# Average Sample Number Function for Pareto Heavy Tailed Distributions

Authors: Boukhalfa El-Hafsi
Journal: ISRN Applied Mathematics
 (2013)

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## Abstract

The main purpose of this work is shortly to give the average sample number function after a sequential probability ratio test on the index parameter alpha of stable densities, which we give a mean of the number of data required to take decision in the case1<α<2, we use the fact that the tails of Levy-stable distributions are asymptotically equivalent to a Pareto law for large data. Stable distributions are a rich class of probability distributions that allow skewness and heavy tails and have many intriguing mathematical properties. The lack of closed formulas for densities and distribution functions for all has been a major drawback to the use of stable distributions by practitioners, but few stable distributions have the analytical formula of their densities functions which are Gauss, Levy, and Cauchy.


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## Body

## 1. Introduction

Because it has a large application in many fields, stable distributions have been proposed as a model for many types of physical and economic systems [1–3]. There are several reasons for using a stable distribution to describe a system. The first is where there are solid theoretical reasons for expecting a non-Gausian stable model. The second reason is the Generlized Central Limit Theorem which states that the only possible non-trivial limit of normalized sums of independent identically distributed terms is stable. The third argument for modeling with stable distributions is empirical: many large data sets exhibit heavy tails and skewness. The strong empirical evidence for these features combined with Generalized Central Limit Theorem is used by many to justify the use of stable models. There are now reliable computer programs to compute stable densities, distribution functions and quantiles. With these programs, it is possible to use stable models in variety of practical problems [1, 2].In Section1, we give some definitions, arithmetic properties of stable laws, and the numerical calculation of stable densities, distribution functions Mittnik et al. [4] and Nolan [1, 2], and simulation algorithm [1, 5]. In Section 2, we propose a method to calculate the likelihood ratio appliying the pareto limit of stable densities for large data, with discuss, Finally we interests of the number of observations needed to take decision about the best value of index parameter by its averange sample number function in the case 1<α<2, when moment of order one exists.There are now reliable computer programs to compute stable densities, distribution functions and quantiles. Nolan [2] creates algorithms for generate general stable densities based on integral representations of the densities which were derived by Zolotarev [6], then Nolan [2, 3] provides a useful program named “STABLE”. With these program, it is possible to use stable models in variety of practical problems. For Wald’s SPRT [7] approximatios, Raghavachari [8] gave exact formulas in the case of an exponential density θ-1exp(-x/θ) for testing θ0 versus θ1, provided θ0/θ1>max(B,B/A). For a sequential probability ratio test, Under other distributions A. Wald’s approximations [7] are often used, but how about stable distributions? there are not closed formula for the densities to calculate the likelihood ratio, we compare the fit of two pareto stable models using sequential probability ratio test.


