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Vol. 14 (2009), Paper no. 87, pages 2492–2526.
Journal URL
http://www.math.washington.edu/~ejpecp/

Asymptotic analysis for bifurcating autoregressive
processes via a martingale approach
Bernard Bercu∗

Benoîte de Saporta†

Anne Gégout-Petit‡

Abstract
We study the asymptotic behavior of the least squares estimators of the unknown parameters
of general pth-order bifurcating autoregressive processes. Under very weak assumptions on
the driven noise of the process, namely conditional pair-wise independence and suitable moment conditions, we establish the almost sure convergence of our estimators together with the
quadratic strong law and the central limit theorem. All our analysis relies on non-standard
asymptotic results for martingales.

Key words: bifurcating autoregressive process ; tree-indexed times series; martingales ; least
squares estimation ; almost sure convergence ; quadratic strong law; central limit theorem.
AMS 2000 Subject Classification: Primary 60F15; Secondary: 60F05, 60G42.
Submitted to EJP on September 9, 2008, final version accepted November 2, 2009.

∗

Université de Bordeaux, IMB, CNRS, UMR 5251, and INRIA Bordeaux, team CQFD, France, bernard.bercu@math.ubordeaux1.fr
†
Université de Bordeaux, GREThA, CNRS, UMR 5113, IMB, CNRS, UMR 5251, and INRIA Bordeaux, team CQFD,
France, saporta@math.u-bordeaux1.fr
‡

Université de Bordeaux, IMB, CNRS, UMR 5251, and INRIA Bordeaux, team CQFD, France, anne.petit@ubordeaux2.fr

2492

1

Introduction

Bifurcating autoregressive (BAR) processes are an adaptation of autoregressive (AR) processes to
binary tree structured data. They were first introduced by Cowan and Staudte [2] for cell lineage
data, where each individual in one generation gives birth to two offspring in the next generation.
Cell lineage data typically consist of observations of some quantitative characteristic of the cells over
several generations of descendants from an initial cell. BAR processes take into account both inherited and environmental effects to explain the evolution of the quantitative characteristic under study.
More precisely, the original BAR process is defined as follows. The initial cell is labelled 1, and the
two offspring of cell n are labelled 2n and 2n + 1. Denote by X n the quantitative characteristic of
individual n. Then, the first-order BAR process is given, for all n ≥ 1, by
¨
X 2n
= a + bX n + ǫ2n ,
X 2n+1 = a + bX n + ǫ2n+1 .

The noise sequence (ǫ2n , ǫ2n+1 ) represents environmental effects while a, b are unknown real
parameters with |b| < 1. The driven noise (ǫ2n , ǫ2n+1 ) was originally supposed to be independent
and identically distributed with normal distribution. However, two sister cells being in the same
environment early in their lives, ǫ2n and ǫ2n+1 are allowed to be correlated, inducing a correlation
between sister cells distinct from the correlation inherited from their mother.
Several extensions of the model have been proposed. On the one hand, we refer the reader to
Huggins and Basawa [10] and Basawa and Zhou [1; 15] for statistical inference on symmetric
bifurcating processes. On the other hand, higher order processes, when not only the effects of
the mother but also those of the grand-mother and higher order ancestors are taken into account,
have been investigated by Huggins and Basawa [10]. Recently, an asymmetric model has been
introduced by Guyon [5; 6] where only the effects of the mother are considered, but sister cells are
allowed to have different conditional distributions. We can also mention a recent work of Delmas
and Marsalle [3] dealing with a model of asymmetric bifurcating Markov chains on a Galton Watson
tree instead of regular binary tree.
The purpose of this paper is to carry out a sharp analysis of the asymptotic properties of the
least squares (LS) estimators of the unknown parameters of general asymmetric pth-order BAR
processes. There are several results on statistical inference and asymptotic properties of estimators
for BAR models in the literature. For maximum likelihood inference on small independent trees, see
Huggins and Basawa [10]. For maximum likelihood inference on a single large tree, see Huggins
[9] for the original BAR model, Huggins and Basawa [11] for higher order Gaussian BAR models,

and Zhou and Basawa [15] for exponential first-order BAR processes. We also refer the reader to
Zhou and Basawa [14] for the LS parameter estimation, and to Hwang, Basawa and Yeo [12] for
the local asymptotic normality for BAR processes and related asymptotic inference. In all those
papers, the process is supposed to be stationary. Consequently, X n has a time-series representation
involving an holomorphic function. In Guyon [5], the LS estimator is also investigated, but the
process is not stationary, and the author makes intensive use of the tree structure and Markov
chain theory. Our goal is to improve and extend the previous results of Guyon [5] via a martingale
approach. As previously done by Basawa and Zhou [1; 14; 15] we shall make use of the strong
law of large numbers [4] as well as the central limit theorem [7; 8] for martingales. It will allow
2493

us to go further in the analysis of general pth-order BAR processes. We shall establish the almost
sure convergence of the LS estimators together with the quadratic strong law and the central limit
theorem.
The paper is organised as follows. Section 2 is devoted to the presentation of the asymmetric
pth-order BAR process under study, while Section 3 deals with the LS estimators of the unknown
parameters. In Section 4, we explain our strategy based on martingale theory. Our main results
about the asymptotic properties of the LS estimators are given in Section 5. More precisely, we shall
establish the almost sure convergence, the quadratic strong law (QSL) and the central limit theorem
(CLT) for the LS estimators. The proof of our main results are detailed in Sections 6 to 10, the more

technical ones being gathered in the appendices.

2

Bifurcating autoregressive processes

In all the sequel, let p be a non-zero integer. We consider the asymmetric BAR(p) process given, for
all n ≥ 2 p−1 , by
Pp
(
n
ǫ2n ,
X 2n
= a0 +
ak X [ k−1
] +
2
P k=1
(2.1)
p

n
X 2n+1 = b0 +
k=1 bk X [ k−1 ] + ǫ2n+1 ,
2

where [x] stands for the largest integer less than or equal to x. The initial states {X k , 1 ≤ k ≤
2 p−1 − 1} are the ancestors while (ǫ2n , ǫ2n+1 ) is the driven noise of the process. The parameters
(a0 , a1 , . . . a p ) and (b0 , b1 , . . . , b p ) are unknown real numbers. The BAR(p) process can be rewritten
in the abbreviated vector form given, for all n ≥ 2 p−1 , by
¨
X2n
= AXn + η2n ,
(2.2)
X2n+1 = BXn + η2n+1 ,
where the regression vector Xn = (X n , X [ n ] , . . . , X [
2

n
]
2 p−1

) t , η2n = (a0 + ǫ2n )e1 , η2n+1 = (b0 + ǫ2n+1 )e1

with e1 = (1, 0, . . . , 0) t ∈ R p . Moreover, A and B are the p × p companion matrices




b1 b2 · · · a p
a1 a2 · · · a p




 1 0 ··· 0 
 1 0 ··· 0 
B=
A=
. .
. ,

.. ..
.. ..


. .. 
.
. .. 
.


 0
 0
0 0
1
0
0 0
1
0
This process is a direct generalization of the symmetric BAR(p) process studied by Huggins, Basawa
and Zhou [10; 14]. One can also observe that, in the particular case p = 1, it is the asymmetric

BAR process studied by Guyon [5; 6]. In all the sequel, we shall assume that E[X k8 ] < ∞ for all
1 ≤ k ≤ 2 p−1 − 1 and that matrices A and B satisfy the contracting property
β = max{kAk, kBk} < 1,
where kAk = sup{kAuk, u ∈ R p with kuk = 1}.

As explained in the introduction, one can see this BAR(p) process as a pth-order autoregressive
process on a binary tree, where each vertex represents an individual or cell, vertex 1 being the
original ancestor, see Figure 1 for an illustration. For all n ≥ 1, denote the nth generation by
Gn = {2n , 2n + 1, . . . , 2n+1 − 1}.
2494

Figure 1: The tree associated with the bifurcating auto-regressive process.
In particular, G0 = {1} is the initial generation and G1 = {2, 3} is the first generation of offspring
from the first ancestor. Let G rn be the generation of individual n, which means that rn = log2 (n).
Recall that the two offspring of individual n are labelled 2n and 2n + 1, or conversely, the mother of
individual n is [n/2]. More generally, the ancestors of individual n are [n/2], [n/22 ], . . . , [n/2 rn ].
Furthermore, denote by
n
[
Tn =

Gk
k=0

the sub-tree of all individuals from the original individual up to the nth generation. It is clear
that the cardinality |Gn | of Gn is 2n while that of Tn is |Tn | = 2n+1 − 1. Finally, we denote by
Tn,p = {k ∈ Tn , k ≥ 2 p } the sub-tree of all individuals up to the nth generation without T p−1 . One
can observe that, for all n ≥ 1, Tn,0 = Tn and, for all p ≥ 1, T p,p = G p .

