Actuarial Statistics – 2006 Paper Page 1 of 16
Actuarial Statistics – 2006 Paper
Question 1
We begin by noting that since the claim sizes cannot be 0,
g 0 = (N = 0) = p0 = 0 .
We also note that for the compound distribution to have a value of 1, we
must have a single claim, with a value of 1. So g1 = f1p1 . This will be the basis
for our recursion formula.
Now, mutiply the condition in the question by zn and sum, to get
¥ ¥ æ bö
å n p z n
= å z n ççèça + n ÷÷ø÷÷ pn-1
n =1 n =1
¥ ¥ ¥
zn
å pn z - p0 = å azz pn-1 + b å n pn-1
n n -1
n =0 n =1 n =1
¥ ¥
zn
GN (z ) - p0 = az å z pn + b å pn -1
n
n =0 n =1 n
¥
zn
(1 - az ) N
G (z ) = p 0
+ b å n pn-1
n =1
Differentiating with respect to z
-aGN (z ) + (1 - az )GN¢ (z ) = bGN (z )
a +b
GN¢ (z ) = G (z )
1 - az N
Now, let
Daniel Guetta, 2010
,Actuarial Statistics – 2006 Paper Page 2 of 16
¥
GS (z ) = å gn z n
n =0
We have M S (u ) = GN (M X (u )) , and we also know that G (z ) = M (log z ) , so
GS (z ) = M S (log z ) = GN (M X (log z )) = GN (GX (z ))
Differentiating, we get
GS¢ (z ) = GN¢ (GX (z ))GX¢ (z )
a +b
= G (G (z ))GX¢ (z )
1 - aGX (z ) N X
a +b
= G (z )GX¢ (z )
1 - aGX (z ) S
So
(1 - aG X
(z ))GS¢ (z ) = (a + b)GS (z )GX¢ (z )
We now feed in the fact that [note: the second sum goes from 1 instead of 0
because f0 = 0]
¥ ¥
GS (z ) = å gn z n GX (z ) = å fk z k
n =0 n =1
And get
æ ¥ öæ ¥ ö
b -1 ÷
æ¥ öæ ¥ ö
çç1 - a
å f z a ÷ç
÷
÷ ç å b g z ÷
÷ = (a + b ) ç
ç å g z ÷÷ çå b fb z b -1 ÷÷÷
a ֍
ççè a
a=1
÷ø çèç b =1
b
÷ø çèç a
a =0
÷ø ççè b =1
÷ø
Now, equate coefficients of zr – 1
rgr - a å b fag b = (a + b) å bg a fb
a +b =r a +b =r
r -1 r
rgr - a å (r - a)fagr -a = (a + b)å b fb gr -b
a =1 b =1
And so
r r -1
rgr = å (a b + b b )fb gr -b + å (ar - a a)fagr -a
b =1 a =1
r -1
= å (ar + b b )fb gr -b + (ar + br )fr g 0
b =1
r
= å (ar + b b )fb gr -b
b =1
Which means that
r æ
bj ö
gr = å çça + ÷÷÷ fj gr -j
ç
j =1 è r ÷ø
This is our recursion formula for the g, starting from g1 = f1p1 .
Let us find a and b when
Daniel Guetta, 2010
, Actuarial Statistics – 2006 Paper Page 3 of 16
e -lln
pn =
(1 - e ) n !
-l
Note that
e -lln -1
pn -1 =
(1 - e )(n - 1)!
-l
And so
b p
a+ = n
n pn -1
e -lln
=
(1 - e ) n !
-l
e -ll n -1
(1 - e )(n - 1)!
-l
l
=
n
And so a = 0 and b = l . Our recursion formula becomes
r
lj
gr = å fg
j =1 r j r -j
Starting from g1 = f1p1 .
Daniel Guetta, 2010
Actuarial Statistics – 2006 Paper
Question 1
We begin by noting that since the claim sizes cannot be 0,
g 0 = (N = 0) = p0 = 0 .
We also note that for the compound distribution to have a value of 1, we
must have a single claim, with a value of 1. So g1 = f1p1 . This will be the basis
for our recursion formula.
Now, mutiply the condition in the question by zn and sum, to get
¥ ¥ æ bö
å n p z n
= å z n ççèça + n ÷÷ø÷÷ pn-1
n =1 n =1
¥ ¥ ¥
zn
å pn z - p0 = å azz pn-1 + b å n pn-1
n n -1
n =0 n =1 n =1
¥ ¥
zn
GN (z ) - p0 = az å z pn + b å pn -1
n
n =0 n =1 n
¥
zn
(1 - az ) N
G (z ) = p 0
+ b å n pn-1
n =1
Differentiating with respect to z
-aGN (z ) + (1 - az )GN¢ (z ) = bGN (z )
a +b
GN¢ (z ) = G (z )
1 - az N
Now, let
Daniel Guetta, 2010
,Actuarial Statistics – 2006 Paper Page 2 of 16
¥
GS (z ) = å gn z n
n =0
We have M S (u ) = GN (M X (u )) , and we also know that G (z ) = M (log z ) , so
GS (z ) = M S (log z ) = GN (M X (log z )) = GN (GX (z ))
Differentiating, we get
GS¢ (z ) = GN¢ (GX (z ))GX¢ (z )
a +b
= G (G (z ))GX¢ (z )
1 - aGX (z ) N X
a +b
= G (z )GX¢ (z )
1 - aGX (z ) S
So
(1 - aG X
(z ))GS¢ (z ) = (a + b)GS (z )GX¢ (z )
We now feed in the fact that [note: the second sum goes from 1 instead of 0
because f0 = 0]
¥ ¥
GS (z ) = å gn z n GX (z ) = å fk z k
n =0 n =1
And get
æ ¥ öæ ¥ ö
b -1 ÷
æ¥ öæ ¥ ö
çç1 - a
å f z a ÷ç
÷
÷ ç å b g z ÷
÷ = (a + b ) ç
ç å g z ÷÷ çå b fb z b -1 ÷÷÷
a ֍
ççè a
a=1
÷ø çèç b =1
b
÷ø çèç a
a =0
÷ø ççè b =1
÷ø
Now, equate coefficients of zr – 1
rgr - a å b fag b = (a + b) å bg a fb
a +b =r a +b =r
r -1 r
rgr - a å (r - a)fagr -a = (a + b)å b fb gr -b
a =1 b =1
And so
r r -1
rgr = å (a b + b b )fb gr -b + å (ar - a a)fagr -a
b =1 a =1
r -1
= å (ar + b b )fb gr -b + (ar + br )fr g 0
b =1
r
= å (ar + b b )fb gr -b
b =1
Which means that
r æ
bj ö
gr = å çça + ÷÷÷ fj gr -j
ç
j =1 è r ÷ø
This is our recursion formula for the g, starting from g1 = f1p1 .
Let us find a and b when
Daniel Guetta, 2010
, Actuarial Statistics – 2006 Paper Page 3 of 16
e -lln
pn =
(1 - e ) n !
-l
Note that
e -lln -1
pn -1 =
(1 - e )(n - 1)!
-l
And so
b p
a+ = n
n pn -1
e -lln
=
(1 - e ) n !
-l
e -ll n -1
(1 - e )(n - 1)!
-l
l
=
n
And so a = 0 and b = l . Our recursion formula becomes
r
lj
gr = å fg
j =1 r j r -j
Starting from g1 = f1p1 .
Daniel Guetta, 2010
