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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
...
So g1 = f1p1
...

Now, mutiply the condition in the question by zn and sum, to get
¥
¥
æ
bö
n
=
p
z
å 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 =0
n =1
n =1
¥
¥
zn
n
GN (z ) - p0 = az å z pn + b å pn -1
n =0
n =1 n
¥
zn
=
+
1
(
)
az
G
z
p
b
(
) N
å n pn-1
0
n =1
¥

¥

n

n -1

¥

Differentiating with respect to z

-aGN (z ) + (1 - az )GN¢ (z ) = bGN (z )
GN¢ (z ) =

a +b
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
n =0

¥

GX (z ) = å fk z k
n =1

And get
¥
ö
ö
æ
öæ ¥
æ¥
öæ ¥
a ֍
b -1 ÷
a ֍
çç1 - a
ç
÷
÷
÷÷ çå b fb z b -1 ÷÷÷
f
z
b
g
z
(
a
b
)
g
z
=
+
ç
ç
å
å
å
÷
÷
a
b
a
ççè
çèç
÷ø
÷ø
÷ø çèç
÷ø ççè
a=1

b =1

a =0

b =1

Now, equate coefficients of zr – 1
rgr - a

å

a +b =r

b fag b = (a + b)

å

a +b =r
r

r -1

bg a fb

rgr - a å (r - a)fagr -a = (a + b)å b fb gr -b
a =1

b =1

And so
r

r -1

b =1
r -1

a =1

rgr = å (a b + b b )fb gr -b + å (ar - a a)fagr -a
= 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
r ÷ø
j =1 è

This is our recursion formula for the g, starting from g1 = f1p1
...
Our recursion formula becomes
=

r

gr = å
j =1

lj
fg
r j r -j

Starting from g1 = f1p1
...
The
PGF of Ni is

( )=e

GN (z ) =  z
i



Ni

li (z -1)

Each claim is of size X ~Exp(m)
...
Clearly, it has a compound distribution
...
of X i

Ni

X

= GN (M X (u ))
i
æ é m
ùö
= exp çççli ê
- 1ú÷÷÷
ú ø÷
èç êë m - u
û

¬ Use functions from
start of question

Daniel Guetta, 2010

Actuarial Statistics – 2006 Paper

Page 5 of 16

Now, consider S = S1 +  + Sn

( )

( e )
=  (e )  (e )

M S (u ) =  e uS =  e

uS1

uSn

uS1

uSn

¬ Indep
...
of X i and li
= exp l1 éê m-m u - 1ùú  exp ln éê m-m u - 1ùú
ë
û
ë
û
m
n
é
ù
é
ù
= exp ê åi =1 li ú ê m-u - 1ú
ë
ûë
û
This is clearly a compound Poisson, with Poisson parameter åni=1 li
...

First, consider each Ni and let x be an integer
 (N i = x ) = ò

¥

l =0
¥

 (N i = x | l ) f (l) dl

l x -l amlm -1e -al
⋅ 1 dl
e ⋅
l =0 x !
(m - 1)!
¥ lx
amlm -1e -al (x + m - 1)!
dl
=ò
⋅
e -l ⋅
l =0 x !
(m - 1)! (x + m - 1)!
(x + m - 1)! ¥ l x -l amlm -1e -al
dl
=
e ⋅
(m - 1)! òl=0 x !
(x + m - 1)!
(x + m - 1)! am ¥ l x +m -1e -(a+1)l
dl
=
(m - 1)! x ! òl=0 G(x + m )
¥ (a + 1)x +m l x +m -1e -(a +1)l
am
(x + m - 1)!
dl
=
(m - 1)! x !(a + 1)x +m òl=0
G(x + m )

=ò

At this point, we remember (somehow) that the PDF of G(k, q) is
f (l) = q k l k -1e -ql / G(k )
...
The integral therefore evaluates to 1 and

(x + m - 1)!
am
 (N i = x ) =
(m - 1)! x !(a + 1)x +m
æ
ö
-(x +m )
çx + m - 1÷÷ m
= çç
÷÷ a (a + 1)
x
÷ø
çè
æ
ö
çx + m - 1÷÷ a m 1 x
= çç
÷÷( a+1 ) ( a+1 )
x
÷ø
çè

