3.1 Proof of Theorem 1.1 a
We start by deriving a recursive representation for µn. For k ∈ N ∪ {0},
E [X
k
n + 1|X n] = 1
n + 1
n
X
i=1
E h
−1l
{Z [i,n]=kZ [i,n+1]}
+ 1l
{Z [i,n]k=Z [i,n+1]}
X n i
+ nX
k
n + 1l
{k=0}
= X
k
n + 1
n + 1 h
−nX
k
n f k
n + nX
k −1
n f k
− 1 n
− X
k
n + 1l
{k=0}
i .
Thus the linearity and the tower property of conditional expectation gives µ
k
n + 1 = µ
k
n + 1
n + 1 f k − 1µ
k −1
n − 1 + f kµ
k
n + 1l
{k=0}
. Now defining Q
∈ R
N ×N
as
Q =
− f 0
f 0 1
− f 1 + 1 f 1
1 − f 2 + 1 f 2
.. .
... ...
11
and conceiving µn as a row vector we can rewrite the recursive equation as
µn + 1 = µn I +
1 n + 1
Q ,
where I = δ
i, j i, j
∈N
denotes the unit matrix. Next we show that µ is a probability distribution with
µQ = 0. By induction, we get that 1
−
k
X
l=0
µ
l
=
k
Y
l=0
f l 1 + f l
for any k ∈ N ∪ {0}. Since
P
∞ l=0
1 f l ¾
P
∞ l=0
1 l + 1 = ∞ it follows that µ is a probability
measure on the set N ∪ {0}. Moreover, it is straightforward to verify that
f 0 µ
= 1 − µ =
∞
X
l=1
µ
l
and that for all k ∈ N ∪ {0}
f k − 1 µ
k −1
= 1 + f k µ
k
, hence
µQ = 0. Now we use the matrices P
n
:= I +
1 n+1
Q to define an inhomogeneous Markov process. The entries of each row of P
n
sum up to 1 but as long as f is not bounded each P
n
contains negative entries... Nonetheless one can use the P
n
as a time inhomogeneous Markov kernel as long as at the starting time m
∈ N the starting state l ∈ N ∪ {0} satisfies l ¶ m − 1. 1237
We denote for any admissible pair l, m by Y
l,m n
n ¾ m
a Markov chain starting at time m in state l having transition kernels P
n n ¾ m
. Due to the recursive equation we now have µ
k
n = PY
0,1 n
= k. Next, fix k
∈ N∪{0}, let m k arbitrary, and denote by ν the restriction of µ to the set {m, m+1, . . . }. Since
µ is invariant under each P
n
we get µ
k
= µP
m
. . . P
n k
=
m −1
X
l=0
µ
l
P Y
l,m n
= k + ν P
m
. . . P
n k
. Note that in the n-th step of the Markov chain, the probability to jump to state zero is
1 n+1
for all original states in
{1, . . . , n − 1} and bigger than
1 n+1
for the original state 0. Thus one can couple the Markov chains Y
l,m n
and Y
0,1 n
in such a way that P
Y
l,m n+1
= Y
0,1 n+1
= 0 | Y
l,m n
6= Y
0,1 n
=
1 n+1
, and that once the processes meet at one site they stay together. Then
P Y
l,m n
= Y
0,1 n
¾ 1 −
n −1
Y
i=m
i i + 1
−→ 1. Thus
ν P
m
. . . P
n k
∈ [0, µ[m, ∞] implies that lim sup
n →∞
µ
k
− PY
0,1 n
= k
m −1
X
l=0
µ
∗ l
¶ µ[m, ∞.
As m → ∞ we thus get that
lim
n →∞
µ
k
n = µ
k
.
In the next step we show that the sequence of the empirical indegree distributions X n
n ∈N
con- verges almost surely to
µ. Note that n X
k
n is a sum of n independent Bernoulli random variables. Thus Chernoff’s inequality Chernoff [1981] implies that for any t
P X
k
n ¶ E[X
k
n] − t ¶ e
−nt
2
2E[X
k
n]
= e
−nt
2
2µ
k
n
. Since
∞
X
n=1
e
−nt
2
2µ
k
n
∞, the Borel-Cantelli lemma implies that almost surely lim inf
n →∞
X
k
n ¾ µ
k
for all k ∈ N ∪ {0}. If
A ⊂ N ∪ {0} we thus have by Fatou’s lemma
lim inf
n →∞
X
k ∈A
X
k
n ¾ X
k ∈A
lim inf
n →∞
X
k
n = µA.
1238
Noting that µ is a probability measure and passing to the complementary events, we also get
lim sup
n →∞
X
k ∈A
X
k
n ¶ µA. Hence, given
ε 0, we can pick N ∈ N so large that µN, ∞ ε, and obtain for the total variation norm
lim sup
n ↑∞
1 2
∞
X
k=0
X
k
n − µ
k
¶ lim sup
n ↑∞
1 2
N
X
k=0
X
k
n − µ
k
+
1 2
lim
n ↑∞
∞
X
k=N +1
X
k
n +
1 2
µN , ∞ ¶ ε. This establishes almost sure convergence of X n to
µ in the total variation norm.
3.2 Proof of Theorem 1.1 b
Kategori