Actuarial Modelling of Claim Counts Risk Classification, Credibility ...

Credibility Models for Claim Counts 147
and let us expand the squared sum to get
[ ( ) ] 2
1 = E N iTi +1 − EN iTi +1 i
[ (
+ 2E N iTi +1 − EN iTi +1 i ) ( )]
T
∑ i
EN iTi +1 i − c i0 − c it N it + 2
The second term in the expansion of 1 vanishes, and the first one does not depend on the
c it s.
Let us determine c as arg min 2 . Recall that EN iTi +1 i = iTi +1 i . Setting equal to 0
the partial derivative of 2 with respect to c i0 allows us to write
which gives
T
∑ i
0 = iTi +1E i − c i0 − c it EN it
t=1
T
∑ i
c i0 = iTi +1 − c it it (3.14)
Now, setting equal to 0 the partial derivatives of 2 with respect to c is for s = 1 2T i ,
yields
Noting that
t=1
T
∑ i
0 = iTi +1EN is i − c i0 EN is − c it EN is N it (3.15)
EN it N is = CN is N it + EN is EN it
[
] [
]
= E CN is N it i + C EN is i EN it i + is it
⎧
⎨ is + ( ) 21
is + Vi if s = t
=
⎩
is it 1 + V i otherwise
t=1
t=1
and that
[
]
]
EN is i = E CN is i i + C
[EN is i i + is
= is 1 + V i
allows us to cast equation (3.15) into the form
(
)
T
∑ i
c is = V i iTi +1 − c it it = V i c i0
t=1