This is a negative binomial distribution with parameters
p=
q = 1-

a
a +1
a
a +1

=

1
a +1

r =m

Daniel Guetta, 2010

Actuarial Statistics – 2006 Paper

Page 6 of 16

The total number of claims is given by
N = N1 +  + Nn

Consider the MGF of N:

( )  (e ) = M

( )

M N (t ) =  e Nt =  e

N 1t

Nn t

N1

(t )M N (t )
n

Furthermore, since Ni has a negative binomial distribution,
m

æ
ö÷
1
ç
÷÷
M N (t ) = çç
i
ççè1 - 1 e t ÷÷ø
a+1

And so
mn

æ
ö÷
1
çç
÷÷
M N (t ) = ç
ççè1 - 1 e t ÷÷ø
a+1

This is also the MGF of a negative binomial, with the same p parameter but
with
p = a+a 1
r = nm

In other words
N ~ NegBin (r = nm, p =

a
a+1

)

This can be viewed as a single Poisson distribution N ~ Po(l) with its
parameter mixed over the following distribution:
f (l) =

anmlnm -1e -al
(nm - 1)!

l>0

(*)

Now, the claims sizes all have exponential distribution with parameter m
...
Now
n

Nk

N

S = å å X i =å X i
k =1 i =1

i =1

So S does indeed have a compound mixed Poisson distribution, and the
mixing distribution is that in equation (*)
...




The number of claims arriving in (0, t] is N(t), it is independent of the
Xi and {N (t ), t ³ 0} is a Poisson process with rate l > 0 , which means

that (a) N (t ) ~ Po(lt ) and (b) the times between consecutive arrivals
are IID exponential variables with mean 1 / l
...


At time t, the risk-reserve is then given by
N (t )

U (t ) = u + ct - å X i
i =1

We note that using the properties of compound Poisson variables

(U (t )) = u + ct - lmt
Where m = (X1 )
...
We write c = (1 + q)lm where q > 0 is called the
premium loading factor
...


We prove this inequality in three steps
1
...


Define g(z ) = M X (z ) - 1 - (1 + q)mz
...

First, assume that z ¥ < ¥
...

If z ¥ = ¥ , we need to make sure that the M term in g grows faster
than the r term – otherwise, the function no longer looks as plotted
above
...
t
...
Thus
g(z ) ³ pe hz - 1 + (1 + q)mz  ¥
So there is indeed one unique strictly positive solution for r
...
We

define

a

new

quantity,

yn (u )

,

such

that

yn (u ) £ e -Ru "n  y(u ) £ e -Ru

The new quantity we define is

(

yn (u ) =  Ruin occurs at or before n th claim

)

Clearly, yn (u )  y(u ) as n  ¥
...
We show that yn (u ) £ e -Ru "n

We do this by induction


n = 1 case
Clearly, ruin can’t occur before the first claim
...
The first claim must exceed this amount for ruin to
occur
¥

y1(u ) = ò

0

=ò

¥

 (X1 > u + ct )le -lt dt
¥

ò

x =u +ct

0

fX (x )le -lt dx dt

Now, note that in the range of the integral, e

-r (u +ct -x )

is

greater than 1 and so
y1(u ) £ ò

¥

ò

¥

x =u +ct

0

e -R(u +ct -x ) fX (x )le -lt dx dt

Further note that the integrand is positive, so
y1(u ) £ ò

¥

0

¥

ò

x =0

¥

e -r (u +ct -x ) fX (x )le -lt dx dt

= ò le -lte -r (u +ct ) ò
0

¥

x =0

¥

e rx fX (x ) dx dt

= e -ru ò le -(rc +l )t M X (r ) dt
0

r is defined as M X (r ) = (1 + q)mr + 1 , so
¥

y1(u ) £ e -ru ò le -(rc +l )t {(1 + q)mr + 1} dt
0

=e

-ru

¥

ò {(1 + q)mlr + l}e

-(rc +l )t

0

dt

Remember that c = (1 + q)ml
y1(u ) £ e -ru ò

0

¥

{rc + l}e

-(rc +l )t

dt

The integral is simply an exponential density that
evaluates to 1, so
y1(u ) £ e -ru



Inductive step
Now assume that yn (u ) £ e -ru , and consider

Daniel Guetta, 2010

Actuarial Statistics – 2006 Paper

Page 11 of 16

(

)

yn +1(u ) =  Ruin @ or before (n + 1)th
æRuin @ or before (n + 1)th ö÷
¥
ç
-lt
= ò  çç
÷÷÷ le dt
st
0
| 1 occurs at t
ççè
ø÷

