Remark 3.3. Note that to obtain the non-asymptotic bounds of the Monte Carlo procedure 3.5, we successively used the concentration properties of the reference measure
γ, the control of the distance W
1
µ, γ given by the variational formulation of the entropy see Lemma 3.1 and the tensorization property of the functional inequality satisfied by
γ. The same arguments can therefore be applied to a reference measure
γ satisfying a Poincaré inequality.
3.2 Gaussian concentration - Lower bound
Concerning the previous deviation rate of Proposition 3.2, a natural question consists in understand- ing whether it is sharp or not. Namely, for a given function f satisfying suitable growth conditions
at infinity, otherwise we cannot see the asymptotic growth, do we have a lower bound of the same order, i.e. with Gaussian decay at infinity? The next proposition gives a positive answer to that
question.
Proposition 3.3. Let f : R
d
→ R
+
be a Lipschitz continuous function satisfying ∇f
∞
¶ 1 and
assumption
G
ρ ,
β
for given ρ ,
β 0. For a
C
2
function V on R
d
such that e
−V
is integrable with respect to λ
d
and s.t.
∃¯λ ¾ 1, ¯λI
d ×d
¾ HessV ¾ 0, let
γdx = e
−V x
Z
−1
dx be the associated Gibbs probability measure. We assume that ∃κ ¾ 1 s.t. for |x| ¾ ρ
the measures µdx = mxdx and γdx satisfy
mx ¾ κ
−1
e
−V x
Z
−1
. Let ¯
Λ :=
¯ λ
2
+
sup
s ∈Sd−1
|V sρ |
ρ
2
+
sup
s ∈Sd−1
|∇V sρ |
ρ
. We have
∀r 0, P
µ
f Y − µ f ¾ r − W
1
µ, γ + δ f , γ ¾
Kd ,A
Z ¯ Λ
d 2
κ
exp −¯
Λ h
r β
∨ ρ i
2
, d even,
arccos θ
−12
Kd,A Z ¯
Λ
d 2
κ
exp −θ ¯
Λ h
r β
∨ ρ i
2
, ∀θ 1, d odd,
with δ f , γ = γ f + βρ
− f , f := inf
s ∈S
d −1
f s ρ
, and Kd, A defined in 2.6 where A ⊂ S
d −1
appears in G
ρ ,
β
. Proof.
Set E := {Y ∈ A × [ρ
, ∞}. Here we use the convention that for d = 1, A × [ρ
, + ∞ ⊂
−∞, −ρ ] ∪ [ρ
, + ∞. Write now
P
µ
f Y − µ f ¾ r − W
1
µ, γ + δ f , γ ¾ P
µ
f Y − µ f ¾ r − W
1
µ, γ + δ f , γ, E ¾
κ
−1
P
γ
h f Y ¾ r
− βρ + f , E
i :=
κ
−1
P . 3.6 Denoting Y
= ρ π
S
d −1
Y , we have P
¾ P
γ
h f Y
+ f Y − f Y ¾
r − βρ
+ f , E i
,
G
ρ0,β
¾ P
γ
h β|Y − Y
| ¾ r − βρ + f − f Y
, E i
¾ P
γ
|Y | ¾ r
− βρ β
+ |Y |, E
¾ P
γ
|Y | ¾ r
β ∨ ρ
, π
S
d −1
Y ∈ A .
