Bayesian Inference for Partially Observed Branching Processes

HAL Id: hal-00776926

https://hal.archives-ouvertes.fr/hal-00776926

Preprint submitted on 16 Jan 2013

HAL

is a multi-disciplinary open access archive for the deposit and dissemination of sci- entific research documents, whether they are pub- lished or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers.

L’archive ouverte pluridisciplinaire

destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

Bayesian Inference for Partially Observed Branching Processes

Sophie Donnet, Judith Rousseau

To cite this version:

Sophie Donnet, Judith Rousseau. Bayesian Inference for Partially Observed Branching Processes.

2013. �hal-00776926�

Bayesian Inference for Partially Observed Branching Processes.

Supplementary material

January 2, 2013

A Proof of identifiability

We prove identifiability in the special case corresponding to n

= 0 and X(τ

) unobserved which is the least favorable case. Let θ = (ν

, k = 0, . . . , K) and θ

= (ν

, k = 0, . . . , K) and consider any distributions π and π

on X(τ

), possibly depending on θ and θ

. For N (τ ) = 0, integrating out X(τ

), we obtain

L(N (τ ), T

, . . . , T

; π, θ) = L(N (τ ), X(τ

), T

, . . . , T

, π

, θ

) ⇔

e

E

[e

] = e

E

[e

] (1)

Similarly considering N (τ ) = 1 and integrating out X(τ

) and Z

we obtain

L(T

; θ, π) = ν

e

e

E

[X(τ

)e

] + e

E

[e

] P

k=1

e

ν

Then L(T

; θ, π) = L(T

; θ

, π

) for all T

∈ [τ

, τ

+τ ] leads to the identifiability of the exponents ν

j

for l = 0, · · · , K since the j

’s are all different (see condition C) so that ν

= ν

and ν

= ν

for k ≥ 1. This proves identifiability of the coefficients ν

even though the prior on X(τ

) may be miss-specified.

Identifiability in the cases where X(τ

) and/or

are observed is deduced directly from the previous result.

B Asymptotic study of (X (t))

: proof of Proposition 4 and Theorem 1

In Proposition 4 we give an integral expression of the generating function of the j

-Yule process

with multi-size immigration described in Section 2 of the paper. More precisely, let X(t) by a

branching process such that each particle gives birth to j

particles at random time distributed with E(ν

) and such that groups of j

immigrants arrive at random times distributed with E(ν

), for k = 1 . . . K. We assume that X(0) = x

.

B.1 Proof of Proposition 4

The proof of Proposition 4 can be divided into two parts. Indeed, from one part, the x

particles present at time 0 will give birth to x

independent j

-Yule process (without immigration). Once the processes derived from those particles have been taken into account, we can reduce the study to a j

-Yule process with multi-size immigration starting from 0 particles.

In lemma 1, we recall the expression of the generating function of a j

-Yule process starting with one particle and study its asymptotic distribution. The generating function of a j

-Yule process with multi-size immigration starting with X(0) = 0 particle is then detailed. The proof is similar to the one given in (1) but adapted to our particular case. In the particular case where the sizes of the immigration groups are proportional to j

we derive a explicit expression of the limit distribution in subsection B.2.

Lemma 1.

Let Y (t) be a branching process starting with Y (0) particle such that each particle gives birth to j

particles at interval time distributed with an exponential distribution of parameter ν

. Then we have Ψ(s, t) = E[s

] =

1 − (1 − s

) exp(ν

j

t)

. Moreover,

t→∞

lim e

Y (t) = Γ

Y (0) j

, 1 j

(L)

Proof of Lemma 1

We first assume that Y (0) = 1. By construction of the process Y (t),

Q

(h) = P (Y (t + h) = j|Y (t) = i) only depends on h, i, j. When h is small, Q

(h) verifies:

Q

(h) =



 

 



 

 



ν

ih + o(h) if j = i + j

1 − ν

ih + o(h) if j = i

o(h) if j / ∈ {i, i + j

}

(2)

We now derive a partial differential equation fulfilled by the probability-generating function.