3

Least-squares estimation

The BAR(p) process (2.1) can be rewritten, for all n ≥ 2 p−1 , in the matrix form
Zn = θ t Yn + Vn
where

‚
Zn =

X 2n
X 2n+1

Œ

‚
,

Yn =

1
Xn

2495

(3.1)

Œ

‚
,

Vn =

ǫ2n
ǫ2n+1

Œ
,

and the (p + 1) × 2 matrix parameter θ is given by


a0 b0


 a1 b1 

θ = .
.. 
.
..
. 

ap bp
Our goal is to estimate θ from the observation of all individuals up to the nth generation that is
the complete sub-tree Tn . Each new generation Gn contains half the global available information.
Consequently, we shall show that observing the whole tree Tn or only generation Gn is almost the
same. We propose to make use of the standard LS estimator θbn which minimizes
1 X
∆n (θ ) =
k Zk − θ t Yk k2 .
2 k∈T
n−1,p−1

Consequently, we obviously have for all n ≥ p
−1
θbn = Sn−1

X

Yk Zkt ,

(3.2)

k∈Tn−1,p−1

where the (p + 1) × (p + 1) matrix Sn is defined as
Sn =

X

Yk Ykt

=

X

‚

k∈Tn,p−1

k∈Tn,p−1

1
Xk

Xkt
Xk Xkt

Œ
.

In the special case where p = 1, Sn simply reduces to
‚
Œ
X
1 Xk
Sn =
.
X k X k2
k∈Tn

In order to avoid useless invertibility assumption, we shall assume, without loss of generality, that
for all n ≥ p − 1, Sn is invertible. Otherwise, we only have to add the identity matrix I p+1 to Sn . In
all what follows, we shall make a slight abuse of notation by identifying θ as well as θbn to




b0,n
a
a0
 . 
 . 
 .. 
 .. 




 a

 ap 
b
p,n 

bn ) = 
and
vec(
θ
vec(θ ) = 
 b
.
 b0 
 b0,n 


 . 
 .. 
 .. 
 . 


bb p,n
bp
The reason for this change will be explained in Section 4. Hence, we readily deduce from (3.2) that
X
€
Š
−1
vec Yk Zkt
θbn = (I2 ⊗ Sn−1
)
k∈Tn−1,p−1

−1
= (I2 ⊗ Sn−1
)

X
k∈Tn−1,p−1

2496







X 2k
X k X2k
X 2k+1
X k X2k+1




,


where ⊗ stands for the matrix Kronecker product. Consequently, it follows from (3.1) that
X
€
Š
−1
vec Yk Vkt
θbn − θ = (I2 ⊗ Sn−1
)
k∈Tn−1,p−1

−1
= (I2 ⊗ Sn−1
)

X
k∈Tn−1,p−1







ǫ2k
ǫ2k Xk
ǫ2k+1
ǫ2k+1 Xk




.


(3.3)

Denote by F = (Fn ) the natural filtration associated with the BAR(p) process, which means that Fn
is the σ-algebra generated by all individuals up to the nth generation, Fn = σ{X k , k ∈ Tn }. In all
the sequel, we shall make use of the five following moment hypotheses.
(H.1) One can find σ2 > 0 such that, for all n ≥ p − 1 and for all k ∈ Gn+1 , ǫk belongs to L 2 with
E[ǫk |Fn ] = 0

and

E[ǫk2 |Fn ] = σ2

a.s.

(H.2) It exists |ρ| < σ2 such that, for all n ≥ p − 1 and for all different k, l ∈ Gn+1 with [k/2] =
[l/2],
E[ǫk ǫl |Fn ] = ρ
a.s.
Otherwise, ǫk and ǫl are conditionally independent given Fn .
(H.3) For all n ≥ p − 1 and for all k ∈ Gn+1 , ǫk belongs to L 4 and
sup

sup E[ǫk4 |Fn ] < ∞

a.s.

n≥p−1 k∈Gn+1

(H.4) One can find τ4 > 0 such that, for all n ≥ p − 1 and for all k ∈ Gn+1 ,
E[ǫk4 |Fn ] = τ4

a.s.

and, for ν 2 < τ4 and for all different k, l ∈ Gn+1 with [k/2] = [l/2]
2 2
E[ǫ2k
ǫ2k+1 |Fn ] = ν 2

a.s.

(H.5) For all n ≥ p − 1 and for all k ∈ Gn+1 , ǫk belongs to L 8 with
sup

sup E[ǫk8 |Fn ] < ∞

a.s.

n≥p−1 k∈Gn+1

Remark 3.1. In contrast with [14], one can observe that we do not assume that (ǫ2n , ǫ2n+1 ) is a
sequence of independent and identically distributed bi-variate random vectors. The price to pay for
giving up this iid assumption is higher moments, namely assumptions (H.3) and (H.5). Indeed we need
them to make use of the strong law of large numbers and the central limit theorem for martingales.
However, we do not require any normality assumption on (ǫ2n , ǫ2n+1 ). Consequently, our assumptions
are much weaker than the existing ones in previous literature.

2497

We now turn to the estimation of the parameters σ2 and ρ. On the one hand, we propose to estimate
the conditional variance σ2 by
b 2n =
σ

X

1

2|Tn−1 | k∈T

bk k2 =
kV

n−1,p−1

X

1

2|Tn−1 | k∈T

2
2
(b
ǫ2k
+ ǫb2k+1
)

(3.4)

n−1,p−1

b t = (b
where for all n ≥ p − 1 and for all k ∈ Gn , V
ǫ2k , ǫb2k+1 ) with
k


 ǫb2k

=

b0,n
− a

X 2k

−

 ǫb2k+1 = X 2k+1 − bb0,n −

Pp

bi,n X [
i=1 a

Pp

i=1

bbi,n X
[

k
]
2i−1
k
]
2i−1

,
.

One can observe that, on the above equations, we make use of only the past observations for the
estimation of the parameters. This will be crucial in the asymptotic analysis. On the other hand, we
estimate the conditional covariance ρ by
bn =
ρ

4

1

X

|Tn−1 | k∈T

ǫb2k ǫb2k+1 .

(3.5)

n−1,p−1

Martingale approach

In order to establish all the asymptotic properties of our estimators, we shall make use of a martingale approach. It allows us to impose a very smooth restriction on the driven noise (ǫn ) compared
with the previous results in the literature. As a matter of fact, we only assume suitable moment
conditions on (ǫn ) and that (ǫ2n , ǫ2n+1 ) are conditionally independent, while it is assumed in [14]
that (ǫ2n , ǫ2n+1 ) is a sequence of independent identically distributed random vectors. For all n ≥ p,
denote


ǫ2k

X 
 ǫ2k Xk 
Mn =
 ∈ R2(p+1) .

ǫ2k+1 

k∈Tn−1,p−1
ǫ2k+1 Xk
−1
Let Σn = I2 ⊗ Sn , and note that Σ−1
n = I2 ⊗ Sn . For all n ≥ p, we can thus rewrite (3.3) as

θbn − θ = Σ−1
n−1 Mn .

(4.1)

The key point of our approach is that (Mn ) is a martingale. Most of all the asymptotic results for
martingales were established for vector-valued martingales. That is the reason why we have chosen
to make use of vector notation in Section 3. In order to show that (Mn ) is a martingale adapted
to the filtration F = (Fn ), we rewrite it in a compact form. Let Ψn = I2 ⊗ Φn , where Φn is the
rectangular matrix of dimension (p + 1) × δn , with δn = 2n , given by
!
1
1
···
1
Φn =
.
X2n X2n +1 · · · X2n+1 −1

2498

It contains the individuals of generations Gn−p+1 up to Gn and is also the collection of all Yk , k ∈ Gn .
Let ξn be the random vector of dimension δn







ξn = 







ǫ2 n
ǫ2n +2
..
.
ǫ2n+1 −2
ǫ2n +1
ǫ2n +3
..
.








