Dispersion measure for symmetric, stable probabili

Dispersion measure for symmetric,stable

probability distributions

Jussi I.Tyhtila February 2,2008

Abstract

In this sudy,a dispersion measure for stable distributions is pro-posed by measuring the ’average curvature’of the characteristic func-tion of a stable random variable.The ’average curvature’is to be called the characteristic curvature and it resembles the concept of Fisher information .The intellectual motivation comes from the un-certainty principle for Fourier transform pairs .Characteristic curva-ture reduces to that of standard deviation in the Gaussian case and possesses great many analytical features,for example the scaling be-haviour of the type t

1

)(7)

a

Proof.We note that

ϕY(u)=E(e iuY)=E(e iu(aX+b))=e iub E(e iuaX)=e iubϕX(au)

and for a>0

x−b

F(Y≤x)=F(aX+b≤x)=F(X≤3Stable probability distributions4

a1≤x)∗F(X−b2

a≤x)(10)

holds.

or,still:

Definition6The distribution function F(X≤x)is called stable if to every a1>0,b1,a2>0,b2there correspond constants a>0and b such that the equation

ϕX−b

a2(u)ϕX−b

13.2Sums of symmetric,stable random variables5

|u|

ω(u,α))(12) whereα,β,γ,c>0are constants(γ∈R,−1≤β≤1,0<α≤2and

ω(u,α)=tan(πα

π

ln|u|ifα=1

The proof is long and can be found in[3].Following is important for our purpose of this study:

Corollary1A stable distribution function F(X≤x)that is symmetric around zero1,has the characteristic function

ϕX(u)=e−c|u|α(13) This results from the properties of characteristic functions discussed above.

3.2Sums of symmetric,stable random variables

Now we examine the summing of independent,identically distributed sym-metric,stable random variables.As the study of stochastic processes and therefore stochastic modelling is based on the properties of sums of random variables,in this chapter we study them in the case of symmetric,stable distributions.

Consider the set of independent,identically distributed random variables X i i∈(1,2,...,t)with X i having the characteristic function

ϕX

i

(u)=e−c|u|α∀i(14) Let us consider the sum of such random variables:

S t=

t

i=1X i=X1+X2+···+X t(15)

We wish to determine the characteristic function of S t.As the convolution theorem states that the characteristic function of the sum of iid.random variables is the product of respective characteristic functions,we shall have:

ϕS t(u)=

t

i=1ϕX i(u)(16)3.3Scaling of symmetric,stable distributions6

σ2|u|2(19)

2

Now we have for the sum S t the characteristic function:

u)=e−1

ϕS

t(

t(22) we will have:

u)=e−1

ϕS

t(

3.3Scaling of symmetric,stable distributions7

αX≤x)(24) or equivalently

F(X≤

x

α

)(25) Proof.We have the characteristic function:

ϕX(u)=e−c|u|α(26) Suppose we scale the random variable X such that X−→sX,where s>0. The characteristic function of the scaled random variable sX will be:

ϕsX(u)=e−c|su|α(27) Which is equivalent to:

ϕsX(u)=e−csα|u|α(28) In the previous section we showed,that the sum of t independent,identically distributed symmetric,stable random variables X has the characteristic func-tion:

ϕS t(u)=e−tc|u|α(29)

by comparing,we note that these two expressions are the same,if t=sαor s=t1α,the scaled random variable t1

αX≤x)(30) or equivalently:

F(X≤

x

α

)(31) where F(X≤x)is the distribution of X.QED.So the probability dis-tribution spreads out proportionally to t1

α.This scaling property will come up later on in an interesting context.3.4Some remarks on the scaling properties8

αfor sums of t iid.symmetric,stable random variables.We have also seen that the scale parameter c>0in the characteristic function grows linearly as a function of the number of summands(or time,if you will).These properties do tell us a whole lot about the sum behaviour of stable random variables. In the Gaussian case one can directly state how the standard deviation grows over time if we consider the sum as a1-dimensional random walk.This is extremely useful,as for example in the context of mathematicalfinance,stan-dard deviation of the underlying price process is called volatility.However, as the Gaussian case is the only case where variance isfinite,we can not give predictions of the growth of standard deviation over time in other cases when α<2.Basically we face the problem that we do know the time-evolution of the dispersion of the sum-distribution,but we do not know how to measure that dispersion in general.This is the topic of the next chapter.

4Characteristic curvature

4.1Motivation

We next proceed to propose a measure for dispersion for symmetric around zero stable distributions withα∈[1,2].The idea is to think of a measure that somehow characterises the dispersive properties of the characteristic functions.We know that characteristic functions determine probability dis-tributions completely[3],whether the moments arefinite or not.So the general dispersive properties of different probability distributions must be somehow incorporated in the properties of characteristic functions.We know from Fourier analysis,that the more spread the original function,the more concentrated its Fourier transform and vice versa[5].Because characteristic function is the Fourier transform of the probability measure,we can apply this idea here.It is therefore natural to think that if the probability dis-tribution is greatly dispersed in space,the characteristic function should be condensed around the origin in the frequency space.This concentration of characteristic functions around the origin is intuitively analogous to that the second derivative in its absolute value should be relatively large around the origin,because the characteristic function satisfiesϕX(0)=1and it should decay fast when moved from the origin into either direction.Hower,be-cause the characteristic function here is symmetric around the origin,we can constrain to analyse the positive part of u.Therefore the negative of the second derivative should be relatively large for small u.To measure this con-√∂u2)ϕX(u)du(32) whereϕX(u)is the characteristic function of the distribution and u≥0.