Using the fact that Y (t) take its values in {j

k + 1, k ∈

} we have:

Ψ(s, t) = E[s

] = X

k∈N

P(Y (t) = j

k + 1|Y (0) = 1)s

= X

k∈N

Q

(t)s

(3)

Using a backward-equation we derive an expression for Q

(t) = P (Y (t) = j

k+ 1). Indeed,

Q

(t+h) = P(Y (t+h) = j

k+1|Y (0) = 1) = Q

(t)(1−ν

h)+Q

(t)ν

h +o(h)

using (2). We directly obtain the following ODE:

Q

(t) = lim

h→0

Q

(t + h) − Q

(t)

h = −ν

Q

(t) + ν

Q

(t) (4)

We now derive from (4) and (3) a partial differential equation for Ψ(s, t)

∂

∂t Ψ(s, t) = s X

k∈N

Q

(t)(s

)

=

∞

X

k=1

s

[−ν

Q

(t) + ν

Q

(t)]

= −ν

Ψ(s, t) + ν

∞

X

k=1

s

Q

(t) (5)

P

k=1

s

Q

(t) is the probability-generating function of a process with the same

division mechanism as Y (t) by which would start with j

+ 1 particles. By properties inherent

to branching processes, this is equivalent to the sum of j

+ 1 independent processes starting

with one particle. As a consequence, let Y

, . . . Y

by independent processes distributed as Z ,

we have:

∞

X

k=1

s

Q

(t) = E[s

] = E[s

]

= Ψ(s, t)

As a consequence the partial differential equation (5) becomes :

Ψ(s, t) = ν

Ψ(s, t)(Ψ

(s, t) − 1) with Ψ(s, 0) = s. The solution of this equation is :

Ψ(s, t) =

1 − e

(1 − s

)

We now study the asymptotic distribution of ˜ Z(t) = e

Y (t) through its moments-generating function :

Φ

(θ) = E[e

] = E[e

] = E[(e

)

] = E[s

] = Ψ(s

, t)

where s

= e

'

1 + θe

. We easily obtain the following limit

t→∞

lim Φ

(θ) = 1 (1 − j

θ)

and recognize the moment generating function of the Γ

1 j0

,

0

.

Now, if the process starts with Y (0) = X(0) particles, each of them initiates a j

-Yule process which is independent of the other ones, leading to :

Ψ(s, t) =

1 − (1 − s

) exp(ν

j

t)

and

t→∞

lim e

Y (t) = Γ

X(0) j

, 1

j

(L)

We now use Lemma 1 to study the distribution of X(t), the number of particles issued from the

multi-immigration j

- Yule process described in the paper. We first assume that X(0) = 0. Let

φ(s, t) denote the probability-generating function function of (X(t))

φ(s, t) = E[s

] =

∞

X

n=0

P

(t)s

where P

(t) = P (X(t) = n) is the probability to have n particles at time t. This probability can be decomposed into :

P

(t) = P

(t)m

(t) +

∞

X

k=1

P

(t)m

(t) (6)

where m

(t) is the probability that k immigration groups arrived in the time interval [0, t) and P

(t) denotes the probability there are n particles at time t given that k immigration groups arrived during [0, t). Moreover P

(t) = δ

because X(0) = 0.

Using the independence of the immigration events, P

(t) can also be decomposed as :

P

(t) = X

i1+···+i_k=n

U

(t) . . . U

(t) (7)

where U

(t) denotes the probability that an immigration group leads (by the branching mech- anism) to m particles at time t given that the group immigrates during the interval [0, t).