.







ǫ2n+1 −1
The vector ξn gathers the noise variables of generation Gn . The special ordering separating odd and
even indices is tailor-made so that Mn can be written as
Mn =

n
X

Ψk−1 ξk .

k=p

By the same token, one can observe that
Sn =

n
X

Φk Φkt

and

Σn =

n
X

Ψk Ψkt .

k=p−1

k=p−1

Under (H.1) and (H.2), we clearly have for all n ≥ 0, E[ξn+1 |Fn ] = 0 and Ψn is Fn -measurable. In
t
addition, it is not hard to see that for all n ≥ 0, E[ξn+1 ξn+1
|Fn ] = Γ ⊗ Iδn where Γ is the covariance
matrix associated with (ǫ2n , ǫ2n+1 )
!
σ2 ρ
.
Γ=
ρ σ2
We shall also prove that (Mn ) is a square integrable martingale. Its increasing process is given for
all n ≥ p + 1 by
n =

n−1
X
k=p−1

Ψk (Γ ⊗ Iδk )Ψkt = Γ ⊗

n−1
X
k=p−1

Φk Φkt = Γ ⊗ Sn−1 .

It is necessary to establish the convergence of Sn , properly normalized, in order to prove the asympbn . One can observe that the sizes of Ψn and ξn
b 2n and ρ
totic results for the BAR(p) estimators θbn , σ
are not fixed and double at each generation. This is why we have to adapt the proof of vector-valued
martingale convergence given in [4] to our framework.

5

Main results

We now state our main results, first on the martingale (Mn ) and then on our estimators.
2499

Proposition 5.1. Assume that (ǫn ) satisfies (H.1) to (H.3). Then, we have
lim

n→∞

Sn
|Tn |

=L

(5.1)

a.s.

where L is a positive definite matrix specified in Section 7.
This result is the keystone of our asymptotic analysis. It enables us to prove sharp asymptotic
properties for (Mn ).
Theorem 5.1. Assume that (ǫn ) satisfies (H.1) to (H.3). Then, we have
Mnt Σ−1
n−1 Mn = O (n)

(5.2)

a.s.

In addition, we also have
lim

n
1X

n→∞

n k=p

Mkt Σ−1
M = 2(p + 1)σ2
k−1 k

(5.3)

a.s.

Moreover, if (ǫn ) satisfies (H.4) and (H.5), we have the central limit theorem
p

1
|Tn−1 |

L

Mn −→ N (0, Γ ⊗ L).

(5.4)

From the asymptotic properties of (Mn ), we deduce the asymptotic behavior of our estimators. Our
first result deals with the almost sure asymptotic properties of the LS estimator θbn .
Theorem 5.2. Assume that (ǫn ) satisfies (H.1) to (H.3). Then, θbn converges almost surely to θ with
the rate of convergence


log |Tn−1 |
2
b
k θn − θ k = O
a.s.
(5.5)
|Tn−1 |
In addition, we also have the quadratic strong law
lim

n
1X

n→∞

n k=1

|Tk−1 |(θbk − θ ) t Λ(θbk − θ ) = 2(p + 1)σ2

(5.6)

a.s.

where Λ = I2 ⊗ L.
Our second result is devoted to the almost sure asymptotic properties of the variance and covariance
bn . Let
b 2n and ρ
estimators σ
σ2n =

X

1

2|Tn−1 | k∈T

2
2
(ǫ2k
+ ǫ2k+1
) and ρn =

n−1,p

1

X

|Tn−1 | k∈T

ǫ2k ǫ2k+1 .

n−1,p

b 2n converges almost surely to σ2 .
Theorem 5.3. Assume that (ǫn ) satisfies (H.1) to (H.3). Then, σ
More precisely,
lim

n→∞

1

X

n k∈T

(b
ǫ2k − ǫ2k )2 + (b
ǫ2k+1 − ǫ2k+1 )2 = 2(p + 1)σ2

n−1,p

2500

a.s.

(5.7)

|Tn |

b 2n − σ2n ) = 2(p + 1)σ2
(σ
n
bn converges almost surely to ρ
In addition, ρ
lim

n→∞

lim

n→∞

1

X

n k∈T

(5.8)

a.s.

(b
ǫ2k − ǫ2k )(b
ǫ2k+1 − ǫ2k+1 ) = (p + 1)ρ

a.s.

(5.9)

n−1,p

lim

n→∞

|Tn |
n

bn − ρn ) = 2(p + 1)ρ
(ρ

a.s.

(5.10)

b 2n and ρ
bn .
Our third result concerns the asymptotic normality for all our estimators θbn , σ
Theorem 5.4. Assume that (ǫn ) satisfies (H.1) to (H.5). Then, we have the central limit theorem
p

L
|Tn−1 |(θbn − θ ) −→ N (0, Γ ⊗ L −1 ).

(5.11)

In addition, we also have
p
and

b 2n
|Tn−1 |(σ
p



L

2

− σ ) −→ N 0,

τ4 − 2σ4 + ν 2 
2

L

bn − ρ) −→ N (0, ν 2 − ρ 2 ).
|Tn−1 |(ρ

(5.12)

(5.13)

The rest of the paper is dedicated to the proof of our main results. We start by giving laws of large
numbers for the noise sequence (ǫn ) in Section 6. In Section 7, we give the proof of Proposition 5.1.
Sections 8, 9 and 10 are devoted to the proofs of Theorems 5.2, 5.3 and 5.4, respectively. The more
technical proofs, including that of Theorem 5.1, are postponed to the Appendices.

6

Laws of large numbers for the noise sequence

We first need to establish strong laws of large numbers for the noise sequence (ǫn ). These results
will be useful in all the sequel. We will extensively use the strong law of large numbers for locally
square integrable real martingales given in Theorem 1.3.15 of [4].
Lemma 6.1. Assume that (ǫn ) satisfies (H.1) and (H.2). Then
lim

X

1

n→+∞

|Tn | k∈T

ǫk = 0

(6.1)

a.s.

n,p

In addition, if (H.3) holds, we also have
lim

n→+∞

and
lim

n→+∞

1

X

1

|Tn | k∈T

(6.2)

a.s.

n,p

X

|Tn−1 | k∈T

ǫk2 = σ2

ǫ2k ǫ2k+1 = ρ

n−1,p−1

2501

a.s.

(6.3)

Proof: On the one hand, let

X

Pn =

ǫk =

n X
X

ǫi .

k=p i∈Gk

k∈Tn,p

We have
∆Pn+1 = Pn+1 − Pn =

X

ǫk .

k∈Gn+1

Hence, it follows from (H.1) and (H.2) that (Pn ) is a square integrable real martingale with increasing process
n
X
|Gk | = (σ2 + ρ)(|Tn | − |T p−1 |).
< P >n = (σ2 + ρ)
k=p

Consequently, we deduce from Theorem 1.3.15 of [4] that Pn = o(< P >n ) a.s. which implies (6.1).
On the other hand, denote
n
X
1 X
Qn =
ei ,
|Gk | i∈G
k=p
k

where en = ǫn2 − σ2 . We have
∆Q n+1 = Q n+1 − Q n =

X

1

|Gn+1 | k∈G

ek .

n+1

First of all, it follows from (H.1) that for all k ∈ Gn+1 , E[ek |Fn ] = 0 a.s. In addition, for all different
k, l ∈ Gn+1 with [k/2] 6= [l/2],
E[ek el |Fn ] = 0
a.s.
thanks to the conditional independence given by (H.2). Furthermore, we readily deduce from (H.3)
that
a.s.
sup sup E[ek2 |Fn ] < ∞
n≥p−1 k∈Gn+1

Therefore, (Q n ) is a square integrable real martingale with increasing process
n ≤ 2
≤ 2
≤ 2

sup

sup E[ei2 |Fk ]

p−1≤k≤n−1 i∈Gk+1

sup

sup E[ei2 |Fk ]

p−1≤k≤n−1 i∈Gk+1

sup

n
X
1

|G j |

j=p
n 
X
j=p

sup E[ei2 |Fk ] < ∞

1j
2

a.s.

