we know the following:
S ∼ CP(λ, F X ) and E[S ] = λE[X] = λ,
S C ∼ CP(λ, FC ) and E[S C ] = λE[C] = λ(1 − e−D ),
S I ∼ CP(λ, FY ) and E[S I ] = λE[Y] = λ{e−D − e−(D+M) },
S R ∼ CP(λ, FZ ) and E[S R ] = λE[Z] = λe−(D+M) .
To illustrate these results, a simulation of 10 000 aggregate losses was carried
out (using R), using X ∼ Exp(1), λ = 100, D = 0.3 and M = 3. In this case
the expected payouts are simply 100 times what they are for individual losses,
that is E[S ] = 100 × 1 = 100, E[S C ] = 100(1 − e−0.3 ) = 25.92, E[S I ] =
100(e−0.3 − e−3.3 ) = 70.39, E[S R ] = 100e−3.3 = 3.688. The simulation results
are given in Table 7.12. We note that the number of losses (N ∼ Poi(100))
ranges from 63 to 139, and that the means of the simulated data are in close
agreement with the theoretical results.
Note

The simulation was carried out by executing a function, here called

cpsimre1 and previously stored as a text file, as given below. The output

Table 7.12. Further simulation results for exponential losses

N
S
SC
SI
SR

Number

Min.

Median

Mean

Max.

10 000
10 000
10 000
10 000
10 000

63
49.65
15.66
30.47
0.000

100.0
99.39
25.80
69.84
3.189

99.81
99.73
25.86
70.18
3.700

139
156.5
36.97
115.1
17.89

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Case studies

from the function is an R object – a list containing five vectors – called
payout.list. The command
case1sim = cpsimre1(10000,100,1,0.3,3)

executes the function once, carrying out 10 000 simulations of the number of
claims and resulting aggregate payout variables in the case λ = 100, μ = 1,
D = 0.3 and M = 3. The results (the list of five vectors) are held in the object
case1sim and can be retrieved using, for example,
summary(case1sim[[3]])

to obtain a summary of the values of S C . The text defining the function is
cpsimre1 = function(n,lam,mu,d,m){
xagg = cagg = yagg = zagg = numcl = (1:n)*0
for (i in 1:n){nc = rpois(1,lam)
if(nc == 0){
numcl[i] = xagg[i] = cagg[i] = yagg[i] = zagg[i] = 0}
else{x = rexp(nc,1/mu)
c = pmin(x,d)
z = pmax(x - d - m,0)
y = x - c - z
numcl[i] = nc
xagg[i] = sum(x)
cagg[i] = sum(c)
yagg[i] = sum(y)
zagg[i] = sum(z) } }
payout.list = list(numcl,xagg,cagg,yagg,zagg) }

7.2.2 Case 2 – Pareto losses
Suppose now that the individual loss variable X has a Pareto distribution, X ∼
Pa(α, λ). This distribution is fat-tailed, whereas the exponential is thin-tailed
(see Chapter 2). For the distribution to have finite mean we require α > 1,
and for it to have finite variance we require α > 2; it should be noted that, in
many applications of the Pareto distribution as a model for insurance losses,
the value of α estimated from data is less than 2.
Before proceeding, the reader should verify that, for X ∼ Pa(α, λ) with α > 1
and for b > 0,

b

x f (x)dx =

x
b

αλα
λ
dx =
α+1
λ+b
(λ + x)

α

λ + bα
.
α−1

7.2 Case study 2: shared liabilities – who pays what?

339

As in Case 1 we will take the mean of the distribution of X to be our monetary unit, that is E[X] = λ/(α − 1) = 1, and so we will take α = λ + 1. We then
have
λ λ+1
Pr(X > k) =
, for k ≥ 0.
λ+k
As in Case 1, we can find expressions for E[C], E[Y] and E[Z] in turn quite
easily by evaluating the integrals which define these expectations. But again it
is easier to find E[Z] first and then deduce the others:
E[Z] =
=

{x − (D + M)} f (x)dx

D+M

x f (x)dx − (D + M) Pr(X > D + M)

D+M
λ+1

λ+D+M
+D+M
λ
λ+1
λ
− (D + M)
λ+D+M
λ
λ
=
.
λ+D+M
It follows (see Case 1), on replacing D + M by D in this result, that
=

λ
λ+D+M

E[Y + Z] =

λ
λ+D

λ

,

and hence
E[Y] =

λ
λ+D

λ

λ
λ+D+M

λ

.

