Poisson processes
A short introduction
Péter Medvegyev
January 2009
Medvegyev (CEU)
Poisson processes
January 2009
1 / 42
De…nition of Poisson processes
De…nition
A counting Lévy process is called Poisson process.
A process is called counting process if every trajectory is increasing and
the image space of the trajectories is the set f0, 1, 2, . . .g . One should
emphasize that all the natural numbers are in the image space, so the
trajectories have jumps with size one. As the domain of de…nition of the
trajectories is R+ the jumps cannot accumulate.
Medvegyev (CEU)
Poisson processes
January 2009
2 / 42
The distribution of the …rst jump is exponential
The time of the …rst jump
τ 1 (ω ) $ inf ft : X (t, ω ) = 1g = inf ft : X (t, ω ) > 0g < ∞
has exponential distribution as
P (τ 1 > t + s ) =
= P (X (t + s ) = 0) = P (X (s ) = 0, X (t + s ) X (s ) = 0) =
= P (X (s ) = 0) P (X (t + s ) X (s ) = 0) =
= P (X (s ) = 0) P (X (t ) = 0) ,
hence for some number 0 < λ < ∞
P (τ 1 > t ) = P (X (t ) = 0) = exp ( λt ) .
Medvegyev (CEU)
Poisson processes
January 2009
3 / 42
Strong Markov property
Theorem
If X is a Lévy process and τ < ∞ is a stopping time then the process
X (t ) $ X (t + τ )
X (τ )
is
1
also a Lévy process,
2
the distribution of X and the distribution of X are equal and
3
X is independent of Fτ .
Medvegyev (CEU)
Poisson processes
January 2009
4 / 42
The time between the jumps is also exponential
By the strong Markov property the distribution of
X1 (t ) $ X (τ 1 + t ) X (τ 1 ) is the same as the distribution of X (t ), so
the time between the …rst and the second jump of X is
τ 2 (ω ) $ inf ft : X (t + τ 1 (ω ) , ω ) = 2g = inf ft : X1 (t, ω ) > 0g < ∞.
τ 1 and τ 2 are independent and they have the same distribution. In the
same way one can prove:
Theorem
If X is a Poisson process then the distribution of the time between the
jumps is exponential with the same parameter λ and the time between the
jumps are independent variables.
Medvegyev (CEU)
Poisson processes
January 2009
5 / 42
Predictable stopping times
De…nition
A stopping time τ > 0 is called predictable, if there is a sequence of
stopping times (ρn ) such that ρn < τ and ρn % τ.
Theorem
The jumps of a Poisson process are unpredictable.
If the …rst jump τ 1 was predictable then
Nn ( t ) $ N ( ρ n + t )
N ( ρn )
would be a Poisson process, with the same parameter λ..The …rst jump of
Nn is τ 1 ρn . But τ 1 and τ 1 ρn have expected value 1/λ > 0. Hence
by the Dominated Convergence Theorem
1
1
= lim
= lim E (τ 1
n !∞ λ
n !∞
λ
which is impossible.
Medvegyev (CEU)
ρn ) = E
Poisson processes
lim (τ 1
n !∞
ρn ) = 0,
January 2009
6 / 42
Gamma and beta distributions
De…nition
The function
Γ (x ) $
Z ∞
tx
1
exp ( t ) dt,
x >0
0
is called gamma function. The function
B (x, y ) =
Z 1
tx
1
(1
t )y
1
dt,
x > 0, y > 0
0
is called beta function.
Medvegyev (CEU)
Poisson processes
January 2009
7 / 42
Gamma and beta distributions
With partial integration one can easily prove that
Lemma
For every x > 0
Γ (x + 1) = x Γ (x ) .
With integration by substitution, with u = λt, one can also prove that
Lemma
If λ > 0 then
Z ∞
0
Medvegyev (CEU)
tx
1
exp ( λt ) dt =
Poisson processes
Γ (x )
.
λx
January 2009
8 / 42
Gamma and beta distributions
De…nition
If λ and a are positive numbers then the distribution with density function
f (x ) $
λa a
x
Γ (a )
1
exp ( λx ) ,
x >0
is called gamma distribution with parameters (a, λ). One can denote this
distribution by Γ (a, λ).