## 2. Stable Distributions

Definition 1.
A random variableX has a distribution infinitely divisible if and only if for all n,∃X1,…,Xn independent with the same even law such as
(1)X=dX1+⋯+Xn,
where =d means equality in distribution.Remark 2.
The random variablesXi did not even law as X. However, as we see in the following examples, they belong to the same family of distributions.Example 3.
The characteristic function of normal distribution𝒩(μ,σ2) is written as
(2)φ𝒩(μ,σ2)(t)=exp{iμt-σ2t22}=[exp{iμnt-σ2/n2t2}]n=[φ𝒩(μ/n,σ2/n)(t)]n,
as nth power of characteristic function of a normal distribution 𝒩(μ/n,σ2/n).Example 4.
The characteristic function of Cauchy distributionC(c) is written as
(3)φC(c)(t)=exp{-c|t|}=[exp(-cn|t|)]n=[φ𝒞(c/n)(t)]n,
as nth power of the characteristic function of a Cauchy distribution 𝒞(c/n).Example 5.
The characteristic function of Poisson distributionP(λ) is written as
(4)φ𝒫(λ)(t)=exp{λ(eit-1)}=[exp{λn(eit-1)}]n=[φ𝒫(λ/n)(t)]n,
as nth power eme of the characteristic function of a Poisson law 𝒫(λ/n).Example 6.
The characteristic function of Gamma distributionΓ(r,λ) is written as
(5)φ-(r,λ)(t)=1(1-it/λ)r=[1(1-it/λ)r/n]n=φ-(r/n,λ)(t),
as nth power eme of characteristic function of a Gamma distribution -(r/n,λ).
The same is true for the exponential law(Exp(λ)=Γ(1,λ)) and the law of χ2=Γ(n/2,1/2)).Remark 7.
A mixture of finished normal distributions is not divisible infinitely.Theorem 8.
A random variableX is the limit of a sum of n random variables i.i.d. if and only if X is divisible indefinitely.Proof.
The demonstration was detailed in Shiryayev [9].Definition 9.
A random variableX is stable if for X1, X2 two independent random variables that have same law as X and any two real positive constants a, b, then
(6)aX1+bX2=cX+d,indistribution
satisfies for some c∈ℝ+ and d∈ℝ.Definition 10.
A random variableX is stable distribution if and only if, for all n and any independent identique family X1,X2,…,Xn which have the same law that X, there is an>0 and bn∈R, as
(7)X1+X2+⋯+Xn=danX+bn.Proposition 11.
IfX is stable, X is divisible indefinitely. The inverse is false (see the example below, the Poisson distribution).Proof.
Just take the random variablesYi=(Xi-bn/n)/an,i=1,…,n.
AsXi are independent, Yi are also independent, and by substitution, we have
(8)Y1+Y2+⋯+Yn=dX.Remark 12.
This definition is coinside of the central limit theorem ifX is following a normal law.Theorem 13.
X has domain of attraction; that is, there is a sequence of random variables i.i.d.(Xi)i∈N, a sequence of positive real numbers (ai)i∈ℕ, and a sequence of real numbers (bi)i∈ℕ, such as
(9)1an∑i=1nXi-bn⟶X,indistribution.Proof.
The demonstration is detailed in Shiryayev [9].Corollary 14 (Levy-Khinchin).
The characteristic function of a stable random variableX admits the following form:
(10)ψX(t)={exp[-σα|t|α(1-iβ·sgn(t)·tg(απ2))+iμt],α≠1exp[-σ|t|(1+2πiβ·sgn(t)·ln|t|)+iμt],α=1,wheresgn(t)={1,t>00,t=0-1,t<0.Proof.
The demonstration is detailed in Gnedenko and Kolmogorov [10].The random variableX is a stable law with parameters α, β, σ, and μ; we note that
(11)X≡Sα(β,σ,μ),where0<α≤2,-1≤β≤1,σ≥0,μ∈ℝ.But this representation of the FC, called parameterization standard, has the disadvantage of not being continued in all of its parameters. In fact, there is discontinuity in the points whereα=1 and β=0; otherwise, there are other parameterization of the FC more adapted to different problems. Consider
(12)ψX(t)={exp[-σα|t|α(1+iβ·sgn(t)·tanαπ2)(|σt|1-α-1)+iμ0t],α≠1exp[-σ|t|(1+2πiβ·sgn(t)·ln|t|+lnσ)+iμ0t],α=1.This representation of Zolotarev’sM parameterization notes X~Sα0(β,σ,μ0). The parameters α,β, and σ are the same as those of the standard parameterization, but μ and μ0 are related by
(13)μ={μ0-βσtanαπ2,α≠1μ0-2πβσlnσ,α=1.This parameterization is very important because the characteristic function and the cumulative distribution function are continuing in relation to the four parameters.S0 is well conditioned numerically by calculation.Another parameterizationS1 is given by the following:
(14)ψX(t)={exp[-σ2α|t|αexp(-iβ2·sgn(t)·π2K(α))+iμt],α≠1exp[-σ2|t|(π2+iβ2·sgn(t)·ln|t|)+iμt],α=1,
where
(15)K(α)=α-1+sgn(1-α)={α,α≤1α-2,α≥1,
where parameters α and μ are the same as for standard parameterization; other parameters meet the following relationships:
(16)σ2=σ(1+β2tan2απ2)1/2α,tan(β2πK(α)2)=βtanαπ2,α≠1,σ2=2σπ,β2=β,α=1.Remark 15.
The main disadvantage is that the densities of stable laws are unknown except in three cases.(1)
The Gaussian distributionS2(0,σ,μ) where f(x,2,0,σ,μ)=(1/2σπ)exp(-(x-μ)2/4σ2), x∈ℝ.(2)
The Cauchy distributionS1(0,σ,μ) where f(x,1,0,σ,μ)=(2σ/π((x-μ)2+4σ2)),x∈ℝ.(3)
The Lévy distributionS1/2(1,σ,μ) where f(x,1/2,σ,μ)=(σ/2π)1/2(x-μ)3/2exp(-σ/2(x-μ))·I]μ,∞[(x).
But since the implementation of the fast Fourier transform, stable densities are easy to calculate. We can approach this method densityf(x)=(1/2π)∫-∞+∞e-itxψX(t)dt.