Combining (7) and (6), we can rewrite the probability-generating function φ(s, t):

φ(s, t) =

∞

X

n=0

P

(t)s

= m

(t) +

∞

X

n=0

s

∞

X

k=1

m

(t) X

i1+···+i_k=n

U

(t) . . . U

(t)

=

∞

X

k=0

m

(t)

∞

X

n=0

s

U

(t)

!

Denoting J(s, t) = P

n=0

s

U

(t), we obtain:

φ(s, t) =

∞

X

k=0

m

(t)J

(s, t) (8)

where m

(t) is the probability that k immigration groups arrived in the time interval [0, t).

Using the Poisson properties of our immigration process we have : m

(t) = e

with

µ = ν

+ · · · + ν

and so by (8):

φ(s, t) =

∞

X

k=0

e

(µt)

k! J

(s, t) = e

exp{µtJ (s, t)} (9)

We now compute J (s, t) and to that purpose we first study U

(t). Recall that U

(t) is the probability that an immigration group leads (by the branching mechanism) to n particles at time t given that the group immigrates during the interval [0, t). As a consequence, using the infinitesimal probabilities, U

(t) can be decomposed into :

U

(t) = Z

0 K

X

k=1

r

(u)Q(n, t|j

, u)d

N (u|t) (10)

where

• d

N (u|t) is the conditional infinitesimal immigration rate i.e. the probability that there is exactly one immigration group during the infinitesimal interval [u, u + du) ⊂ [0, t) given there is exactly one immigration group in the interval [0, t). In the case of a Poisson process, d

N (u|t) =

• r

(u) is the probablity that the immigration group is of size j

given that it arrived at time u, k = 1, . . . , K . Using the Poisson properties of our immigration process we have :

r

(u) = ν

ν

+ · · · + ν

= ν

µ

and so is independent of u.

• Q(n, t|j

, u) denotes the probability that an immigration occuring at time u and consisting of j

particles leads to n particles at time t. Q(n, t|j

, u) only relies on the branching part of the process and can be decomposed as previously:

Q(n, t|j

, u) = X

i1+···+i_jk=n

Q

(t − u) . . . Q

(t − u) (11)

where Q

(t) is the probability that one particle leads to i particles by division process in a period of length t. Indeed, once arrived, each particle is the initial point of a branching process which evolves independently during the remaining time t − u.

As a consequence, we can express J (s, t) as:

J (s, t) =

∞

X

n=0

s

U

(t) =

∞

X

n=0 K

X

k=1

ν

µ s

Z

Q(n, t|j

, u) du t

=

K

X

k=1

ν

µ

Z

∞

X

n=0

s

X

i1+···+i_jk=n

Q

(t − u) . . . Q

(t − u) du t

=

K

X

k=1

ν

µ

Z

"

X

n=0

s

Q

(t − u)

#

du t .

Let Ψ(s, t) = P

n=0

s

Q

(t) be the probability-generating function of a j

-Yule process without immigration, starting with one particle, then

J(s, t) =

K

X

k=1

ν

µ

Z

(Ψ(s, t − u))

du

t (12)

which combined with Lemma 1 leads to :

J (s, t) =

K

X

k=1

ν

µ

Z

1 − (1 − s

) exp(ν

j

(t − u))

du t

Substituting v = (1 − s

) exp(ν

j

(t − u)) into the integral gives :

Z

1 − (1 − s

) exp(ν

j

(t − u))

du

t = 1

tν

j

Z

sj0) exp(ν0j0t) 1− ¹

sj0

1

(1 − v)

v dv

and proposition 4 is proved.

B.2 Proof of Theorem 1

Recall that assumption A states that ∀k ∈ {1 . . . K},

0

= r

∈

. The usual decomposition of

1

(1−v)^rkv

into fractions leads to

Z

1 − (1 − s

) exp(ν

j

(t − τ ))

dτ

t = 1

tν

j

"

log(v) − log(1 − v) +

rk−1

X

l=1

1/l (1 − v)

#

sj0) exp(ν0j0t)

1− ¹

sj0

.