De…nition
If α and β are positive numbers then the distribution with density function
f (x ) $
1
xα
B (α, β)
1
(1
x )β
1
,
x 2 (0, 1)
is called beta distribution with parameters (α, β). One can denote this
distribution by B (a, β).
Medvegyev (CEU)
Poisson processes
January 2009
9 / 42
Examples
Example
Γ (1, λ) is the exponential distribution with parameter λ.
Example
The distribution χ21 is Γ (1/2, 1/2) .
Let ξ = N (0, 1) and let us calculate the distribution of η $ ξ 2 . If x
then Fη (x ) = P ξ 2 < x = 0. If x > 0 then
p
p
Fη (x ) = P ξ 2 < x = P
x<ξ< x =
Z p
x
1
= p
p exp
x
2π
di¤erentiating, if x > 0 then
2
fη (x ) = p
exp
2π
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y2
2
t2
2
2
dt = p
2π
p
y= x
Z px
exp
0
1
1
p =p
exp
2 x
2πx
Poisson processes
t2
2
0,
dt.
x
.
2
January 2009
10 / 42
The beta-gamma relation
Example
Prove that
B (x, y ) =
Γ (x ) Γ (y )
!
Γ (x + y )
Let us de…ne
Γ (x, λ) $
Z ∞
tx
1
0
exp ( λt ) dt,
x, λ > 0.
Substituting u = tλ
Γ (x, λ) =
Medvegyev (CEU)
Z ∞
0
u
λ
x 1
exp ( u )
Poisson processes
du
Γ (x )
= x .
λ
λ
January 2009
11 / 42
Calculating one side
With substitution s = t/ (1
I
$
Z ∞
0
t)
Γ (x + y , 1 + s ) s
= Γ (x + y )
= Γ (x + y )
= Γ (x + y )
= Γ (x + y )
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Z ∞
x 1
ds =
Z ∞
0
(x +y ) x 1
(1 + s )
s
Γ (x + y )
sx 1
ds =
(1 + s )x +y
ds =
0
Z 1
1+
0
Z 1
Z 1
0
1
1
tx
1
t
1
(1
t )y
1
t
t
1
t )2
(1
x 1
t
1
x 1
t
t
(x +y )
1
0
(x +y )
t
1
(1
t )2
dt =
dt =
dt $ Γ (x + y ) B (x, y ) .
Poisson processes
January 2009
12 / 42
Calculating the other side
I
$
$
=
=
Z ∞
0
Γ (x + y , 1 + s ) s x
Z ∞Z ∞
0
0
0
0
Z ∞Z ∞
Z ∞
$
1 x 1
dtds =
exp ( t (1 + s )) t x +y
1 x 1
dsdt =
t x +y
1
exp ( t )
=
s
exp ( ts ) s x
t x +y
1
exp ( t ) Γ (x, t ) dt =
t x +y
1
exp ( t )
0
$ Γ (x )
Medvegyev (CEU)
Z ∞
s
0
0
Z ∞
ds $
exp ( t (1 + s )) t x +y
0
Z ∞
1
Z ∞
0
ty
1
1
dsdt $
Γ (x )
dt $
tx
exp ( t ) dt = Γ (x ) Γ (y ) .
Poisson processes
January 2009
13 / 42
The density function of the Gaussian distribution
Example
Calculate the integral
Z ∞
∞
I
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$
Z ∞Z ∞
=
1
2
exp
x 2 dx !
x 2 1 + t 2 dxdt =
"
#∞
Z ∞
exp x 2 1 + t 2
=
dt =
2 (1 + t 2 )
0
0
x exp
0
Z ∞
0
0
π
1
dt = .
2
1+t
4
Poisson processes
January 2009
14 / 42
The density function of the Gaussian distribution
π
4
=
Z ∞Z ∞
0
=
Z ∞
x2 1 + t2
x exp
0
x exp
x2
0
=
Z ∞
Z ∞
exp
dtdx =
(xt )2 dtdx =
0
x exp
0
hence
x2
Z ∞
0
Z ∞
∞
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exp
exp
u2
du
dx =
x
x 2 dx =
Poisson processes
p
Z ∞
2
exp
x 2 dx
,
0
π.