Finally, we have
λ λ
.
λ+D
We give some numerical values follow for the case α = 12 (X ∼ Pa(12, 11),
with mean 1 and variance 1.2 – which gives a model with variance fairly close
to that of the exponential model used in Case 1), and using the same values for
D and M, in Table 7.13.
These values are quite close to those obtained using X ∼ Exp(1) in Case 1.
An explanation may be found in the choice of the parameter values: α = 12
corresponds to a much less fat-tailed distribution than is the case for lower
values of α, and is very much higher than would normally be appropriate for
modelling insurance losses. Tables 7.14 and 7.15 follow for the more realistic
cases α = 2.5 (X ∼ Pa(2.5, 1.5), with mean 1 and variance 5) and α = 1.8
(X ∼ Pa(1.8, 0.8), with mean 1 but without a finite variance).
E[C] = 1 −

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Case studies
Table 7.13. Probabilities when X ∼ Pa(12, 11)

Pr(X ≤ D)
Pr(D < X ≤ D + M)
Pr(X > D + M)

M=3
D = 0.1

M=3
D = 0.3

M=5
D = 0.1

M=5
D = 0.3

0.1029
0.8463
0.0508

0.2759
0.6811
0.0429

0.1029
0.8867
0.0103

0.2759
0.7151
0.0089

Table 7.14. Probabilities when X ∼ Pa(2.5, 1.5)

Pr(X ≤ D)
Pr(D < X ≤ D + M)
Pr(X > D + M)

M=3
D = 0.1

M=3
D = 0.3

M=5
D = 0.1

M=5
D = 0.3

0.1490
0.7903
0.0607

0.3661
0.5793
0.0546

0.1490
0.8264
0.0246

0.3661
0.6111
0.0229

Table 7.15. Probabilities when X ∼ Pa(1.8, 0.8)

Pr(X ≤ D)
Pr(D < X ≤ D + M)
Pr(X > D + M)

M=3
D = 0.1

M=3
D = 0.3

M=5
D = 0.1

M=5
D = 0.3

0.1910
0.7512
0.0578

0.4363
0.5109
0.0528

0.1910
0.7815
0.0274

0.4363
0.5379
0.0258

Within each table we can note the same behaviour as we found in Case 1 –
but drawing general conclusions from a comparison of the tables must be
approached with care, bearing in mind that having fixed the mean loss at 1,
the loss distribution is constrained to be of the form Pa(λ + 1, λ). For example,
under this constraint,
λ+1
λ
,
λ+D+M
and for fixed D+ M this probability is not monotonic (increasing or decreasing)
in λ.
Table 7.16 of tail probabilities shows the values of Pr(X > 3.1) again, alongside columns for smaller tails, namely Pr(X > 6), Pr(X > 8) and Pr(X > 10) –
in the last three columns the probabilities are increasing as α decreases. The
expected sizes of the components of the loss under X ∼ Pa(1.8, 0.8) are given
in Table 7.17. In the case that our unit (E[X]) represents £5000, we have
Table 7.18 (entries in £).

Pr(X > D + M) =

7.2 Case study 2: shared liabilities – who pays what?

341

Table 7.16. Tail probabilities when X ∼ Pa(α, α − 1)

α = 12
α = 2.5
α = 1.8

Pr(X > 3.1)

Pr(X > 6)

Pr(X > 8)

Pr(X > 10)

0.0508
0.0607
0.0578

0.0054
0.0179
0.0212

0.0014
0.0099
0.0134

0.0004
0.0061
0.0092

Table 7.17. Expected values when X ∼ Pa(1.8, 0.8)

E[C]
E[Y]
E[Z]

M=3
D = 0.1

M=3
D = 0.3

M=5
D = 0.1

M=5
D = 0.3

0.0899
0.6285
0.2816

0.2249
0.5046
0.2705

0.0899
0.7079
0.2022

0.2249
0.5782
0.1969

Table 7.18. Expected values (£) when one monetary unit is £5000

E[C]
E[Y]
E[Z]

M = 15 000
D = 500

M = 15 000
D = 1500

M = 25 000
D = 500

M = 25 000
D = 1500

450
3142
1408

1124
2523
1353

450
3539
1011

1124
2891
984

When we compare these results with those for Case 1 we see that, over all
losses, the policyholder has a similar expected payout (to within £200), while
that for the direct insurer is considerably reduced and that for the reinsurer is
considerably increased (hugely increased in relative terms).
To illustrate these results, a simulation of 10 000 losses was carried out
(using R), using X with a Pareto distribution with α = 1.8, mean = 1, D = 0.3
and M = 3. As in Case 1, the loss vector x was manipulated to produce vectors c, y, z and zstar, and the data were then summarised. Vectors containing
the losses less than 0.3, between 0.3 and 3.3, and greater than 3.3 were again
constructed and summarised. The simulation results are given in Table 7.19.
The fat-tailed nature of the Pa(1.8, 0.8) distribution used here as our model
for the losses results in a maximum observed loss of 194.5 (compare this with
E[X] = 1): in repeated simulations, we find that the maximum observed loss
and maximum payout by the reinsurer exhibit high variation.

