Preliminaries
We are interested in the compound random variable
S
=
∑
i
=
1
N
X
i
{\displaystyle S=\sum _{i=1}^{N}X_{i}\,}
where
N
{\displaystyle N\,}
and
X
i
{\displaystyle X_{i}\,}
fulfill the following preconditions.
Claim size distribution
We assume the
X
i
{\displaystyle X_{i}\,}
to be i.i.d. and independent of
N
{\displaystyle N\,}
. Furthermore the
X
i
{\displaystyle X_{i}\,}
have to be distributed on a lattice
h
N
0
{\displaystyle h\mathbb {N} _{0}\,}
with latticewidth
h
>
0
{\displaystyle h>0\,}
.
f
k
=
P
[
X
i
=
h
k
]
.
{\displaystyle f_{k}=P[X_{i}=hk].\,}
Claim number distribution
N
{\displaystyle N\,}
is the "claim number distribution", i.e.
N
∈
N
0
{\displaystyle N\in \mathbb {N} _{0}\,}
.
Furthermore,
N
{\displaystyle N\,}
has to be a member of the Panjer class . The Panjer class consists of all counting random variables which fulfill the following relation:
P
[
N
=
k
]
=
p
k
=
(
a
+
b
k
)
⋅
p
k
−
1
,
k
≥
1.
{\displaystyle P[N=k]=p_{k}=(a+{\frac {b}{k}})\cdot p_{k-1},~~k\geq 1.\,}
for some
a
{\displaystyle a\,}
and
b
{\displaystyle b\,}
which fulfill
a
+
b
≥
0
{\displaystyle a+b\geq 0\,}
.
the value
p
0
{\displaystyle p_{0}\,}
is determined such that
∑
k
=
0
∞
p
k
=
1.
{\displaystyle \sum _{k=0}^{\infty }p_{k}=1.\,}
Sundt proved in the paper [ 2] that only the binomial distribution , the Poisson distribution and the negative binomial distribution belong to the Panjer class, depending on the sign of
a
{\displaystyle a\,}
. They have the parameters and values as described in the following table.
W
N
(
x
)
{\displaystyle W_{N}(x)\,}
denotes the probability generating function .
Distribution
P
[
N
=
k
]
{\displaystyle P[N=k]\,}
a
{\displaystyle a\,}
b
{\displaystyle b\,}
p
0
{\displaystyle p_{0}\,}
W
N
(
x
)
{\displaystyle W_{N}(x)\,}
E
[
N
]
{\displaystyle E[N]\,}
V
a
r
(
N
)
{\displaystyle Var(N)\,}
Binomial
(
n
k
)
p
k
(
1
−
p
)
n
−
k
{\displaystyle {\binom {n}{k}}p^{k}(1-p)^{n-k}\,}
−
p
1
−
p
{\displaystyle {\frac {-p}{1-p}}}
p
(
n
+
1
)
1
−
p
{\displaystyle {\frac {p(n+1)}{1-p}}}
(
1
−
p
)
n
{\displaystyle (1-p)^{n}\,}
(
p
x
+
(
1
−
p
)
)
n
{\displaystyle (px+(1-p))^{n}\,}
n
p
{\displaystyle np\,}
n
p
(
1
−
p
)
{\displaystyle np(1-p)\,}
Poisson
e
−
λ
λ
k
k
!
{\displaystyle e^{-\lambda }{\frac {\lambda ^{k}}{k!}}\,}
0
{\displaystyle 0\,}
λ
{\displaystyle \lambda \,}
e
−
λ
{\displaystyle e^{-\lambda }\,}
e
λ
(
s
−
1
)
{\displaystyle e^{\lambda (s-1)}\,}
λ
{\displaystyle \lambda \,}
λ
{\displaystyle \lambda \,}
negative binomial
Γ
(
r
+
k
)
k
!
Γ
(
r
)
p
r
(
1
−
p
)
k
{\displaystyle {\frac {\Gamma (r+k)}{k!\,\Gamma (r)}}\,p^{r}\,(1-p)^{k}\,}
1
−
p
{\displaystyle 1-p\,}
(
1
−
p
)
(
r
−
1
)
{\displaystyle (1-p)(r-1)\,}
p
r
{\displaystyle p^{r}\,}
(
p
1
−
x
(
1
−
p
)
)
r
{\displaystyle \left({\frac {p}{1-x(1-p)}}\right)^{r}\,}
r
(
1
−
p
)
p
{\displaystyle {\frac {r(1-p)}{p}}\,}
r
(
1
−
p
)
p
2
{\displaystyle {\frac {r(1-p)}{p^{2}}}\,}
Recursion
The algorithm now gives a recursion to compute the
g
k
=
P
[
S
=
h
k
]
{\displaystyle g_{k}=P[S=hk]\,}
.
The starting value is
g
0
=
W
N
(
f
0
)
{\displaystyle g_{0}=W_{N}(f_{0})\,}
with the special cases
g
0
=
p
0
⋅
exp
(
f
0
b
)
if
a
=
0
,
{\displaystyle g_{0}=p_{0}\cdot \exp(f_{0}b){\text{ if }}a=0,\,}
and
g
0
=
p
0
(
1
−
f
0
a
)
1
+
b
/
a
for
a
≠
0
,
{\displaystyle g_{0}={\frac {p_{0}}{(1-f_{0}a)^{1+b/a}}}{\text{ for }}a\neq 0,\,}
and proceed with
g
k
=
1
1
−
f
0
a
∑
j
=
1
k
(
a
+
b
⋅
j
k
)
⋅
f
j
⋅
g
k
−
j
.
{\displaystyle g_{k}={\frac {1}{1-f_{0}a}}\sum _{j=1}^{k}\left(a+{\frac {b\cdot j}{k}}\right)\cdot f_{j}\cdot g_{k-j}.\,}
Example
The following example shows the approximated density of
S
=
∑
i
=
1
N
X
i
{\displaystyle \scriptstyle S\,=\,\sum _{i=1}^{N}X_{i}}
where
N
∼
NegBin
(
3.5
,
0.3
)
{\displaystyle \scriptstyle N\,\sim \,{\text{NegBin}}(3.5,0.3)\,}
and
X
∼
Frechet
(
1.7
,
1
)
{\displaystyle \scriptstyle X\,\sim \,{\text{Frechet}}(1.7,1)}
with lattice width h = 0.04. (See Fréchet distribution .)
References