a.s.
a.s.

p−1≤k≤n−1 i∈Gk+1

Consequently, we obtain from the strong law of large numbers for martingales that (Q n ) converges
almost surely. Finally, as (|Gn |) is a positive real sequence which increases to infinity, we find from
Lemma A.1 in Appendix A that
n X
X
k=p i∈Gk

ei = o(|Gn |)

2502

a.s.

leading to

n X
X
k=p i∈Gk

ei = o(|Tn |)

a.s.

as |Tn | − 1 = 2|Gn |, which implies (6.2). We also establish (6.3) in a similar way. As a matter of
fact, let
n
X
X
1
Rn =
(ǫ2i ǫ2i+1 − ρ).
|Gk−1 | i∈G
k=p
k−1

Then, (R n ) is a square integrable real martingale which converges almost surely, leading to (6.3). ƒ
Remark 6.2. Note that via Lemma A.2
1

lim

n→+∞

lim

|Gn | k∈G
1

n→+∞

X
X

|Gn | k∈G

ǫ2k = 0,

lim

n→+∞

n

2
ǫ2k
= σ2 ,

lim

n→+∞

n

X

1

|Gn | k∈G
|Gn | k∈G

a.s.

n

X

1

ǫ2k+1 = 0
2
ǫ2k+1
= σ2

a.s.

n

In fact, each new generation contains half the global available information, observing the whole tree Tn
or only generation Gn is essentially the same.
For the CLT, we will also need the convergence of higher moments of the driven noise (ǫn ).
Lemma 6.3. Assume that (ǫn ) satisfies (H.1) to (H.5). Then, we have
lim

n→+∞

and
lim

n→+∞

X

1

|Tn | k∈T

1

ǫk4 = τ4

X

|Tn−1 | k∈T

a.s.

n,p

2 2
ǫ2k
ǫ2k+1 = ν 2

a.s.

n−1,p−1

Proof : The proof is left to the reader as it follows essentially the same lines as the proof of
Lemma 6.1 using the square integrable real martingales
Qn =
and
Rn =

n
X

n
X
1

X

|Gk | i∈G
k=p
1

X

|Gk−1 | i∈G
k=p

(ǫi4 − τ4 )

k

2 2
(ǫ2i
ǫ2i+1 − ν 2 ).

k−1

Remark 6.4. Note that again via Lemma A.2
lim

n→+∞

1

X

|Gn | k∈G

4
ǫ2k
= τ4 and

lim

n→+∞

n

2503

1

X

|Gn | k∈G

n

4
ǫ2k+1
= τ4

a.s.

7

Proof of Proposition 5.1

Proposition 5.1 is a direct application of the two following lemmas which provide two strong laws
of large numbers for the sequence of random vectors (Xn ).
Lemma 7.1. Assume that (ǫn ) satisfies (H.1) and (H.2). Then, we have
1 X
Xk = λ = a(I p − A)−1 e1
lim
a.s.
n→+∞ |Tn |
k∈T

(7.1)

n,p

where a = (a0 + b0 )/2 and A is the mean of the companion matrices
A=

1
2

(A + B).

Lemma 7.2. Assume that (ǫn ) satisfies (H.1) to (H.3). Then, we have
1 X
Xk Xkt = ℓ,
a.s.
lim
n→+∞ |Tn |
k∈T

(7.2)

n,p

where the matrix ℓ is the unique solution of the equation
1
ℓ = T + (AℓAt + BℓB t )
2
1
T = (σ2 + a2 )e1 e1t + (a0 (Aλe1t + e1 λ t At ) + b0 (Bλe1t + e1 λ t B t ))
2
with a2 = (a02 + b02 )/2.
Proof : The proofs are given in Appendix A.

ƒ

Remark 7.3. We shall see in Appendix A that
ℓ=

∞
X
1

X

2k
k=0

C∈{A;B}k

CTCt

where the notation {A; B}k means the set of all products of A and B with exactly k terms. For example,
we have {A; B}0 = {I p }, {A; B}1 = {A, B}, {A; B}2 = {A2 , AB, BA, B 2 } and so on. The cardinality of
{A; B}k is obviously 2k .
Remark 7.4. One can observe that in the special case p = 1,
lim

n→+∞

lim

n→+∞

1

X

|Tn | k∈T
1

=

n

X

|Tn | k∈T

Xk

X k2 =

a

a.s.

1− b

a2 + σ2 + 2λab

a.s.

1 − b2

n

where
ab =

a0 a1 + b0 b1
2

,

b=

a1 + b1

2504

2

,

b2 =

a12 + b12
2

.

8

Proof of Theorems 5.1 and 5.2

Theorem 5.2 is a consequence of Theorem 5.1. The first result of Theorem 5.1 is a strong law of
large numbers for the martingale (Mn ). We already mentioned that the standard strong law is
useless here. This is due to the fact that the dimension of the random vector ξn grows exponentially
fast as 2n . Consequently, we are led to propose a new strong law of large numbers for (Mn ), adapted
to our framework.
Proof of result (5.2) of Theorem 5.1: For all n ≥ p, let Vn = Mnt Σ−1
n−1 Mn where we recall that
−1
−1
Σn = I2 ⊗ Sn , so that Σn = I2 ⊗ Sn . First of all, we have
t
t −1
Vn+1 = Mn+1
Σ−1
n Mn+1 = (Mn + ∆Mn+1 ) Σn (Mn + ∆Mn+1 ),
t −1
t
−1
= Mnt Σ−1
n Mn + 2Mn Σn ∆Mn+1 + ∆Mn+1 Σn ∆Mn+1 ,

−1
t −1
t
−1
= Vn −Mnt (Σ−1
n−1 −Σn )Mn +2Mn Σn ∆Mn+1 +∆Mn+1 Σn ∆Mn+1 .

By summing over this identity, we obtain the main decomposition
Vn+1 + An = V p + Bn+1 + Wn+1 ,

(8.1)

where
An =
Bn+1 = 2

n
X

n
X
k=p

Mkt (Σ−1
− Σ−1
)Mk ,
k−1
k

Mkt Σ−1
∆Mk+1
k

and

k=p

Wn+1 =

n
X

t
∆Mk+1
Σ−1
∆Mk+1 .
k

k=p

The asymptotic behavior of the left-hand side of (8.1) is as follows.
Lemma 8.1. Assume that (ǫn ) satisfies (H.1) to (H.3). Then, we have
lim

n→+∞

Vn+1 + An
n

= (p + 1)σ2

a.s.

(8.2)

Proof: The proof is given in Appendix B. It relies on the Riccation equation associated to (Sn ) and
the strong law of large numbers for (Wn ).
ƒ
Since (Vn ) and (An ) are two sequences of positive real numbers, we infer from Lemma 8.1 that
Vn+1 = O (n) a.s. which ends the proof of (5.2).
ƒ
Proof of result (5.5) of Theorem 5.2: It clearly follows from (4.1) that
Vn = (θbn − θ ) t Σn−1 (θbn − θ ).
Consequently, the asymptotic behavior of θbn − θ is clearly related to the one of Vn . More precisely,
we can deduce from convergence (5.1) that
lim

n→∞

λmin (Σn )
|Tn |

= λmin (Λ) > 0
2505

a.s.

since L as well as Λ = I2 ⊗ L are definite positive matrices. Here λmin (Λ) stands for the smallest
eigenvalue of the matrix Λ. Therefore, as
kθbn − θ k2 ≤

Vn

λmin (Σn−1 )

,

we use (5.2) to conclude that

kθbn − θ k2 = O



n


=O

|Tn−1 |

log |Tn−1 |



|Tn−1 |

a.s.

which completes the proof of (5.5).

ƒ

We now turn to the proof of the quadratic strong law. To this end, we need a sharper estimate of the
asymptotic behavior of (Vn ).
Lemma 8.2. Assume that (ǫn ) satisfies (H.1) to (H.3). Then, we have for all δ > 1/2,
k Mn k2 = o(|Tn−1 |nδ )

(8.3)

a.s.

Proof: The proof is given in Appendix C.

