Latala's inequality
Dmitriy M. Stolyarov
April 25, 2015
1 Formulation and a simplication
Let
1
X1 , X2 , . . . , Xn be i.i.d non-negative random variables with the distribution X .
E |X − 1| = α 6= 0. Dene their products
We suppose that
EX =
and
R0 = 1;
Rk =
k
Y
Xj , k = 1, 2, . . . , n.
j=1
p<1
The following theorem is proved in [2] (see also [1] for the
Theorem 1 (Latala's theorem).
version).
The inequality
n
X
n
n
X
X
vj Rj 6
kvj k
kvj k . E j=0
j=0
j=0
holds independently of the norm k · k and n (the constant depends only on α).
Here the
vj
are vectors in some Banach space
B.
Surely, only the left hand-side inequality is inter-
esting. Latala's theorem is more general, we have stated a particular case (in the original version, the
random variables
Xj
may have dierent distributions; however, the discussed case already contains all
the diculties). Our aim here is to explain the proof and provide a small simplication (which will lead
to some drop of the constant in the main inequality, but, maybe, make the proof more transparent for
an analyst).
Consider the i.i.d random variables Yj such that Yj = 1 ± α with probability 21 . Dene R̃j to
be their products. Then,
Lemma 1.
n
n
X
X
vj R̃j 6 E vj Rj .
E
j=0
j=0
Proof.
We may suppose that there exist independent events Ej such that E(Xj | Ej ) = 1+α and P (Ej ) =
1
(to do this, represent the probability space as an innite direct product of unit intervals such that
2
each Xj depends on the j th coordinate only). Let S be the algebra of sets generated by the Ej . Then,
by Jensen's inequality:
n
n
X
X
E vj R j S 6 v j Rj j=0
j=0
almost surely. It remains to notice that the expression on the left coincides with
n
X
vj R̃j ,
j=0
if
Yj
equals
1+α
on
Ej
and
1−α
on the complement of
1
Ej .
This lemma shows that we may work with the case
the Rademacher representation of the
Xj = Yj .
It may seem convenient to work with
Yj
and proceed analytically. However, the author does not know
n,
we will drop
how to prove Theorem 1 in this way.
The proof will be by induction in
to get a bound for the sum with
v0
v0 ,
use the induction assumption, and then try
present there. However, this will be rather delicate. There will be
two principles. The rst one is the statement of the theorem for
by
E kv0 + X1 v1 k),
n=1
(one can estimate
kv0 k
and
kv1 k
the second one is that if one adds a big constant vector to a random vector, then the
mathematical expectation of the norm of the latter increases in a reasonable way.
2 Preliminary lemmas
We begin with the variant of the case
sharp (take
B = R).
Lemma 2.
For any v0 , v1 ∈ B,
n=2
of Theorem 1. One can easily see that these inequalities are
E v0 + X1 v1 > αkv1 k;
Proof.
E v0 + X1 v1 >
α
kv0 k.
α+1
The rst inequality may be rewritten as
1
v0 + (1 + α)v1 + 1 v0 + (1 − α)v1 > αkv1 k,
2
2
which is nothing but the triangle inequality. The second inequality also follows from the triangle inequality:
1
v0 + (1 + α)v1 + 1 v0 + (1 − α)v1 = 1 + α v0 + v1 + 1 − α v0 + v1 >
2
2
2
1+α
2
1−α
1
1 − α
1 α
v0 + v1 + 1 − α v0 + v1 > 1 − α
−
kv0 k >
kv0 k.
2
1+α
2
1−α
2
1−α 1+α
1+α
Lemma 3.
Let Y be a random vector in B. Suppose v ∈ B be such that P (kY k >
kv0 k
4 )
6 41 . Then,
kv0 k
E Y + v > E kY k +
.
8
Proof.
We use Bayes's formula and the triangle inequality:
kv0 k kv0 k kv0 k kv0 k E Y + v = P kY k >
E Y + v kY k >
+ P kY k 6
E Y + v kY k 6
>
4
4
4
4
kv0 k kv0 k kv0 k kv0 k
kv0 k P kY k >
E Y kY k >
− kv0 k + P kY k 6
+ E Y kY k 6
>
4
4
4
2
4
P kY k 6 kv0 k
kv0 k kv0 k
4
E kY k +
− P kY k >
kv0 k > E kY k +
.
2
4
8
We used the inequality
kY + v0 k > kv0 k − kY k >
kv0 k
2
+ kY k
for the second summand, when we were
passing from the rst line to the second one.