3.7 1656
Write P
¾ Z
A
σds Z
+∞ ρ
∨
r β
d ρρ
d −1
exp −V sρZ
−1
, where
σds stands for the Lebesgue measure of S
d −1
. Now, HessV ¶ ¯ λI
d ×d
yields ∀ρ ¾ ρ
∨
r β
, |V sρ|ρ
2
¶ ¯ Λ, ¯
Λ :=
¯ λ
2
+
sup
s ∈Sd−1
|V sρ |
ρ
2
+
sup
s ∈Sd−1
|∇V sρ |
ρ
and therefore P
¾ |A|
Z
+∞ ρ
∨
r β
d ρρ
d −1
exp −¯
Λρ
2
Z
−1
¾ |A|
Z 2 ¯
Λ
d 2
Z
+∞ ρ
∨
r β
2 ¯ Λ
1 2
d ρρ
d −1
exp −
ρ
2
2 =
|A| Z
2 ¯ Λ
d 2
Q
d
ρ ∨
r β
2 ¯ Λ
1 2
. 3.8
We now have the following explicit expression: ∀x 0, Q
d
x := exp− x
2
2 M d, x,
M d , x :=
d 2
−1
X
i=
x
2i
d 2
−1
Y
j=i+ 1
2 j, d even,
d −1
2
−1
X
i=
x
2i+1
d −1
2
−1
Y
j=i+ 1
2 j + 1 +
d −1
2
−1
Y
j=
2 j + 1 exp x
2
2 Z
+∞ x
exp −
ρ
2
2 dρ, d odd,
with the convention that P
−1 i=
= 0, ∀k ∈ N, Q
k −1
j=k
j = 1.
Observe now that R
∞ x
exp −ρ
2
2dρ = 2π
1 2
P [N 0, 1 ¾ x] ¾ 2π
1 2
P [Y ∈ K , |Y | ¾
x cos ˜
θ ] := 2π
1 2
Qx, where Y ∼ N 0
2 ×1
, I
2 ×2
is a standard bidimensional Gaussian vec- tor and
K := {z ∈ R
2
, 〈z, e
1
〉 ¾ cos ˜ θ |z|}, ˜
θ ∈ 0,
π 2
, e
1
= 1, 0. Since Qx =
˜ θ
π
exp −
x
2
2 cos
2
˜ θ
, we derive that
Q
d
x ¾
2
d 2−1
d2 − 1 exp−
x
2
2
, d even,
˜ θ 2
d 2
π
1 2
Q
d −1
2
j= 1
j −
1 2
exp−
x
2
2 cos
2
˜ θ
, d odd, which plugged into 3.8 yields:
P ¾
Kd ,A
Z ¯ Λ
d 2
exp −¯
Λ h
r β
∨ ρ i
2
, d even,
˜ θ Kd,A
Z ¯ Λ
d 2
exp −
¯ Λ
cos
2
˜ θ
h
r β
∨ ρ i
2
, d odd.
Corollary 3.2. Under the assumptions of Proposition 3.3, let Y
1
, · · · , Y
M
be i.i.d. R
d
-valued random variables with law
µ. We have ∀r 0, ∀M ¾ 1, P
1 M
M
X
k= 1
f Y
k
− E
µ
f Y
1
¾
r − W
1
µ, γ + δ f , γ
¾ 2 exp
−Mθ ¯ Λ + χ
r β
∨ ρ
2
1657
where χ =
1 ρ
2
log
Z ¯ Λ
d 2
κ Kd
,A +
, θ = 1 for d even, and χ =
1 ρ
2
log
Z ¯ Λ
d 2
κ Kd
,A arccos θ
−12
+
, θ ∈ 1, +∞ for
d odd, and with Kd , A defined in 2.6.
Proof. We only consider d even. By independence of the Y
k k
∈[[1,M]]
, exploiting T
M k=
1
{ f Y
k
− E
µ
f Y
1
¾
r − W
1
µ, γ + δ f , γ} ⊂ {
1 M
P
M k=
1
f Y
k
− E
µ
f Y
1
¾
r − W
1
µ, γ + δ f , γ}, we have
P 1
M
M
X
k= 1
f Y
k
−E
µ
f Y
1
¾
r −W
1
µ, γ+δ f , γ ¾
Kd , A
Z ¯ Λ
d 2
κ
M
exp
−M ¯ Λ
r β
∨ ρ
2
.
For χ =
1 ρ
2
log
Z ¯ Λ
d 2
κ Kd
,A +
, we thus obtain
P
1
M
M
X
k= 1
f Y
k
− E
µ
f Y
1
¾
r − W
1
µ, γ + δ f , γ
¾ exp
−M¯ Λ + χ
r β
∨ ρ
2
,
which completes the proof. .
3.3 Proofs of Theorem 2.2 and Corollary 2.1