January 2009
15 / 42
The convolution
Theorem
If ξ and η are independent and the density function of ξ is f and the
density function of η is g , then the density function of ξ + η exists and it
is equal to the convolution
Z ∞
∞
f (z ) g (x
Fξ + η (x ) = P ( ξ + η < x ) =
=
Z ∞
∞
Z ∞
∞
z ) dz.
P (ξ + η < x j ξ = z ) dF (z ) =
P (z + η < x j ξ = z ) dF (z ) =
Z ∞
∞
P (η < x
z ) f (z ) dz
Then it is su¢ cient to di¤erentiate.
Medvegyev (CEU)
Poisson processes
January 2009
16 / 42
A remark about the di¤erentiation under the integral sign
A di¤erentiation under the integral sign is a bit problematic, so perhaps
the next calculation is better
Z x Z ∞
∞
∞
f (z ) g (u
z ) dzdu =
=
=
Z ∞
∞
Z ∞
∞
Z ∞
∞
f (z )
f (z )
Z x
∞
g (u
Z x z
∞
P (η < x
z ) dudz =
g (v ) dvdz =
z ) f (z ) dz
And this implies that the density function of Fξ +η is
R∞
f z g u z ) dz.
∞ ( ) (
Medvegyev (CEU)
Poisson processes
January 2009
17 / 42
The convolution of non-negative random variables
If ξ
0 and η
0, then the convolution integral is equal to
Z x
f (z ) g (x
z ) dz.
0
Medvegyev (CEU)
Poisson processes
January 2009
18 / 42
The sum of independent gamma distributions
Theorem
If the distribution of independent variables τ i is Γ (ai , λ) then the
distribution of the sum ∑ni=1 τ i is Γ (∑ni=1 ai , λ) .
Z x
0
λa
(x
Γ (a )
t )a
1
exp ( λ (x
λ a +b
=
exp ( λx )
Γ (a ) Γ (b )
λ a +b
=
exp ( λx )
Γ (a ) Γ (b )
Z 1
t ))
Z x
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(x
t )a
1
exp ( λt ) dt =
1 b 1
dt =
(xz )b
1
t
0
xz )a
(x
1
xdz =
0
λ a +b
=
exp ( λx ) x a +b
Γ (a ) Γ (b )
=
λb b
t
Γ (b )
1
Z 1
(1
1
zb
1
dz =
0
λ a +b
exp ( λx ) x a +b
Γ (a + b )
Poisson processes
z )a
1
.
January 2009
19 / 42
The distribution of the nth jump of a Poisson process
Corollary
The distribution of the time of the n-th jump of a Poisson process is
Γ (n, λ) .
Corollary
The distributions χ2n and Γ (n/2, 1/2) are equal.
Medvegyev (CEU)
Poisson processes
January 2009
20 / 42
The Poisson process and the Poisson distribution
Corollary
If X is a Poisson process and λ is the parameter of the exponential
distribution describing the distribution of the time intervals between the
jumps then
(λt )n
exp ( λt ) .
P (X (t ) = n ) =
n!
Let σn $ ∑nk =1 τ k . The distribution of σn +1 is Γ (n + 1, λ).
Z ∞
λ n +1
P (X (t ) < n + 1 ) = P ( σ n +1 > t ) =
x n exp ( λx ) dx =
Γ (n + 1)
t
"
#∞ Z
∞
λn x n 1
λn +1 x n exp ( λx )
+
n
exp ( λx ) dx =
=
Γ (n + 1)
λ
Γ (n + 1)
t
t
(λt )n
=
exp ( λt ) + P (X (t ) < n) .
n!
Medvegyev (CEU)
Poisson processes
January 2009
21 / 42
The Poisson process and the Poisson distribution
P (X (t ) = n ) = P (X (t ) < n + 1)
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P (X (t ) < n ) =
Poisson processes
(λt )n
exp ( λt ) .
n!
January 2009
22 / 42
Simulation of Poisson distribution
L $ exp ( λ),p $ 1.k=0.
do:
Generate uniform random number u in [0, 1] and p $ p u
k + +.
while p L.
return k 1.