This lemma is a formalization of our second principle that says that when perturbs a random vector
with a big constant vector, the average norm increases.
2
3 First attempts
Suppose that we are going to prove the inequality
c
n
X
n
X
kvj k 6 E vj Rj ,
j=0
where
c
(1)
j=0
is suciently small constant, by induction in
n.
Suppose that it holds for
n − 1.
For that we
write
j
j
n
n
n
X
1 1 X
Y
X
Y
E
vj Rj = E v0 +(1−α)v1 +(1−α)
vj
Xk + E v0 +(1+α)v1 +(1+α)
vj
Xk 2
2
j=0
j=2
j=2
k=1
(2)
k=1
v0 + (1 − α)v1 , {(1 − α)vj }nj=2
v0 + (1 + α)v1 , {(1 + α)vj }nj=2 and
and apply the induction hypothesis to each of the summands (with vectors
X2 , X3 , . . . , Xn in the rst summand and vectors
X2 , X3 , . . . , Xn in the second). This will lead us to
and variables
variables
n
n
X
1
X
1
E
vj Rj > c kv0 + (1 − α)v1 k + kv0 + (1 + α)v1 k +
ckvj k.
2
2
j=0
j=2
So, we will have to bound
It is impossible.
ckv0 k + ckv1 k by
1
1
c kv0 + (1 − α)v1 k + kv0 + (1 + α)v1 k .
2
2
However, this reasoning gives us hope that if we pass from inequality (1) to a more
general one, the induction may work. Let us try to prove the inequality
n
n
X
X
v j Rj >
cj kvj k,
E
j=0
where
cj
(3)
j=0
are some coecients that are uniformly bounded from below. Of course, inequalities (3) and (1)
are equivalent, but (3) seems to be more induction-friendly. Indeed, let us try to do the same reasoning
again (i.e., use formula (2) and apply the induction hypothesis (3) to the same vectors and random
variables). This will lead us to the inequality
n
n
X
1
X
1
E
vj Rj > c0 kv0 + (1 − α)v1 k + kv0 + (1 + α)v1 k +
cj−1 kvj k.
2
2
j=0
j=2
So, to deduce inequality (3) for
c0
n−1
summands, we have to show that
n
X
1
kv0 + (1 − α)v1 k + kv0 + (1 + α)v1 k +
cj−1 − cj kvj k > c0 kv0 k + c1 kv1 k.
2
2
j=2
1
cj (however,
v0 → ∞; but
cj − cj−1 > 0. Again,
This seems to be not as hopeless as the previous attempt if we make a proper choice of the
this inequality also seems to be untrue if
v0
is big, because it turns into an equality when
for this case we have the second principle, Lemma 3). It seems reasonable to take
we use Lemma 2 to estimate the sum of linear combinations on the left hand-side:
c0
n
n
X
X
1
kv0 + (1 − α)v1 k + kv0 + (1 + α)v1 k +
cj−1 − cj kvj k > c0 αkv1 k +
cj−1 − cj kvj k.
2
2
j=2
j=2
1
3
So, we can make prove the induction step when
c0 kv0 k 6 (c0 α − c1 )kv1 k +
n
X
cj−1 − cj kvj k.
(4)
j=2
Again, it is convenient to assume that
c0 α > c1 .
How can we deal with the case when condition (4) is not satised? We are going to apply Lemma 3
with the constant vector
v0
and
Y =
need to estimate the probability of
inequlity (4) does not hold (to link
|X1 |
n
X
vj
j=1
Pn
vj Rj in the role of the random vector. Therefore, we will
1
the event kY k > kv0 k. Of course, we will have to use the fact that
4
Y and v0 ). So, we will have to estimate the probability of the event
j
Y
j=1
Xk > (c0 α − c1 )kv1 k +
n
X
cj−1 − cj kvj k.
(5)
j=2
k=1
4 Tails estimates
A brief examination of our assumptions show that the value
cj − cj−1
However, we have not used the assumption that the
Xj
Pn j → ∞. Therefore,
P
kvj k.
j=1 vj Rj k 6
tends to zero as
we cannot derive any information about the event (5) from the trivial estimate
Ek
are independent seriously yet. It appears, that
to get the desired estimates, we have to leave the convex world.
Lemma 4.
Suppose the variables Xj be as above. Then,
n
12 2
X
6
vj Rj E
j=0
n
1 X j
β kvj k.
