some remarks on the tails of the gaussian distribution

8/14/2019 Some remarks on the tails of the Gaussian distribution

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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,

Pagina 2 di5


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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

Pagina 5 di5

some remarks on the tails of the gaussian distribution

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