Taking the logarithm ∑ ln ui
distribution of 1/λ ln ui is
P
1
ln ui < x
λ
= P (ln ui >
= 1
that is exponential. So p
waiting time in [0, 1] .
Medvegyev (CEU)
λ that is ∑
1
λ
ln ui
1.The
λx ) = P (ui > exp ( λx )) =
exp ( λx ) ,
1 is the number of the jumps with exponential
Poisson processes
January 2009
23 / 42
The CreditRisk+ model
Let T > 0 be a …xed time interval. The question is that how much loss
can occur during the time period [0, T ]? Assume that
1
the frequency of the events is given distribution, very often a Poisson
distribution and
2
the severity of the losses is given by random variables with common
distribution,
3
the loss events are independent.
The goal is to asses how risky is the portfolio? The risk measure of the
portfolio is some α-quantile, that is the number
Xα $ inf fX : P (Loss > X )
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Poisson processes
αg $ Varα
January 2009
24 / 42
The Laplace transform and the generating function
De…nition
If ξ
0 then the function
L (s ) $ E (exp ( sξ )) =
Z ∞
exp ( sx ) dF (x ) ,
s>0
0
is called the Laplace transform of ξ. If ξ is concentrated on the
f0, 1, 2, . . .g then
G (z ) $ E z ζ
∞
=
∑ z n pn ,
n =0
jz j
1
is called the generating function of ξ.
In both cases the distribution of ξ is well-de…ned by the function.
Medvegyev (CEU)
Poisson processes
January 2009
25 / 42
The loss process
The loss process is a compound process Loss = ∑k∞=0 χ (σk
T ) ξk
where (σk ) are the jump times of the underlying counting process and ξ k
is the loss at time σk . One can calculate the Laplace transform of the total
loss. If P (N = k ) $ pk then
LLoss (s ) $ E (exp ( s Loss )) = E (E (exp ( s Loss ) j N = k )) =
∞
∑ E (exp (
=
k =0
s Loss ) j N = k ) pk =
∞
∑E
=
k =0
∞
=
k
exp
∑ (E (exp (
s
∑ ξi
i =0
!!
pk =
s ξ )))k pk = G Lξ (s )
k =0
where G (z ) $ ∑k∞=0 z k pk and we used that the number of the loss event
is independent of the actual losses.
Medvegyev (CEU)
Poisson processes
January 2009
26 / 42
Laplace transform of the losses
Example
If number of the loss events follows a Poisson distribution then
G (z ) $
∞
∑
k =0
zk
∞
λk
(zλ)k
exp ( λ) = exp ( λ) ∑
=
k!
k!
k =0
= exp ( λ) exp (zλ) = exp (λ (z
1)) .
If the size of the losses has exponential distribution with parameter µ then
Lξ (s ) =
Z ∞
0
= µ
µ exp ( µx ) exp ( sx ) dx = µ
exp ( µ + s ) x
µ+s
∞
=
0
Z ∞
exp (
0
(µ + s ) x ) dx =
µ
.
µ+s
Hence
LLoss (s ) = exp λ
Medvegyev (CEU)
µ
µ+s
1
Poisson processes
= exp
λs
µ+s
January 2009
27 / 42
The Poisson parameter is random
The parameter of the Poisson distribution depends on many external
factors. Hence of course we do not know the exact value of λ. Assume that
P (N = n j λ ) =
λn
exp ( λ) .
n!
Assume that λ has a gamma distribution with parameters (α, β) , where
(α, β) depend on the external factors. What is the distribution of N?
P (N = n) = E (P (N = n j λ)) =
Medvegyev (CEU)
Poisson processes
1
(E (λn exp ( λ)))
n!
January 2009
28 / 42
Mixing with gamma distribution
Using the formula for the density function of the gamma distribution
γ
E (λ exp (λz )) =
=
βα Γ (α + γ)
Z ∞
0
λγ exp (λz )
Z ∞
(β
βα α
λ
Γ (α)
z )α+γ α+γ
λ
Γ (α + γ)
1
exp ( βλ) d λ =
1
exp ( ( β z ) λ) d λ =
0
z)
βα Γ (α + γ)
Γ (α + γ)
=
α+γ 1 = γ
Γ (α) ( β z )
β Γ (α) (1 z/β)α+γ
Γ (α) ( β
α+γ
whenever β > z and α + γ > 0.