The constant2

2πis suitable forfixing the proper scale,as we shall see later

on.Note that this integral resembles the definition of Fisher information, differences being that now the likelihood function is the characteristic func-tion,and the parameter is u.By performing the derivation and simplifying, the characteristic curvature can be formulated equivalently as:

Φ=2

2π

∞

(

ϕ′X(u)2

−(ϕ′X(0))2−ϕ′′X(0)(34)

see eg.[3].

4.2Characteristic curvature for symmetric,stable dis-

tributions

Theorem7Given symmetric around zero stable distribution with charac-teristic functionϕX(u)=e−c|u|αits characteristic curvature satisfies

Φ(c;α)=2

2π

∞

{−∂2lnϕX(u)√α)c1H√

α

)is the Euler Gamma function and H is the Hurst exponent.∂u2

=−α(α−1)cuα−2(39) Which is the same as

−∂2lnϕX(u)

√

√

cαuα−1The integral will

become: ∞

uα−2e−t

cα ∞0u−1e−t dt(45)α

α

the integral becomes

1α

α

e−t dt(46) Taking the constant outside the integral gives:

1α ∞

t−1

αc1−1α

)−1e−t dt(48)

Recalling that the Euler Gamma function is defined as

Γ(z)= ∞0t z−1e−t dt z∈C(49) We will obtain the elegant result:

∞0uα−2e−cuαdu=

c1

α

Γ(1−

1

√α−1α)(51)

Simplifying,it will become:

Φ(c;α)=2

2π

(α−1)c1

α

)(52)

Because the Gamma function satisfies

Γ(z+1)=zΓ(z)(53) We have

Γ(2−1

α

)Γ(1−

1

α

)=

α

α

)(54)

Substituting this in the characteristic curvature:

Φ(c;α)=2

2π

(α−1)c1

α−1

Γ(2−

1

√αΓ(2−1

√α)c14.3Properties of characteristic curvature for stable distributions and some restrictions12

√α)c1

2.If we restrict that1≤α≤2,

we will have following desirable properties for characteristic curvature: Theorem8Given1≤α≤2and∀c>0,the characteristic curvature Φ(α,c)satisfies the following properties:

(1)Φ(c;α)>0

(2)∂Φ(c;α)

∂c =

2

2π

Γ(2−

1

α>0

QED.In other words,characteristic curvature is strictly increasing in c,which is the scale factor,and is always positive.

4.4Characteristic curvature for Gaussian distribution Let us calculate the characteristic curvature for centered Normal distribution with varianceσ2.The characteristic function is

ϕX(u)=e−1

2

σ2.By substituting these values,we will get:

Φ(2,1

√12)

BecauseΓ(3π

2σ2)=

4

2π

2

σ2

√

2

=

2

√

√1 2

σ2)=

√1

4.5Characteristic curvature for Cauchy distribution13

2

σ2)=σ2(57) ,which is obvious,of course.

4.5Characteristic curvature for Cauchy distribution Let us calculate the characteristic curvature for Cauchy distribution,that is α=1and c is unspecified.The characteristic function is:

ϕX(u)=e−c|u|

Substituting these values we will have:

Φ(1,c)=2c

2π

Γ(1)

BecauseΓ(1)=1,we will have simply:

Φ(1,c)=2c 2π

4.6Sums of symmetric,stable random variables and

characteristic curvature

We proceed to consider the characteristic curvature for sums of independent, identically distributed symmetric,stable random variables X i with charac-teristic functionϕX

i

(u)=e−|u|α∀i∈(1,2,3,...,t).We recall that the sum

S t=

t

i=1X i=X1+X2+···+X t(58)

has the characteristic function

ϕS

t(

u)=e−tc|u|α(59) We calculate the respective characteristic curvature:

Φ(tc;α)=2

2π

αΓ(2−

1

α(60)4.6Sums of symmetric,stable random variables and characteristic curvature14

√α)c1α(61) which equals

Φ(tc;α)=Φ(c;α)t1

α, which is exactly the same scaling behaviour that stable random variables possess in general.

4.6.1Example:X i∼N(0,σ2)

Given that X i∼N(0,σ2)∀i.That is,we have c=1

2σ2;2)=Φ(

1

2(63)

We recall from earlier results thatΦ(1

2σ2;2)=σ

√

α=σ2t(65)

4.6.2Example:X i∼Cauchy(c)

Let X i have a Cauchy distribution with some scale factor c.We haveα=1, so for the sum of t such random variables,we have:

Φ(tc;1)=Φ(c;1)t1

√5Conclusions15