1 − β j=0
for some β < 1.
Proof.
We use the inequality
P
P
1
1
( |xk |) 2 6
|xk | 2
and the independence of the
Xj :
n
n
n
n
n
12
21
X
j X
√1 + α + √1 − α j X
X
X
1
1
1
1
2
2
2
kvj k 2 β j .
E vj Rj =
kvj k E |X|
6
vj R j 6
kvj k
E
=
2
j=0
j=0
j=0
j=0
j=0
√
Here
β=
√
1+α+ 1−α
2
n
X
< 1.
It remains to use the Cauchy inequality:
1
kvj k 2 β j 6
j=0
n
X
βj
j=0
n
12 X
12
β j kvj k
=
j=0
n
12
1 21 X j
β kvj k .
1−β
j=0
By Chebyshev's inequality, Lemma 4 leads to
n
X
P vj R j >
j=0
n
t X j
1
β kvj k 6 √ .
1 − β j=0
t
There are two things to be noticed. First, it is reasonable to take
cj − cj−1 = cβ j
(6)
of something like that.
Second, the event from formula (5) diers from the one estimated in (6) (we have to multiply by
This gives us some more freedom. Indeed, with probability
use estimate (6) with bigger
t.
4
1
2,
X1
X1 ).
is smaller than one, which allows to
5 Final reasoning
We take
N
to be a big natural number. We dene the coecients
c0 = a;
Here
a, b,
and
d
cj = b −
c1 = c2 = . . . = cN = b;
d
cj
by the rule
Pj
k=N +1
βk
1−β
, j > N + 1.
are small positive constants to be specied later. In order the
requirements (they are
c0 > αc1
and
cj > 0
b < aα,
We assume inequality (3) for
n−1
cj
satisfy all the above
uniformly), these parameters must be under the conditions
d < b.
and try to prove its version for
n.
It remains to deal with the case
when inequality (4) does not hold, which turns into
kv0 k > α −
n
X
b
d
kv1 k +
β (j−N −1) kvj k.
a
a(1 − β)
(7)
j=N +1
Consider the event
Ω:
Ω = {X2 = X3 = . . . = XN = 1 − α},
X̃j be Xj conditioned on Ω. Then,
j
n
n
n
X
Y
X
X
1 +
1 −
N
N −1
vj
X̃j + E w
+(1+α)(1−α)
vj
Ek
vj Rj k = E w0 +(1−α)
0
2
2 j=0
we have left
X1
free for further use. Let
j=N +1
where
w0−
= v0 + (1 − α)(
j=N +1
k=N +1
N
X
j
(1 − α) vj );
v0 + (1 + α)(
j=1
N
X
j
Y
k=N +1
X̃j ,
(1 − α)j vj ).
j=1
kw0− k + kw0+ k >
α
α
α+1 kv0 k, so at least one of these vectors is bigger than 2(α+1) kv0 k. In
order to apply Lemma 3, we need to estimate the probability of the event
By Lemma 2,
n
X
α
N −1
kv0 k 6 (1 + α)(1 − α)
vj
2(α + 1)
j=N +1
j
Y
By assumption (7), this event leads to (we also killed the summand with
n
X
j=N +1
vj
j
Y
X̃j >
X̃j .
k=N +1
kv1 k)
n
X
αd
(j−N −1)
β
kv
k
.
j
2(1 + α)2 (1 − α)N −1 a(1 − β)
j=N +1
k=N +1
By inequality (6), the probability of this event does not exceed
r
2(1 + α)2 (1 − α)N −1 a
.
αd
So, we require
r
which seems to be true when
N
2(1 + α)2 (1 − α)N −1 a
1
< ,
αd
4
is suciently big. So, we can use Lemma 3 for the bigger of the vectors
−
and w0 to get
n
X
±
N −1
E
w
+(1±α)(1−α)
vj
0
j=N +1
j
Y
k=N +1
n
X
α
N −1
X̃j >
kv0 k+E (1±α)(1−α)
vj
8(1 + α)
j=N +1
5
j
Y
k=N +1
w0+
X̃j .
References
Two-sided bounds for Lp -norms of combinations
of products of independent random variables, Stochastic Process. Appl. 125 (2015), 16881713.
[1] E. Damek, R. Latala, P. Nayar and T. Tkocz,
[2] R. Latala,
L1 -norm of combinations of products of independent random variables, Israel J. Math.
203 (2014), 295308.
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