Medvegyev (CEU)
Poisson processes
January 2009
29 / 42
Mixing with gamma distribution
If z =
1, γ = n then as Γ (x + 1) = x Γ (x )
1
Γ (α + n) 1
=
n
n!Γ (α) β (1 + 1/β)α+n
1) Γ ( α + n 1) 1
1
=
n
n!Γ (α)
β (1 + 1/β)α+n
P (N = n ) =
=
(α + n
=
n 1
∏ (α + k ) Γ (α)
=
k =0
n!Γ (α)
1
1
$
n
β (1 + 1/β)α+n
α+n
n
1
pα qn
where p = β/ (1 + β) , q = 1/ (1 + β) . The distribution is called negative
binomial distribution. If α = 1 then the gamma distribution is exponential.
In this case
P (N = n) = pq n .
Medvegyev (CEU)
Poisson processes
January 2009
30 / 42
Negative binomial distribution
If n is a natural number then Γ (n) = (n
Therefore by de…nition
1) Γ (n
1) = . . . = (n
1) !
n 1
α+n
n
1
Γ (α + n)
=
$
n!Γ (α)
∏ (α + k )
k =0
n!
.
On the other hand
n 1
α
n
$
α( α
1)
α( α
n!
n
1)
∏ (α + k )
= ( 1)
n k =0
n!
.
Therefore
P (N = n ) = ( 1)n
Medvegyev (CEU)
Poisson processes
α α n
p q .
n
January 2009
31 / 42
Generator function of the negative binomial distribution
Using the formula for the generator function of the Poisson distribution
and the relation between the gamma and the generalized gamma function
G (z ) $ E z N
=
=
Z ∞
0
=
Z ∞
0
Gλ (z )
exp (λ (z
βα
Γ (α)
Z ∞
βα
=
Γ (a, β + 1
Γ (α)
Medvegyev (CEU)
= E E zN j λ
0
λα
$ E (Gλ (z )) =
βα α 1
λ
exp ( βλ) d λ =
Γ (α)
βα α 1
λ
exp ( βλ) d λ =
1))
Γ (α)
1
exp ( λ ( β + 1
z )) d λ =
Γ (α)
βα
=
z) =
Γ (α) ( β + 1 z )α
Poisson processes
β
β+1
α
z
January 2009
32 / 42
The Laplace transform of the losses
Example
If the losses are exponential with parameter µ then
LLoss (s ) = G (Lθ (z )) =
Medvegyev (CEU)
Poisson processes
β
β+1
µ
µ +s
!α
.
January 2009
33 / 42
Generator function of the losses
Assume that the losses are discrete and …nite. Assume that the
distribution of the individual losses is (rj )M
j =0 .
G (z ) $
∞
∑
n =0
P (L = n ) z n $
= E E
N
z ∑ k =1 ξ k
∞
∑ An z n = E
n =0
zL = E E zL j N = k =
j N = k = ∑ E z ∑ i =1 ξ i
k
pk =
k
=
∑
E zξ
k
pk = F (P (z )) ,
k
j
where P (z ) $ ∑M
j =0 z rj is the generator function of the individual losses
and F is the generator function of the loss frequency.
Medvegyev (CEU)
Poisson processes
January 2009
34 / 42
Generator function of the losses
Example
If the possible losses are in millions 1, 2 and 4, and the frequencies are
r1 = 0, 1, r2 = 0, 4 and r4 = 0, 5 are the probabilities then
P (z ) = 0, 1z + 0, 4z 2 + 0z 3 + 0, 5z 4 .
Medvegyev (CEU)
Poisson processes
January 2009
35 / 42
Generator function of the losses
If the frequencies are Poisson then
F (z ) $
∞
∑
k =0
λk
exp ( λ) z k = exp (λ (z
k!