ƒ

A direct application of Lemma 8.2 ensures that Vn = o(nδ ) a.s. for all δ > 1/2. Hence, Lemma 8.1
immediately leads to the following result.
Corollary 8.3. Assume that (ǫn ) satisfies (H.1) to (H.3). Then, we have
lim

n→+∞

An
n

= (p + 1)σ2

(8.4)

a.s.

Proof of result (5.3) of Theorem 5.1: First of all, An may be rewritten as
An =
1/2

n
X
k=p

Mkt (Σ−1
− Σ−1
)Mk =
k−1
k

n
X

−1/2

−1/2

Mkt Σk−1 ∆k Σk−1 Mk

k=p

1/2

where ∆n = I2(p+1) − Σn−1 Σ−1
n Σn−1 . In addition, via Proposition 5.1
lim

n→∞

Σn

=Λ

|Tn |

a.s.

(8.5)

which implies that
lim ∆n =

1

I2(p+1)
a.s.
(8.6)
2
Furthermore, it follows from Corollary 8.3 that An = O (n) a.s. Hence, we deduce from (8.5) and
(8.6) that


n
An  1 X

a.s.
(8.7)
M t Σ−1 Mk  + o(1)
=
n
2n k=p k k−1
n→∞

2506

and convergence (5.3) directly follows from Corollary 8.3.

ƒ

We are now in position to prove the QSL.
Proof of result (5.6) of Theorem 5.2: The QSL is a direct consequence of (5.3) together with the
fact that θbn − θ = Σ−1
n−1 Mn . Indeed, we have
n
1X

n k=p

Mkt Σ−1
M
k−1 k

n
1X

=

n k=p
n
1X

=

n k=p
n
1X

=

n k=p

(θbk − θ ) t Σk−1 (θbk − θ )
|Tk−1 |(θbk − θ ) t

Σk−1
|Tk−1 |

(θbk − θ )

|Tk−1 |(θbk − θ ) t Λ(θbk − θ ) + o(1)

which completes the proof of Theorem 5.2.

9

a.s.
ƒ

Proof of Theorem 5.3

bn − Vn .
bn is strongly related to that of V
b 2n and ρ
The almost sure convergence of σ
Proof of result (5.7) of Theorem 5.3: We need to prove that
lim

n→∞

1

X

n k∈T

bk − Vk k2 = 2(p + 1)σ2
kV

a.s.

(9.1)

n−1,p−1

bn − Vn k and the processes (An ) and
Once again, we are searching for a link between the sum of kV
(Vn ) whose convergence properties were previously investigated. For all n ≥ p, we have
X
X
bk − Vk k2 =
(b
ǫ2k − ǫ2k )2 + (b
ǫ2k+1 − ǫ2k+1 )2 ,
kV
k∈Gn

k∈Gn

= (θbn − θ ) t Ψn Ψnt (θbn − θ ),
=

t −1
Mnt Σ−1
n−1 Ψn Ψn Σn−1 Mn ,

=

Mnt Σn−1 ∆n Σn−1 Mn ,

−1/2

−1/2

where
−1/2

−1/2

−1/2

−1/2

∆n = Σn−1 Ψn Ψnt Σn−1 = Σn−1 (Σn − Σn−1 )Σn−1 .
Now, we can deduce from convergence (8.5) that
lim ∆n = I2(p+1)

n→∞

2507

a.s.

which implies that

X
k∈Gn



bk − Vk k2 = M t Σ−1 Mn 1 + o(1)
kV
n n−1

a.s.

Therefore, we can conclude via convergence (5.3) that
lim

n→∞

X

1

n k∈T

bk − Vk k2 = lim
kV

n
1X

n→∞

n−1,p−1

n k=p

Mkt Σ−1
M = 2(p + 1)σ2
k−1 k

a.s.

Proof of result (5.8) of Theorem 5.3: First of all,

2|Tn−1 | k∈T

Pn =

2|Tn−1 | k∈T
X

k∈Tn−1,p−1



bk k2 − kVk k2 ,
kV

n−1,p−1

X

1

=
Set

X

1

b 2n − σ2n =
σ



bk − Vk k2 + 2(V
bk − Vk ) t Vk .
kV

n−1,p−1

bk − Vk ) t Vk =
(V

We clearly have
∆Pn+1 = Pn+1 − Pn =

n
X
X
k=p i∈Gk−1

X
k∈Gn

bi − Vi ) t Vi .
(V

bk − Vk ) t Vk .
(V

bk − Vk = (I2 ⊗ Yk ) t (θ − θbn ) which implies that V
bk − Vk is
One can observe that for all k ∈ Gn , V
Fn -measurable. Consequently, (Pn ) is a real martingale transform. Hence, we can deduce from the
strong law of large numbers for martingale transforms given in Theorem 1.3.24 of [4] together with
(9.1) that


X

bk − Vk )||2 
||V
a.s.
Pn = o 
 = o(n)
k∈Tn−1,p−1

It ensures once again via convergence (9.1) that
lim

n→∞

|Tn |
n

b 2n − σ2n ) = lim
(σ

1

n→∞

X

n k∈T

bk − Vk k2 = 2(p + 1)σ2
kV

a.s.

n−1,p−1

bn . We have
We now turn to the study of the covariance estimator ρ
bn − ρ n =
ρ
=

where
Qn =

X

1

X

|Tn−1 | k∈T
1

X

|Tn−1 | k∈T

(b
ǫ2k ǫb2k+1 − ǫ2k ǫ2k+1 ),

n−1,p−1

(b
ǫ2k − ǫ2k )(b
ǫ2k+1 − ǫ2k+1 ) +

n−1,p−1

(b
ǫ2k − ǫ2k )ǫ2k+1 + (b
ǫ2k+1 − ǫ2k+1 )ǫ2k =

X

2508

|Tn−1 |

Q n,

bk − Vk ) t J2 Vk
(V

k∈Tn−1,p−1

k∈Tn−1,p−1

1

with

‚
J2 =

0 1
1 0

Œ
.

Moreover, one can observe that J2 ΓJ2 = Γ. Hence, as before, (Q n ) is a real martingale transform
satisfying


X

bk − Vk )||2 
||V
a.s.
Qn = o 
 = o(n)
k∈Tn−1,p−1

We will see in Appendix D that
lim

n→∞

1

X

n k∈T

(b
ǫ2k − ǫ2k )(b
ǫ2k+1 − ǫ2k+1 ) = (p + 1)ρ

a.s.

(9.2)

n−1,p−1

Finally, we find from (9.2) that
lim

n→∞

|Tn |
n

bn − ρn ) = 2(p + 1)ρ
(ρ

a.s.

which completes the proof of Theorem 5.3.

ƒ

10 Proof of Theorem 5.4
In order to prove the CLT for the BAR(p) estimators, we will use the central limit theorem for
martingale difference sequences given in Propositions 7.8 and 7.9 of Hamilton [8].
Proposition 10.1. Assume that (Wn ) is a vector martingale difference sequence satisfying
(a) For all n≥1, E[Wn Wnt ]=Ωn where Ωn is a positive definite matrix and
lim

n→∞

n
1X

n k=1

Ωk = Ω

where Ω is also a positive definite matrix.
(b) For all n ≥ 1 and for all i, j, k, l, E[Win W jn Wkn Wln ] < ∞ where Win is the ith element of the
vector Wn .
(c)

n
1X

n k=1

P

Wk Wkt −→ Ω.

Then, we have the central limit theorem
n
1 X
L
Wk −→ N (0, Ω).
p
n k=1

2509

We wish to point out that for BAR(p) processes, it seems impossible to make use of the standard CLT
for martingales. This is due to the fact that Lindeberg’s condition is not satisfied in our framework.
Moreover, as the size of (ξn ) doubles at each generation, it is also impossible to check condition (c).
To overcome this problem, we simply change the filtration. Instead of using the generation-wise
filtration, we will use the sister pair-wise one. Let
Gn = σ{X 1 , (X 2k , X 2k+1 ), 1 ≤ k ≤ n}
be the σ-algebra generated by all pairs of individuals up to the offspring of individual n. Hence
(ǫ2n , ǫ2n+1 ) is Gn -measurable. Note that Gn is also the σ-algebra generated by, on the one hand, all
the past generations up to that of individual n, i.e. the rn th generation, and, on the other hand, all
pairs of the (rn + 1)th generation with ancestors less than or equal to n. In short,


Gn = σ F rn ∪ {(X 2k , X 2k+1 ), k ∈ G rn , k ≤ n} .
2
2
− 2σ2 ) and
+ ǫ2n+1
Therefore, (H.2) implies that the processes (ǫ2n , Xn ǫ2n , ǫ2n+1 , Xn ǫ2n+1 ) t , (ǫ2n
(ǫ2n ǫ2n+1 − ρ) are Gn -martingales.