We now split this integral into two options:
The ruin happening at the first claim (ie: first



claim greater than u + ct)
The ruin not happening at the first claim, in



which case, after the first claim, we “reset the
timer” with capital u + ct – x1
– the ruin happening at the first claim, and the ruin not
happening at the first claim:
¥

yn +1(u ) = ò le -lt
0

{ò

¥

x1 =u +ct

fX (x ) dx + ò

u +ct

x1 = 0

yn (u + ct - x 1 )fX (x ) dx

} dt

Now:
-r (u +ct -x1 )

>1



In the first situation, e



In the second situation, the inductive hypothesis
implies that yn (u + ct - x 1 ) £ e

-r (u +ct -x1 )


...
This makes sense – the exponential distribution is highly positively
skewed, and this implies that it places greater weight on claim sizes above the
mean than below
...
The probability of ruin is
therefore lower
...
If anyone can think of a better
way, let me know ]

Daniel Guetta, 2010

Actuarial Statistics – 2006 Paper

Page 13 of 16

Question 4

We write
n

qˆ = a 0 + å ai X i
i =1

We need to choose this estimator to minimise
2
ìïæ
ï
n
ö÷ ü
ïïç
ï
L =  íççq - a 0 - å ai X i ÷÷ ï
ý
ïïçè
ø÷ ï
i =1
ï
ï
îï
þ

This implies that
n
ïì
ïü
¶L
=  ïíq - a 0 - å ai X i ïý = 0
ïîï
ïþï
¶a 0
i =1
n
ìï æ
öüï
¶L
=  ïíX r çççq - a 0 - å ai X i ÷÷÷ïý = 0
÷øïï
ïï çè
¶ar
i =1
î
þ
First consider (2) - (X r )(1)

(1)
"r

n
n
öïü
ïì æ
ïì
ïü
 ïíX r çççq - a 0 - å ai X i ÷÷÷ïý = (X r ) ïíq - a 0 - å ai X i ïý
÷øïï
ïï çè
i =1
i =1
îïï
þïï
î
þ
n

(2)

n

 (X r q ) - a 0  (X r ) - å  {ai X r X i } = (X r )(q) - a 0 (X r ) -  (X r ) å  {ai X i }
i =1

n

i =1

n

 (X r q ) - (X r )(q) = å  {ai X r X i } -  (X r ) å  {ai X i }
i =1

i =1

n

ov (X r , q ) = å ai ov (X r , X i )

(3)

i =1

We now use the conditional variance formula on both sides of (3)

Daniel Guetta, 2010

Actuarial Statistics – 2006 Paper

Page 14 of 16

ov (X r , q ) =  éêov (X r , q | q )ùú + ov éê  (X r | q ),  (q | q )ùú
ë
û
ë
û
=  éêqov (X r ,1 | q )ùú + ov éêëq, q ùúû
ë
û
= ar (q )
ov (X r , X i ) =  éê ov (X r , X i | q )ùú + ov éê  (X r | q ),  (X i | q )ùú
ë
û
ë
û
=  éêdri ar (X i | q )ùú + ov éêëq, q ùúû
ë
û
é 1
ù
=  êdri m 2 ar (Yi | q )ú + ar (q )
i
ë
û
é 1
ù
=  êdri m 2 mi q(1 - q)ú + ar (q )
i
ë
û
é
ù
1
=  êdri m q(1 - q)ú + ar (q )
i
ë
û
Feeding this back into (3), we get

{

n

}

ar (q ) = å ai  éêdri m1 q(1 - q)ùú + ar (q )
i
ë
û
i =1
n
a
ar (q ) = r  (q(1 - q)) + ar(q)å ai
mr
i =1
n

mr ar (q ) = ar  (q(1 - q)) + mr ar(q)å ai

(4)

i =1

Re-arranging (4), we get
ar =

n
ì
ü
mr ar (q ) ï
ï
ï
1
ai ï
í
ý
å
ï
 (q(1 - q)) ï
i =1
ï
ï
î
þ

(5)

Summing (4) from 1 to n, we get
n

åa
i =1

Where m+ = å

n
i =1

i

=

m+
m+ +

(6)

(q (1-q ))
ar(q )

mi
...
Thus, in this particular case, exact Bayes’
credibility is possible, and the resulting estimate is identical to the Buhlman
credibility estimate

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