1))
where P (z ) is the generator function of the loss, F is of the loss
frequency. Let use de…ne µj $ λrj
M
P (z ) =
j =0
b (z ) $ λP (z ) =
P
Medvegyev (CEU)
M
µj
j =0
λ
∑ rj z j $ ∑
Poisson processes
zj .
M
∑ µj z j
j =0
January 2009
36 / 42
Leibnitz formula
Theorem
(uv )(n ) = ∑nk =0 (kn )u (k ) v (n
k ).
One can prove it using inducting. For n = 0 and n = 1 the formula holds.
(uv )(n +1 ) =
n
∑
k =0
n
=
∑
k =0
n
n
k
u (k +1 ) v (n
n (n +1 )
v
+ ∑ u (k ) v (n +1
0
k =1
=
n
=
n
k
k)
n + 1 (n +1 )
v
+ ∑ u (k ) v (n +1
0
k =1
n +1
=
∑
k =0
Medvegyev (CEU)
u (k ) v (n
k)
k)
+ u (k ) v (n +1
n
0
=
k)
=
n
k
+
k)
n+1
n + 1 (n +1 )
+
u
=
k
n+1
k
n + 1 (k ) (n +1
u v
k
Poisson processes
1
k)
+
n (n +1 )
u
=
n
.
January 2009
37 / 42
Panjer recursion
Using the formula for the generator function of the Poisson distribution if
1
An $ P (L = n) = G (n ) (0)
n!
then
1 (n )
1 dn 1
1 dn 1
d
d
G (0) =
G
0
=
λG (z ) P (z ) =
(
)
n
1
n
1
n!
n! dz
dz
n! dz
dz
n
1
n
1
1 d
d
1 d
d b
=
G (z ) λP (z ) =
G (z ) P
(z ) =
n
1
n
1
n! dz
dz
n! dz
dz
=
1 n 1 n 1
G (n
n! k∑
k
=0
k 1)
n 1
=
1 n 1
An
k
n!
k =0
∑
k 1
(n
k
b (k +1 ) ( 0 ) =
(0) P
1)! (k + 1)!µk +1 =
n 1
=
Medvegyev (CEU)
k +1
An
n
k =0
∑
k 1 µ k +1 .
Poisson processes
January 2009
38 / 42
Panjer recursion constants
As with simple calculation
1 n 1
(n k 1) ! (k + 1) ! =
n!
k
1
(n 1) !
=
(n k 1) ! (k + 1) ! =
n! k! (n 1 k )!
k +1
=
.
n
Medvegyev (CEU)
Poisson processes
January 2009
39 / 42
Generalization
What does happen if the distribution of the events is not Poisson? Assume
that the generator function of the losses G has the property
d
A (z )
∑r a z k
ln (G (z )) =
= sk =0 k k .
dz
B (z )
∑k =0 bk z
That is
B (z )
Medvegyev (CEU)
d
G (z ) = G (z ) A (z ) .
dz
Poisson processes
January 2009
40 / 42
The case of the negative binomial distribution
Example
Using the generator function of the negative binomial distribution
d
ln
dz
=
α
β
β+1
P (z )
=
d
dz
α ( β + 1 P (z )) α
( β + 1 P (z ))
Medvegyev (CEU)
α
β
β +1 P (z )
β
β +1 P (z )
1
α
P 0 (z )
=
Poisson processes
α
=
d
dz
( β + 1 P (z )) α
( β + 1 P (z )) α
αP 0 (z )
A (z )
$
β + 1 P (z )
B (z )
January 2009
41 / 42
Generalization
s
∑ bk z
k
k =0
For n
are
!
∞
∑ (n + 1) An +1 z
n =0
n
!
r
=
∑ ak z
k
k =0
∞
∑ An z
n =0
n
!
.
0 the terms in z n on the left hand and right hand side respectively
min (s ,n )
∑
min (r ,n )
bj (n + 1
∑
j ) An +1 j ,
j =0
ai An
i
i =0
which implies that
min (r ,n )
b0 (n + 1) An +1 =
∑
min (s ,n )
ai An
i
i =0
∑
bj (n + 1
j ) An +1
j
j =1
which is a recursion for (An ).
Medvegyev (CEU)
Poisson processes
January 2009
42 / 42