342

Case studies
Table 7.19. Simulation results when X ∼ Pa(1.8, 0.8)

x
c
y
z
zstar
x: 0 < x <= 0.3
x: 0.3 < x <= 3.3
x: x > 3.3

Number

Min.

Median

Mean

Max.

10 000
10 000
10 000
10 000
561
4344
5095
561

0.000
0.000
0.000
0.000
0.001
0.000
0.300
3.301

0.377
0.300
0.077
0.000
1.932
0.115
0.732
5.232

1.009
0.225
0.511
0.273
4.859
0.127
0.974
8.159

194.5
0.300
3.000
191.2
191.2
0.300
3.300
194.5

Note The simple R code used to produce the required vectors of data is the
same as that used in Case 1, except for the third line, in which the command
x = rexp(10000,1)

is replaced by
x = 0.8*(runif(10000)^(-1/1.8) -1)

As in Case 1 we extend the situation to one in which S has a compound Poisson distribution with claim rate now given the symbol ν and individual claim
size variable X ∼ Pa(λ + 1, λ), with mean 1. Each loss arises on a policy with
deductible D in force. In addition, an excess of loss reinsurance arrangement
is in place whereby the direct insurer pays a maximum of M on each loss. Let
S C , S I and S R be as before. Then we know the following:
S ∼ CP(ν, F X ) and E[S ] = νE[X] = ν,

λ
S C ∼ CP(ν, FC ) and E[S C ] = νE[C] = ν 1 − λ +λ D ,
λ
S I ∼ CP(ν, FY ) and E[S I ] = νE[Y] = ν λ +λ D − λ + Dλ + M
λ
S R ∼ CP(ν, FZ ) and E[S R ] = νE[Z] = ν λ + Dλ + M .

λ

,

To illustrate these results, a simulation of 10 000 aggregate losses was carried
out using X ∼ Pa(1.8, 0.8), ν = 100, D = 0.3 and M = 3. The expected
payouts are simply 100 times what they are for individual losses calculated
over all losses, that is E[S ] = 100, E[S C ] = 100 × 0.2249 = 22.49, E[S I ] =
100 × 0.5046 = 50.46 and E[S R ] = 100 × 0.2705 = 27.05.
The simulation results are given in Table 7.20. We note that the number
of losses (N ∼ Poi(100)) ranges from 66 to 141, and that the means of the
simulated data are in close agreement with the theoretical results.

7.2 Case study 2: shared liabilities – who pays what?

343

Table 7.20. Further simulation results when X ∼ Pa(1.8, 0.8)
Number

Min.

Median

Mean

Max.

10 000
10 000
10 000
10 000
10 000

66
36.05
14.01
18.92
0.000

100.0
92.68
22.39
50.10
18.08

99.86
100.6
22.45
50.45
27.69

141
2225
31.74
94.37
2144

N
S
SC
SI
SR

The fat-tailed nature of the Pa(1.8, 0.8) distribution used here as our model
for the individual losses results in a maximum observed aggregate loss of 2225
(compare this with E[X] = 100): in repeated simulations, we find that the maximum observed aggregate loss and maximum aggregate payout by the reinsurer
exhibit high variation.
Note The simulation was carried out by executing a function as in Case 1 –
the function, called ‘cpsimre2, is the same as cpsimre1, except that the
function is defined as
cpsimre2 = function(n,rate,lambda,d,m)

and, in line 3, the statement
nc = rpois(1,lam)

is replaced by
nc = rpois(1,rate)

and, in line 5, the statement
x = rexp(nc,1/mu)

is replaced by
x = lambda*(runif(nc)^(-1/(lambda + 1)) -1)

The command
case2sim = cpsimre2(10000,100,0.8,0.3,3)

executes the function once, carrying out 10 000 simulations of the number of
claims and resulting aggregate payout variables in the case ν = 100, λ = 0.8,
D = 0.3 and M = 3. The results (the list of five vectors) are held in the object
case2sim and can be retrieved using, for example,
summary(case2sim[[4]])

to obtain a summary of the values S I .