GAUSSIAN COMPARISON LEMMA
1. Gaussian Comparision Lemma
Lemma 1.1. Let G : Rn → R be a bounded, twice continuously differentiable function with
bounded derivatives
∂G(x)
∂G(x)
Gi (x) =
1 6 i 6 n and Gij =
1 6 i, j 6 n.
∂xi
∂xi ∂xj
If X ∼ Nn (0, ΣX ) and Y ∼ Nn (0, ΣY ) are normal random vectors then
Z 1
n
1 X
E G(Y) − E G(X) =
∆ij
E Gij (Xt ) dt
2
0
i,j=1
where ∆ij = E Yi Yj −E Xi Xj = (ΣY − ΣX )ij and Xt ∼ Nn (0, Σt ) with Σt := (1−t) ΣX +t ΣY .
Proof: Assume without loss of generality that X and Y are independent. For each t ∈ [0, 1]
define the random vector
Xt = (1 − t)1/2 X + t1/2 Y
and the associated function ϕ(t) = E G(Xt ). Note that X0 = X, X1 = Y, and that Xt ∼
Nn (0, Σt ), where Σt is defined as in the statement of the lemma. Thus
Z 1
E G(Y) − E G(X) = ϕ(1) − ϕ(0) =
ϕ0 (t) dt,
0
and it suffices to show that for each t ∈ (0, 1)
phipr0
ϕ0 (t) =
(1.1)
n
1 X
∆ij E Gij (Xt ).
2
i,j=1
1/2
To this end, fix t ∈ (0, 1) and note that Xt is distributed as Σt Z where Z ∼ N (0, I) is
a standard normal random vector with independent components. To simplify notation, let
1/2
At := Σt . It follows from our regularity assumptions and the chain rule that
" n
#
X
d
d
d
E G(At Z) = E
G(At Z) = E
Gi (At Z) (At Z)i
ϕ0 (t) =
dt
dt
dt
i=1
phipr1
(1.2)
=
n
X
(A0t )ij E (Zj Gi (At Z)) ,
i,j=1
A0t
where
denotes the entry-by-entry derivative of the matrix At . Fix i, j for the moment and
define the function
Hij (s) := E Gi (At Zs )
where
Zs := (Z1 , · · · , Zj−1 , s, Zj+1 , · · · , Zn ).
Date: April 2017.
1
2
GAUSSIAN COMPARISON LEMMA
It follows from a simple conditioning argument and Gaussian integration by parts that
GIP
(1.3)
E [Zj Gi (At Z)] = E [Zj Hij (Zj )] = E Hij0 (Zj ).
By another application of the chain rule,
n
X
d
d
Hij0 (s) = E
Gi (At Zs ) =
E Gik (At Zs ) (At Zs )k
ds
dt
k=1
=
n
X
(At )jk E Gik (At Zs ).
k=1
Thus, as Z1 , . . . , Zn are independent,
E Hij0 (Zj ) =
n
X
(At )jk E Gik (At Z).
k=1
Combining this last equation with (1.2), we find that
n
n
X
X
0
ϕ (t) =
E Gik (At Z) ·
(A0t )ij (At )jk
phipr2
=
(1.4)
i,k=1
n
X
j=1
E Gik (Xt ) · (A0t At )ik .
i,k=1
atsq
Recalling that At =
and A0t are symmetric,
1/2
Σt ,
(1.5)
(A2t )0 = A0t At + At A0t = A0t At + (A0t At )T .
it is easy to see that (A2t )0ik = (Σt )0ik = ∆ik . Furthermore, as At
Fix 1 6 i < k 6 n. Continuity of the second partial derivatives ensures that Gik = Gki , and
therefore
E Gik (Xt ) · (A0t At )ik + E Gki (Xt ) · (A0t At )ki
= E Gik (Xt ) (A0t At )ik + (A0t At )ki
= E Gik (Xt ) (A2t )0ik = E Gik (Xt ) ∆ik ,
where the penultimate equality follows from (1.5). A similar argument shows that (A0t At )ii =
∆ii /2. Thus (1.1) follows from (1.2), and the proof is complete.
GAUSSIAN COMPARISON LEMMA
eq:g1
:biv-norm-tail
eq:zzr0
eq:zzr1
eq:zzr2
3
1.1. Further Reading: Gaussian Tail Bounds. Let Φ̄(x) = 1 − Φ(x) where Φ(x) is the
cumulative distribution function of the standard Gaussian distribution. Recall that for x > 0,
1
2
Φ̄(x) 6 √
e−x /2 .
(1.6)
2πx
The proof of Theorem ?? requires an inequality for the probability that two correlated Gaussian random variables each exceeds a common threshold.
Lemma 1.2. Let (Z, Zρ ) be jointly Gaussian random variables with mean 0, variance 1, and
correlation E(ZZρ ) = ρ ∈ (−1, 1). Then for any u > 0,
(1 + ρ)2
p
(1.7)
P(Z > u, Zρ > u) 6
exp −u2 /(1 + ρ) .
2πu2 1 − ρ2
Proof of Lemma 1.2. Fix u > 0. When ρ > 0 the proof follows from known inequalities in
the literature (see R. Willink, Bounds on the bivariate normal distribution function, Comm.
Statist. Theory Methods 33 (2004),
pp.2281-2297). Here we consider the case ρ < 0. Note
p
that we may write Zρ = ρZ + 1 − ρ2 Z 0 , where Z 0 is a standard Gaussian random variable
independent of Z. By conditioning on the value of Z, it is easy to see that
Z ∞
u − ρt
.
(1.8)
P(Z > u, Zρ > u) =
Φ̄ (g(t)) φ(t) dt where g(t) = p
1 − ρ2
u
Now define
r
1−ρ
2
η=
and h(x) = ex /2 Φ̄(x).
1+ρ
√
2 /2
0
x
As h (x) = x e
Φ̄(x) − 1/ 2π, inequality (1.6) implies that h(x) is decreasing for x > 0.
It follows from equation (1.8) that
Z ∞
2
(1.9)
P(Z > u, Zρ > u) =
e−g(t) /2 h(g(t)) φ(t) dt
u
Z
∞
6 h(g(u))
2 /2
e−g(t)
φ(t) dt
u
Z
∞
= h(ηu)
2 /2
e−g(t)
φ(t) dt,
u
where in the last step we have used the fact that g(u) = ηu. Routine algebra and a change of
variables establishes that
Z ∞
Z ∞
1
(t − ρu)2
−g(t)2 /2
−u2 /2
√ exp −
e
φ(t) dt = e
(1.10)
dt
2(1 − ρ2 )
2π
u
u
p
2
1 − ρ2 e−u /2 Φ̄(ηu).
=
Combining (1.9), (1.11), and inequality (1.6) yields the bound (1.7), as desired.