非寿险精算Loss number distributionPPT课件

an (a,b,0) dis. + a degenerated dis.
ETNB distribution
pTk

1 k
k ln 1
,
k 1, 2,3,
a

,b r 1

r0


1
1 1
Ch2.3 Counting Process
X n , n 1,2,...are i.i.d. exponential random
variable having mean 1/
– Mixing N | L ~ Poisson(L ), L ~ Gamma( , )
N ~ Negative Binomial(, )
– Compound N ~ Poisson(),M ~ Logarithmic( )
•

Limiting
N
S M
leaid1s
i~
to
Negative Binomial( ,
Binomial distribution
Ch2.1.1 Poisson Distribution
N~Poisson( )
pk

ek k!
,
k 0,1, 2,...
• E[N]=Var[N]=
• Pgfቤተ መጻሕፍቲ ባይዱP (z) ez1
• Additivity Let N 1,N 2,...,N n be independent Poisson
非寿险精算Loss number distribution
Questions
Suppose the numbers of payment for some medical insurance policy follows Poisson distribution, now modified it with deductible of ￥100，which distribution can be used to describe the numbers of payment for the modified policy? And which distribution for the case that the insurer issued a reinsurance program for this policy?
p1, , pm corresponds to N 1, ,N m respectively.
Then
N
1
~
Poisson(
p 1
),
,N m ~Poisson(pm )
Ch2.1.2 Negative Binomial Dis.
Negative Binomial(r, )
pk

k

r k
Poisson Distrilbn(u1tion)

)
r , 0,and : r be a constant
Ch2.1.2 Geometric Distribution
• Special case of negative binomial dis. when
r=1, i.e.
Ch2.1 The (a,b,0) Class
Let N be a random variable representing the number of some events. Denote its probability function as The (a,b,0) class: If there exists constants a
Ch2.3 Counting Process
Nt t s, Nt Ns
Nt Ns Nt Ns
Ns Nu,u s
pk,kn (s,t) Pr(Nt Ns n | Ns k),
0 s t , k, n 0,1,... Let N0 0, pn (t) p0,n (0, t) Pr( Nt n)
and b such that pk pk1 a b k k 1, 2,3,......
Then N is a member of the (a,b,0) class. Include and only include Poisson distribution
Negative Binomial distribution
1 1
1
r

1
k

,
k 0,1, 2,
,r 0, 0
• E[N]= r ＜ Var[N]= r (1 )
• Pgf P (z) 1 z 1r
• Generalization of Poisson distribution
pk

1
1


1



k

,
k 0,1,2,..., 0
• Memoryless property
Given that there are at least m claims, the probability distribution of the number of claims in excess of m does not depend on m.
The process is homogeneous Poisson process
•
Transition Probability pk,k (s,t) e(ts) pk,kn (s, t)
(t s) en (ts) , n 0,1,...
n!
• Dis. of increment (t s)n e(ts)
ak
b,
k 0,1, 2,
,
k pˆk pˆk 1

k
nk n k1
linear in k
Ch2.2 The (a,b,1) class
The (a,b,1) class: If there exists constants a
and b such that pk pk1 a b k , k 2,3,
Ch2.1 The (a,b,0) Class
Poisson Binomial
Negative Binomial Geometric
0 q
1 q

1
1

(m 1) q 1 q
(r 1) 1
0
e (1 q)m
(1 )r
(1 )1
k
pk pk 1
variables with parameters 1, 2,..., n . Then N N 1 N 2 ... N n has a Poisson
•
distribution
Decomposability
with
parameter
1

2

...

n
.
If N ~ Poisson(), N N 1 N m with probabilities
A stochastic process is a sequence of r.v.s
{Nt ;t [0,T ]} or a sequence of r.v.s {Ni ;i 1,..., m}
for s t, Nt Ns is increment of time interval [s,t]
t s
k
n
1
(y
)pk
,k
n
1
(s,
y
)
exp

t
s kn
(x
)dx

dy
Notes
to
ensure

pk,kn (s,t) 1,
n 0

k0
mta0x
k
(t)

1

is
required
Ch2.3.2 Poisson Process
• Let transition intensity k (t) constant


t s

(
x)dx
n
exp n!
t s
k (t) (t)
(x)dx , n 0,1,...
Ch2.3.2 Poisson Process
Let X 1 denote the time of the first event and X n the time between (n-1)st and the nth event {X n ; n 1}is the sequence of inter-arrival time
Pr( Nt N s n)
, stationary n!
• Dis. of each time
(t)n et
pn (t) Pr( Nt n) p0,n (0, t) n!
Nt ~ Poisson(t)
•
nonhomogeneous Poisson process
pk,kn (s,t)
pk,k1 (t, t pk,kn (t, t

h) h)

k (t)h
o(h), n
o(h),
2 k
(t)depends
on
t，homogeneous
pk,k (s,t)

exp

t
s
k
(x
)dx

,
and
for
n
1，2，
pk,kn (s,t)
pk,kn (s,t)

Pr( Nt N s n) pk,kn (s, t)pk (s) k 0
Ch2.3.1 Birth Process
Birth process
pk,kn (t,t h) k (t)h, n 1 pk,kn (t, t h) 0, n 2 pk,k (t,t h) 1 k (t)h o(h) k (t) k homogeneous
Pr(N m k N m ) Pr(N k)
Ch2.1.3 Binomial Distribution
Binomial(m,q)
pk

m k
qk

1q m
k
,
k 0,1, 2,
,m ,0 q 1
• E[N]= mq ＞ Var[N]= mq(1 q)
Then N is a member of the (a,b,1) class.
zero-truncated
pT 0
0,
pTk 1,
k 1
pTk

1
pk

p 0
zero-modified
pM 0
0,
pkM

1

pM 0
,
k 1
pkM

1

pM 0
1

p 0
pk
A zero-truncated (a,b,1) dis. + a degenerated dis.
Pr( X 1 t) Pr( Nt 0) et Pr( X 2 t | X1 s) Pr(0 event in (s, s t) | X1 s)
(byindependent increments) Pr(0 event in (s, s t)) (by stationary increments ) Pr(0 event in (0,t)) et Pr( X 2 t) Pr( X 2 t | X1 s) et
the number of payments N1, N2 ,..., Nm in time intervals
0,
t 1

,

0,
t 2

,
,0,tm ,
0

t 1

tm T respectively
N1 N2 N3
Nt
Nm
For atn1 yt2t itn3 [0,T], the numbetr of payments istm N t
• Pgf P (z) 1q z 1m
• Random variable with binomial distribution has the finite maximum value, so it can be used for the case with finite support.
Stationary increments: Nt Ns depends on t and s only through the difference t-s.
Independent increments: increment of any set of disjoint intervals are independent.