some remarks on the tails of the gaussian distribution
TRANSCRIPT
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Tight bounds for gaussian tails
Jacopo DAurizio, classe A049
17 Giugno 2013
I Abstract
In this article, we will prove the following inequalities for the gaussian distribution N(0, 1).
LetXbe a random variable with density:
f(x) = 12 ex
2
/2.
For any k >0,
P[X > k ]> 2 ek
2/2
2
k+
k2 + 4 (1)
follows from the conditional moment inequality.
From the refined version of the conditional moment inequality, we will prove the stronger:
P[X > k]>
ek
2/2
2( 1)k+ 2x2 + 4 . (2)
II The double conditional moment inequalityWe know that for any random variable with a square-integrable, continuous and positive density over
R+,
E
(X E[X])2
>0
holds, from which the moment inequality:
E
X2
> E [X]2 (3)
follows. Letk be a positive real number.
By considering the conditional version of the latter inequality, we have:
E (X k)
2
|X > k > E [(X k)|X > k]2 , (4)that can be proved by the Cauchy-Schwarz/Bunyakowsky inequality, too.
We will prove that ifXhas a gaussian distribution with density given by f(x),
2 k]E[(X k)|X > k]2
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d
dkA(0) = 1 +k A(0). (6)
It follows thatA(0) may be represented as the Laplace continued fraction:
A(0) = 1
k+ 1k+ 2
k+ 3k+...
. (7)
From now on, let
A(n) = n
k+ n+1k+ n+2
k+...
.
We have that: A(n) = n
k+A(n+1),
A(n) = n1A(n1)
k. (8)
Theorem II.1. For any natural numbern, the function
A(n)
A(n+1)(k)
is positive, convex and decreasing.
Proof.d
dkA(0) = 1 +kA(0).
Since, by induction:d
dkA(n) =A(n)
2+kA(n) n= n 1
n A
(n)
A(n1) n,
the convexity ofA(n) is equivalent to the fact that A(n)2
+ kA(n) is an increasing function, or A(n1)
A(n) is a
decreasing function. By defining:
M(0) =
+0
ex2/2kx dx= ek
2/2 A(0),
M(1) =
+0
x ex2/2kx dx= ek
2/2 (1 k A(0)) =ek2/2A(0)A(1),
M(n) = +
0
xn ex2/2kx dx= (n 1)M(n2) kM(n1)
by induction we have:
M(n) =ek2/2
nj=0
A(j).
By the Cauchy-Schwarz inequality we haveM(n+2)M(n) > M(n+1)2
, from whichA(n+2) > A(n+1), follows.
M(n)(k) =+j=0
(1)jj!
2n+j1
2
n+j+ 1
2
kj
Now we have: A(0)(k)k
= 1
2
+0
1
x+k e
x/2
x dx,
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IV THE LOWER BOUND FROM THE REFINED CONDITIONAL MOMENT INEQUALITY
A(0)(k) = 1
2
+0
k
x+k2 e
x/2
x dx,
A(0)(k) = 1
2 +
0
2k
x2 +k2ex
2/2 dx,
A(0)(k) =
2
+0
ek2x2/2
x2 + 1 dx,
III The lower bound from the conditional moment inequality
Assuming thatXhas a gaussian distribution with density given by f(x), the latter is equivalent to:
that can be proved by the Cauchy-Schwarz/Bunyakowsky inequality, too. As a consequence of the
integration by parts, the inequality (4) can be rewritten in terms ofAk as:
(1 k Ak)2 < Akk+ (1 +k2)Ak ,
or:
A2k+k Ak 1 0.SinceAk trivially belongs to the interval
0, 1k
, the latter inequality gives:
Ak 2k+
k2 + 4
.
IV The lower bound from the refined conditional moment in-
equality
In this section we will prove the remarkable property of the gaussian distribution to satisfy a tighter
version of the conditional moment inequality:
EX>k
[(X k)]2 < 2 EX>k
(X k)2 . (9)
That inequality is equivalent to:
+
k
(x k) ex2/2 dx2
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V Convexity
We say that f : R+ R+ is reciprocally convex iff(x) is concave and f(1/x) is convex,i.e. x f(x) is convex. We denote byFthe set of such functions, and observe that:
f F,x, y R+, f
2xyx+y
f(x) +f(y)
2 f
x+y
2
xf(x) +yf(y)
x+y .
Ak = ek2/2
+k
ex2/2 dx=
+0
ex2/2kx dx= k
+0
ek2(x2/2+x) dx
x2/2 +x= u, (x+ 1) dx= du, dx= du
x+ 1 =
du2u+ 1
+
0
eu
2u+k2du=
1
k +
0
eu2u/k2 + 1
du= 1
k +
0
+
j=0
(1)j (2j 1)!! uj eu
k2jj! =
+
j=0
(1)j(2j 1)!!k2j+1
From here or fromdAkdk
= 1 +k Akthe continued fraction expansion
Ak = 1
k+ 1k+ 2
k+ 3...
.
+j=0
(1)j(2j 1)!!k2j+1
=+j=0
(1)jk2j+1
2
+0
(2u)j1/2 eu du= 1
k
1
+0
1
1 + 2u/k2eu
udu
Ak = +0
eu2u+k2
du= 1k
2
+0
ev2
/2
1 +v2/k2dv =
2
+0
ek2
w
2
/2
1 +w2 dw
Ak =
1
2
+0
ek2w/2
(1 +w)
wdw
The same result can be obtained by considering the Fourier transform ofe|x|; in conclusion, we havethatA(0) can be expressed as a Stijelties transform:
A(0) =
2
+0
ek2x2/2
x2 + 1 dx.
If we prove that An+1
An is and increasing function, equivalent to the convexity ofA(n) and the inequality
A(n1)+A(n+1) 2A(n), by considering the limits in 0 and + and the properties of the Wallis productwe have that:
n
n+ 1 A
(n)
A(n+1) n
n+ 1
0
sinn x dx0
sin(n+1) x dx=
2
n
n+12
n2
2
1.
SinceA(n) is a decreasing function and A(n+1) = nA(n)
k, we have that:
A(n)(0) =
2
0 sin(n1) x dx
=n2
+ 12
n2
2 = (n).Now using the Gurlands inequality we get that:
A(n)(0)
n
n+ 1
n+ 3/2,
2n2
2n+ 1
.
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VII CONTINUED FRACTIONS
Using the inequality: 2n
n
2 (2n+ 1)
2 16n 1 + 2
8n+ 3
we get the tighter inequality:
1
2
4n(4n 1)
4n+ 1 A(n)(0) 1
2
(4n 4)(4n 3)
(4n 5) .
VI Convexity
VII Continued fractions
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