Proof of result (5.4) of Theorem 5.1: First, recall that Yn = (1, Xn ) t . We apply Propositions 10.1
to the Gn -martingale difference sequence (Dn ) given by


ǫ2n


 Xn ǫ2n 
Dn = vec(Yn Vnt ) = 
.
 ǫ2n+1 
Xn ǫ2n+1
We clearly have

‚
Dn Dnt

=

2
ǫ2n
ǫ2n ǫ2n+1
2
ǫ2n+1 ǫ2n ǫ2n+1

Œ
⊗ Yn Ynt .

Hence, it follows from (H.1) and (H.2) that
E[Dn Dnt ] = Γ ⊗ E[Yn Ynt ].
Moreover, we can show by a slight change in the proof of Lemmas 7.1 and 7.2 that
lim

n→∞

1

X

|Tn | k∈T

E[Dk Dkt ] = Γ ⊗ lim

n→∞

n−1,p−1

1
|Tn |

E[Sn ] = Γ ⊗ L,

which is positive definite, so that condition (a) holds. Condition (b) also clearly holds under (H.3).
We now turn to condition (c). We have
X
Dk Dkt = Γ ⊗ Sn + R n
k∈Tn−1,p−1

where
Rn =

X
k∈Tn−1,p−1

‚

2
ǫ2k
− σ2
ǫ2k ǫ2k+1 − ρ
2
ǫ2k+1 ǫ2k − ρ ǫ2k+1
− σ2

2510

Œ
⊗ Yk Ykt .

Under (H.1) to (H.5), we can show that (R n ) is a martingale transform. Moreover, we can prove that
R n = o(n) a.s. using Lemma A.6 and similar calculations as in Appendix B where a more complicated
martingale transform (Kn ) is studied. Consequently, condition (c) also holds and we can conclude
that
X
1
1
L
Dk = p
Mn −→ N (0, Γ ⊗ L).
(10.1)
p
|Tn−1 | k∈Tn−1,p−1
|Tn−1 |
Proof of result (5.11) of Theorem 5.4: We deduce from (4.1) that
p

|Tn−1 |(θbn − θ ) = |Tn−1 |Σ−1
n−1 p

Mn

.

|Tn−1 |

Hence, (5.11) directly follows from (5.4) and convergence (8.5) together with Slutsky’s Lemma. ƒ
Proof of results (5.12) and (5.13) of Theorem 5.4: On the one hand, we apply Propositions 10.1
to the Gn -martingale difference sequence (vn ) defined by
2
2
vn = ǫ2n
+ ǫ2n+1
− 2σ2 .

Under (H.4), one has E[vn2 ] = 2τ4 − 4σ4 + 2ν 2 which ensures that
lim

X

1

n→∞

|Tn | k∈T

E[vk2 ] = 2τ4 − 4σ4 + 2ν 2 > 0.

n,p−1

Hence, condition (a) holds. Once again, condition (b) clearly holds under (H.5), and Lemma 6.3
together with Remark 6.4 imply condition (c),
lim

n→∞

X

1

|Tn | k∈T

vk2 = 2τ4 − 4σ4 + 2ν 2

a.s.

n,p−1

Therefore, we obtain that
p

1

X

p
L
vk = 2 |Tn−1 |(σ2n − σ2 ) −→ N (0, 2τ4 − 4σ4 + 2ν 2 ).

(10.2)

|Tn−1 | k∈Tn−1,p−1

Furthermore, we infer from (5.8) that
p
b 2n − σ2n ) = 0
lim |Tn−1 |(σ
n→∞

a.s.

(10.3)

Finally, (10.2) and (10.3) imply (5.12). On the other hand, we apply again Proposition 10.1 to the
Gn -martingale difference sequence (w n ) given by
w n = ǫ2n ǫ2n+1 − ρ.
Under (H.4), one has E[w n2 ] = ν 2 − ρ 2 which implies that condition (a) holds since
lim

n→∞

1

X

|Tn | k∈T

E[w k2 ] = ν 2 − ρ 2 > 0.

n,p−1

2511

Once again, condition (b) clearly holds under (H.5), and Lemmas 6.1 and 6.3 yield condition (c),
lim

n→∞

X

1

|Tn | k∈T

w k2 = ν 2 − ρ 2

a.s.

n,p−1

Consequently, we obtain that
p

X

1

|Tn−1 | k∈Tn−1,p−1

wk =

p

L

|Tn−1 |(ρn − ρ) −→ N (0, ν 2 − ρ 2 ).

Furthermore, we infer from (5.10) that
p
bn − ρ n ) = 0
lim |Tn−1 |(ρ

a.s.

n→∞

Finally, (5.13) follows from (10.4) and (10.5) which completes the proof of Theorem 5.4.

(10.4)

(10.5)
ƒ

Appendices
A

Laws of large numbers for the BAR process

We start with some technical Lemmas we make repeatedly use of, the well-known Kronecker’s
Lemma given in Lemma 1.3.14 of [4] together with some related results.
Lemma A.1. Let (αn ) be a sequence of positive real numbers increasing to infinity. In addition, let (x n )
be a sequence of real numbers such that
∞
X
|x n |
n=0

Then, one has
lim

n→∞

αn

< +∞.

n
1 X

αn

x k = 0.

k=0

Lemma A.2. Let (x n ) be a sequence of real numbers. Then,
lim

n→∞

1

X

|Tn | k∈T

x k = x ⇐⇒ lim

1

n→∞

n

X

|Gn | k∈G

x k = x.

n

Proof: First of all, recall that |Tn | = 2n+1 − 1 and |Gn | = 2n . Assume that
lim

n→∞

We have the decomposition,

X
k∈Tn

1

X

|Tn | k∈T

xk =

X

x k = x.

n

xk +

k∈Tn−1

2512

X
k∈Gn

xk.

(A.1)

Consequently,
lim

n→∞

X

1

|Gn | k∈G

xk

=

lim

n→∞

n

X

2

|Tn | + 1 k∈T

X

1

x k − lim

|Tn−1 | + 1 k∈T

n→∞

n

xk,

n−1

= 2x − x = x.
Conversely, suppose that

X

1

lim

n→∞

|Gn | k∈G

x k = x.

n

A direct application of Toeplitz Lemma given in Lemma 2.2.13 of [4]) yields
lim

n→∞

1

X

|Tn | k∈T

=

xk

lim

n→∞

n

=

lim

n→∞

n X
1 X

|Tn | k=0 i∈G
n
1 X

|Tn | k=0

2k

xi,

k

X

1

|Gk | i∈G

Lemma A.3. Let (An ) be a sequence of real-valued matrices such that
n
X

lim

n→∞

x i = x.

k

P∞

n=0 kAn k

< ∞ and

Ak = A.

k=0

In addition, let (X n ) be a sequence of real-valued vectors which converges to a limiting value X . Then,
lim

n
X

n→∞

An−k X k = AX .

(A.2)

k=0

Proof: For all n ≥ 0, let
Un =

n
X

An−k X k .

k=0

We clearly have for all integer n0 with 1 ≤ n0 < n,
∞
n
n

X
X
X


Ak X
,
Ak X −
An−k X k −
kUn − AX k =

≤

n0
X
k=0

≤

k=n+1

k=0

k=0

n
X

kAn−k kkX k − X k +

k=0n
X

kAn−k kkX k − X k +

k=n0 +1

∞
X

kAk kkX k,

k=n+1
∞
X

kAn−k kkX k − X k +

k=n+1

kAk kkX k.

We assume that (X n ) converges to a limiting value X . Consequently, we can choose n0 such that for
all k > n0 , kX k − X k < ǫ. Moreover, one can find M > 0 such that for all k ≥ 0, kX k − X k ≤ M and
kX k ≤ M . Therefore, we obtain that
kUn − AX k ≤ (n0 + 1)M sup kAk k + ǫ
k≥n−n0

2513

n
X
k=n0 +1

kAn−k k + M

∞
X
k=n+1

kAk k.