Introducing the following notations :

J

(s, t) = 1 tν

j

[log(v) − log(1 − v)]

1

sj0) exp(ν0j0t) 1− ¹

sj0

= − 1 tν

j

log

1 − s

(1 − e

)

And for 1 ≤ l ≤ R

− 1

J

(s, t) = 1 tν

j

1/l (1 − v)

sj0) exp(ν0j0t)

1− ¹

sj0

= 1

tlν

j

"

1 − (1 − 1

s

)e

− s

#

we can write : J (s, t) =

P

k=1

ν

P

l=0

J

(s, t). A rearrangement is the sums (using the fact that r

< r

< · · · < r

) leads to :

J(s, t) =

(ν

+ · · · + ν

)(J

(s, t) + J

(s, t) + · · · + J

(s, t)) +

(ν

+ · · · + ν

)(J

(s, t) + · · · + J

(s, t))

+ · · ·

+

ν

(J

(s, t) + · · · + J

(s, t))

Setting

α

=



 

 

1 for 0 ≤ l ≤ r

− 1

νk+···+ν_K

µ

for any r

≤ l ≤ r

− 1 and for any k = 2 . . . K we have the following expression for the probability-generating function of X(t):

φ(s, t) = e

exp{µtJ (s, t)} = exp

"

−µt + µt

rK−1

X

l=0

α

J

(s, t)

#

We can now study the asymptotic comportment of ˜ X(t) = e

X(t). Let Φ

(θ) = E[e

] be the moment generating function of ˜ X(t). We have :

Φ

(θ) = E[(e

)

] = φ(e

, t) = exp

"

−µt + µt

rK−1

X

l=0

α

J

(s

, t)

#

(13)

with s

= exp

θe

'

1 + θe

. We study the convergence of each term of the product.

J

(s

, t) = −

0j0

log h

1 − s

(1 − e

) i

and so exp [−µt + µtJ

(s

, t)] = h

e

− s

(e

− 1) i

ν0j0

. Using s

= e

≈

1 + j

θe

, we obtain :

exp [−µt + µtJ

(s

, t)] ≈

(1 − j

θ)

−µ ν0j0

.

We know consider the terms of the form φ

(s

, t) = exp [µtα

J

(s

, t)]. We have

φ

(s

, t) = exp µα

lν

j

1 − (1 − s

)e

− s

≈

exp µα

lν

j

h

(1 − j

θ)

− 1 i

Finally for all θ ≤ 0,

t→∞

lim E[e

] = (1 − j

θ)

−µ ν0j0

rK−1

Y

l=1

exp µα

lν

j

h

(1 − j

θ)

− 1 i

= E[e

]

rK−1

Y

l=1

E[e

]

where Y

∼ Γ

ν0j0

,

0

with moment generating function equal to M(θ) = (1 − j

θ)

−µ

ν0j0

. More- over Y

is such that its moment generating function is exp

h

(1 − j

θ)

− 1 i

which can be

rewritten as

exp µα

lν

j

h

(1 − j

θ)

− 1 i

=

∞

X

k=0

exp

− µα

lν

j

µα

lν

j

1 k!

1 (1 − j

θ)

=

∞

X

k=0

ρ

1 (1 − j

θ)

where ρ

= exp h

−

0j0

i

lν0j0

k!

is the probability that a Poisson random variable of pa- rameter

0j0

is equal to k. So Y

is distributed with an infinite mixture of Γ kl,

0

with Poisson weights. Finally, e

X(t) converges in distribution towards a sum of r

inde- pendent variables P

l=0

Y

where the Y

∼ Γ

µ ν0j0

,

0

and Y

∼ P

k=0

ρ

Γ kl,

0

with ρ

= exp

h

−

0j0

i

.

References

[1] Ron Shonkwiler. On age-dependent branching processes with immigration. Comput. Math.

Appl., 6(3):289–296, 1980.