On the one hand
sup kAk k

∞
X

and

k≥n−n0

k=n+1

kAk k

both converge to 0 as n tends to infinity. On the other hand,
n
X
k=n0 +1

kAn−k k ≤

∞
X
n=0

kAn k < ∞.

Consequently, kUn − AX k goes to 0 as n goes to infinity, as expected.

ƒ

Lemma A.4. Let (Tn ) be a convergent sequence of real-valued matrices with limiting value T . Then,
lim

n→∞

n
X
1

X

2k
k=0

C∈{A;B}k

where the matrix
ℓ=

C Tn−k C t = ℓ

∞
X
1

X

2k
k=0

C∈{A;B}k

CTCt

is the unique solution of the equation
1
ℓ = T + (AℓAt + BℓB t ).
2

(A.3)

Proof: First of all, recall that β = max{kAk, kBk} < 1. The cardinality of {A; B}k is obviously 2k .
Consequently, if
n
X
1 X
Un =
C(Tn−k − T )C t ,
k
2
k
k=0
C∈{A;B}

it is not hard to see that
kUn k ≤

n
X
1
k=0

2k

n


X





β 2(n−k)
Tk − T
.
× 2k β 2k
Tn−k − T
=
k=0

Hence, (Un ) converges to zero which completes the proof of Lemma A.4.

ƒ

We now return to the BAR process. We first need an estimate of the sum of the kXn k2 before being
able to investigate the limits.
Lemma A.5. Assume that (ǫn ) satisfies (H.1) to (H.3). Then, we have
X
kXk k2 = O (|Tn |)
a.s.
k∈Tn,p

2514

(A.4)

Proof: In all the sequel, for all n ≥ 2 p−1 , denote A2n = A and A2n+1 = B. It follows from a recursive
application of relation (2.2) that for all n ≥ 2 p−1
Xn =

n −p
 rY


A[

k=0

n
2k

X[

]

n
]
2 rn −p+1

+

rX
k−1
n −p  Y


A[ ni ] η[

i=0

k=0

2

n
2k

(A.5)

]

with the convention that an empty product equals 1. Then, we can deduce from Cauchy-Schwarz
inequality that for all n ≥ 2 p−1

2
rY
k−1

2
rX

n −p
n −p  Y







n
n
n
n
A
−
=
A
X
η
Xn

[ k]
[ i]
[ rn −p+1 ]
[ k ]
2
2
2
2


i=0
k=0
k=0
!2
rX
k−1
n −p  Y

≤
kA[ ni ] k kη[ nk ] k
rX
n −p

≤

k=0
rX
n −p

≤

β



k

η[

!
βk

≤

rX
n −p

1−β

n
2k

]

2




!2

rX
n −p
k=0

k=0

1

2

i=0

k=0

k=0

!
β k kη[

k

β kη[

n
2k

]k

2

!
n
2k

]k

2

.

p

Hence, we obtain that for all n ≥ 2 ,
2



n −p
n −p


 rY
 rY


n
n
n
X
kXn k2 =
Xn −
A
+
A[ nk ] X[ r −p+1
[ rn −p+1 ]
[ k]
]
2 n
2
2
2


k=0
k=0
!
rX
n −p
2
2
n
β k kη[ nk ] k2 + 2β 2(rn −p+1) kX[ r −p+1
≤
]k .
2 n
2
1 − β k=0
Denote α = max{|a0 |, |b0 |} and X 1 = max{kX k k, k ≤ 2 p−1 }. Summing up over the sub-tree Tn,p , we
find that
!
rX
k −p
X
X 2
X
2
i
2
kXk k ≤
β kη[ k ] k
+
2β 2(rk −p+1) kX[ k ] k2
i
1
−
β
2
2 rk −p+1
i=0
k∈T
k∈T
k∈T
n,p

n,p

≤
≤

n,p

X

4

1 − β k∈T

+2X 1

β i (α2 + ǫ 2 k ) +

rX
k −p

k∈Tn,p i=0
2

[

n,p i=0

X

4
1−β

rX
k −p

X

β i ǫ2 k
[ i]

2i

+

2

]

X

2β 2(rk −p+1) kX [

k∈Tn,p

4α2
1−β

k −p
X rX

k
2 rk −p+1

]k

2

βi

k∈Tn,p i=0

β 2(rk −p+1) ,

k∈Tn,p

≤

4Pn
1−β

+

4α2Q n
1−β

2

+ 2X 1 R n ,
2515

(A.6)

where
Pn =

k −p
X rX

k −p
X rX

β i ǫ2 k , Q n =
[

k∈Tn,p i=0

2i

]

X

β i , Rn =

k∈Tn,p i=0

β 2(rk −p+1) .

k∈Tn,p

The last two terms of (A.6) are readily evaluated by splitting the sums generation-wise. As a matter
of fact,
n X
n
X
X
1 − βk
1
Qn =
2k = O (|Tn |),
(A.7)
≤
1
−
β
(1
−
β)
k=p i∈G
k=p
k

and
Rn =

n X
X
k=p i∈Gk

β k−p+1 ≤

n
X
k=p

(2β)k = O (|Tn |).

(A.8)

It remains to control the first term Pn . One can observe that ǫk appears in Pn as many times as it has
descendants up to the nth generation, and its multiplicative factor for its ith generation descendant
is (2β)i . Hence, one has
Xk
X n−r
(2β)i ǫk2 .
Pn =
k∈Tn,p i=0

The evaluation of Pn depends on the value of 0 < β < 1. On the one hand, if β = 1/2, Pn reduces to
Pn =

X
k∈Tn,p

Hence,

(n + 1 −

rk )ǫk2

=

n
X
k=p

(n + 1 − k)

X

ǫi2 .

i∈Gk




n 
X
(n + 1 − k)  1 X 2 
=
ǫ .

|Tn | + 1 k=p
|Gk | i∈G i
2n+1−k
Pn

k

However, it follows from Remark 6.2 that
lim

n→+∞

X

1

|Gn | k∈G

In addition, we also have
lim

n→∞

ǫk2 = σ2

a.s.

n

n
X
k
k=1

2k

= 2.

Consequently, we infer from Lemma A.3 that
lim

n→+∞

Pn
|Tn |

= 2σ2

a.s.

(A.9)

On the other hand, if β 6= 1/2, we have
Pn =

X 1 − (2β)n−rk +1
k∈Tn,p

1 − 2β

ǫk2 =

1

n
X

1 − 2β

k=p

2516

(1 − (2β)n−k+1 )

X
i∈Gk

ǫi2 .

Thus,
Pn
|Tn | + 1

=

n  
X
1 n−k+1

1
1 − 2β

2

k=p

Furthermore,
lim

n→∞

1 − 2β



X



 1

− β n−k+1 
ǫi2  .
|Gk | i∈G
k

n  
X
1 k

1



2

k=1


−β

k

=

1
1−β

.

As before, we deduce from Lemma A.3 that
lim

n→+∞

Pn
|Tn |

=

σ2

.

1−β

a.s.

(A.10)

Finally, Lemma A.5 follows from the conjunction of (A.6), (A.7), (A.8) together with (A.9) and
(A.10).
ƒ
Proof of Lemma 7.1 : First of all, denote
X
Xk
Hn =

Pn =

and

X

ǫk ,

k∈Tn,p

k∈Tn,p−1

As |Tn | = 2n+1 − 1, we obtain from Equation (2.2) the recursive relation
‹
X 
H n = H p−1 +
A k X[ k ] + η k ,
2

k∈Tn,p

= H p−1 + 2AH n−1 + 2a(2n − 2 p−1 )e1 + Pn e1
where e1 = (1, 0, . . . , 0) t ∈ R p , a = (a0 + b0 )/2 and the matrix
A=

A+ B
2

.

By induction, we deduce from (A.11) that
Hn
2n+1

=

H p−1
2n+1

+A

= (A)n−p+1

H n−1
2n

H p−1
2p

+a

 2n − 2 p−1 

n
X

+

2n

(A)n−k

k=p

‚

e1 +

H p−1
2k+1

Pn

e1 ,
2n+1
Œ
 2k − 2 p−1 
Pk
+a
e1 + k+1 e1 .
2k
2

We have already seen via convergence (6.1) of Lemma 6.1 that
lim

n→+∞

Finally, as kAk < 1,

∞
X
n=0

k(A)n k < ∞

Pn
2n+1

=0

and

2517

a.s.

(I p − A)−1 =

∞
X

(A)n ,

n=0

(A.11)

it follows from Lemma A.3 that
lim

n→∞

Hn
2n+1

= a(I p − A)−1 e1

a.s.

which ends the proof of Lemma 7.1.

ƒ

Proof of Lemma 7.2 : We shall proceed as in the proof of Lemma 7.1 and use the same notation.
Let
X
X
Kn =
Xk Xkt
and
Ln =
ǫk2 .
k∈Tn,p−1

k∈Tn,p

We infer again from (2.2) that
X 

Kn = K p−1 +

‹
A k X[ k ] + η k

k∈Tn,p

X

= K p−1 +
+

A k X[ k ] + η k

2

2

X 

ǫk2 e1 e1t +

AXk Xkt At + BXk Xkt B t



k∈Tn−1,p−1

k∈Tn,p

X 

‹t

(a0 + ǫ2k )Uk (A) + (b0 + ǫ2k+1 )Uk (B) + 2(a2 + ζ2k )e1 e1t



k∈Tn−1,p−1

where Uk (A) = AXk e1t + e1 Xkt At and Uk (B) = BXk e1t + e1 Xkt B t . In addition, a2 = (a02 + b02 )/2 and
ζ2k = (a0 ǫ2k + b0 ǫ2k+1 ). Therefore, we obtain that


Kn−1 t
Kn−1 t
Kn
1
(A.12)
A n A + B n B + Tn
=
2
2
2
2n+1
where

 2n − 2 p−1 
X
L
1

 n
ζ2k  e1 e1t
=  n+1 + a2
+ n
2n
2 k∈T
2
n−1,p−1
‚
Œ
t
t

 H


H n−1 t
H n−1
1
H n−1 t
n−1 t
t
+
a0 A n e1 + e1 n A + b0 B n e1 + e1 n B
2
2
2
2
2


X
1
ǫ2k Uk (A) + ǫ2k+1 Uk (B) .
+
n+1
2 k∈T


Tn

n−1,p−1

The two first results (6.1) and (6.2) of Lemma 6.1 together with Remark 6.2 and Lemma A.2 readily
imply that
Ln
a.s.
lim n+1 = σ2
n→+∞ 2
and
X
1
ζ2k = 0
a.s.
lim n
n→+∞ 2
k∈T
n−1,p−1

In addition, Lemma 7.1 gives
lim

n→+∞

H n−1
2n

=λ

2518

a.s.

Furthermore, denote

X 

Un =


ǫ2k Uk (A) + ǫ2k+1 Uk (B) .

k∈Tn−1,p−1



For all u ∈ R p , let Un (u) = u t Un u. The sequence Un (u) is a real martingale transform. Moreover,
it follows from Lemma A.5 that
2 
2
X 



a.s.
u t Uk (A)u + u t Uk (B)u = O (|Tn |)
k∈Tn−1,p−1

Consequently, we deduce from the strong law of large numbers for martingale transforms given in
Theorem 1.3.24 of [4] that Un (u) = o(|Tn |) a.s. for all u ∈ R p which leads to Un = o(|Tn |) a.s.
Therefore, we obtain that (Tn ) converges a.s. to T given by
T = (σ2 + a2 )e1 e1t +

1€
2

Š
Aλa0 e1t + a0 e1 λ t At + Bλb0 e1t + b0 e1 λ t B t .

Finally, iteration of the recursive relation (A.12) yields
Kn
2n+1

=

X

1
2n−p+1

C

C∈{A;B}n−p+1

K p−1
2p

Ct +

n−p
X
1
k=0

2k

X

C Tn−k C t .

C∈{A;B}k

On the one hand, the first term on the right-hand side converges a.s. to zero as its norm is bounded
β 2(n−p+1) kK p−1 k/2 p . On the other hand, thanks to Lemma A.4, the second term on the right-hand
side converges to ℓ given by (A.3), which completes the proof of Lemma 7.2. .
ƒ
We now state a convergence result for the sum of kXn k4 which will be useful for the CLT.
Lemma A.6. Assume that (ǫn ) satisfies (H.1) to (H.5). Then, we have
X
kXk k4 = O (|Tn |)
a.s.

(A.13)

k∈Tn,p

Proof : The proof is almost exactly the same as that of Lemma A.5. Instead of Equation (A.6), we
have
X
64Pn
64α4Q n
4
kXk k4 ≤
+
+ 8X 1 R n
3
3
(1 − β)
(1 − β)
k∈T
n,p

where
Pn =

k −p
X rX

k∈Tn,p i=0

β i ǫ4 k ,
[

2i

]

Qn =

k −p
X rX

k∈Tn,p i=0

β i,

Rn =

X

β 4(rk −p+1) .

k∈Tn,p

We already saw that Q n = O (|Tn |). In addition, it is not hard to see that R n = O (|Tn |). Therefore,
we only need a sharper estimate for un . Via the same lines as in the proof of Lemma A.5 together
with the sharper results of Lemma 6.3, we can show that Pn = O (|Tn |) a.s. which leads to (A.13). ƒ

2519

B

On the quadratic strong law

We start with an auxiliary lemma closely related to the Riccation Equation for the inverse of the
matrix Sn .
Lemma B.1. Let hn and l n be the two following symmetric square matrices of order δn
hn = Φnt Sn−1 Φn

and

−1
l n = Φnt Sn−1
Φn .

Then, the inverse of Sn may be recursively calculated as
−1
−1
−1
Sn−1 = Sn−1
− Sn−1
Φn (Iδn + l n )−1 Φnt Sn−1
.

(B.1)

In addition, we also have (Iδn − hn )(Iδn + l n ) = Iδn .
Remark B.2. If f n = Ψnt Σ−1
n Ψn , it follows from Lemma B.1 that
−1
−1
t −1
Σ−1
n = Σn−1 − Σn−1 Ψn (I2δn − f n )Ψn Σn−1 .

(B.2)

Proof : As Sn = Sn−1 + Φn Φnt , relation (B.1) immediately follows from Riccati Equation given e.g. in
[4] page 96. By multiplying both side of (B.1) by Φn , we obtain
−1
−1
Sn−1 Φn = Sn−1
Φn − Sn−1
Φn (Iδn + l n )−1 l n ,

−1
−1
= Sn−1
Φn − Sn−1
Φn (Iδn + l n )−1 (Iδn + l n − Iδn ),
−1
= Sn−1
Φn (Iδn + l n )−1 .

Consequently, multiplying this time on the left by Φnt , we obtain that
hn = l n (Iδn + l n )−1 = (l n + Iδn − Iδn )(Iδn + l n )−1 ,
= Iδn − (Iδn + l n )−1

leading to (Iδn − hn )(Iδn + l n ) = Iδn .

ƒ

In order to establish the quadratic strong law for (Mn ), we are going to study separately the asymptotic behaviour of (Wn ) and (Bn ) which appear in the main decomposition (8.1).
Lemma B.3. Assume that (ǫn ) satisfies (H.1) to (H.3). Then, we have
lim

n→+∞

1
n

Wn = 2σ2

(B.3)

a.s.

Proof : First of all, we have the decomposition Wn+1 = Tn+1 + Rn+1 where
Tn+1 =
Rn+1 =

n
X
∆M t

Λ−1 ∆Mk+1

k=p

|Tk |

k+1

n
X
∆M t

k+1

,

(|Tk |Σ−1
− Λ−1 )∆Mk+1
k
|Tk |

k=p

2520

.

We claim that

1

lim

n→+∞

n

Tn = (p + 1)σ2

a.s.

It will ensure via (8.5) that Rn = o(n) a.s. leading to (B.3). One can observe that Tn+1 =
t r(Λ−1/2 H n+1 Λ−1/2 ) where
t
n
X
∆Mk+1 ∆Mk+1
H n+1 =
.
|Tk |
k=p
Our goal is to make use of the strong law of large numbers for martingale transforms, so we start by
adding and subtracting a term involving the conditional expectation of ∆H n+1 give