### 2.1. Interpretation of the Parameters of the Characteristic Function of a Stable Law

(1)
The parameterα known as an exponent characteristic or a stability index describes the form of distribution or the degree of thickness of the tail distribution (0<α≤2). If α is smaller, then the tails of the distribution are thick. In other words, the more small α is, the more we see the existence of very large fluctuations (see Figure 1). A Gaussian distribution that has the maximum value of α is α=2.
(2)
The parameterβ gives an idea of asymmetric distribution. This is the parameter of asymmetry. If β is equal to -1 (resp., 1), distribution is totally asymmetrical left or right. When β=0, then the distribution is symmetric around μ (see Figure 2).
(3)
The parameterσ is called a scale factor. σ is more grand, if more data are volatile (see Figure 3). The parameter σ has the possibility to bend more or less the body of the distribution.
(4)
The parameterμ location corresponds for α>1 to the mean of the distribution. If β=0, then μ is the median. In other cases, μ cannot be interpreted.Figure 1
pdf plot for different values of alpha.Figure 2
pdf plot for different values of beta.Figure 3
pdf plot for different values of sigma.


### 2.2. Arithmetic Properties

Stable random variablesSα(β,σ,μ) have the following properties.(i)
X1,X2 are two independent random variables that follow, respectively, stable laws Sα(β1,σ1,μ1), Sα(β2,σ2,μ2); then, X1+X2 follows a stable law Sα(β,σ,μ) with
(17)σ=(σ1α+σ2α)1/α,β=β1σ1α+β2σ2ασ1α+σ2α,μ=μ1+μ2.
Note that, if β1=β2, then β=β1=β2.(ii)
IfX~Sα(β,σ,μ), a∈ℝ, then X+a~Sα(β,σ,μ+a).(iii)
IfX1,X2 follows a stable law Sα(β,σ,μ),a∈ℝ+,b∈ℝ+,c∈ℝ, then
(18)aX1+bX2+c=dSα(β,σ(aα+bα)1/α,μ(aα+bα)1/α+c),
where “=d” means “has the same distribution.”(iv)
X is a symmetric random variable around μ if and only if the law of X is Sα(0,σ,μ). We note that SαS is a symmetric stable random variable with μ=0.(v)
LetX be a random variable law Sα(β,σ,μ) with α∈(0,2], then
(19)E(|X|p)<∞ifp∈]0;α[,E(|X|p)=∞ifp∈]α;∞[.(vi)
LetX be a random variable law Sα(β,σ,μ) with α∈]1,2[, then
(20)E(X)=μ.In this section, it is assumed that the samplesX1,X2,…,Xn are independent and equally distributed stable random variables, such that Xi⇝Sα(β,σ,μ), i=1,n.¯


### 2.3. The Asymptotic Behavior

Theorem 16 (see [11]).
For0<α<2, -1≤β≤1, σ≥0, μ∈ℝ, If X~Sα(β,σ,μ), then
(21)limx→∞xαP(X>x)=Cα(1+β)σα,whereCα=(2∫0+∞x-αsin(x)dx)-1=1πΓ(α)sin(απ2),
by derivation, for large data,
(22)f(x,α,β,σ,μ)~ασα(1+β)sin(απ2)Γ(α)πx-(α+1),
or Γ denotes the Gamma function defined by Γ(α)=∫0+∞e-ttα-1dt.Asx→∞,P(X>x)=Cα(1+β)σαx-α⇒F(x)=P(x≤x)=1-P(X>x)=1-Cα(1+β)σαx-α, then by derivation f(x)=(F(x))′=αCα(1+β)σαx-α-1=αCα(1+β)σαx-(α+1).


### 2.4. Parameter Estimation

#### 2.4.1. Tail Exponent Estimation

There are many methods to give estimators of the index exponent of stable distributions, but we describe here the regression method, which is based on the effect that, for large enough values ofX, the tails of stable laws come in a power of alpha; see (22); then, we consider a linear regression as
(23)lnP(X>x)~lnCα(1+β)σα-αlnx,
where the slope of the line will be an estimator of -α.


### 2.5. Simulation of anα-Stable Distribution

To simulate the stable laws, there is an algorithm developed by Chambers et al. [5]. It can generate a law Sα(β,1,0); then to obtain a Sα(β,σ,μ) stable law, just make a change of variables.Step 1.
Generate two random variables as follows.
LetU be uniform on ]0,1[. Then, W=πU-(π/2) is uniform on ]-π/2,π/2[, and Z=-ln(1-U) is exponentially distributed with mean 1.Step 2.
GenerateX a Sα(β,1,0) law.(i)
Forα≠1(24)X=ζ+sinαW-ζcosαW(cosW)1/α×(cos(1-α)W-ζsin(1-α)WZ)((1-α)/α).(ii)
Forα=1(25)X=2π[(π2+βW)tanY-βln(π/2)cosW(π/2)+βW],Step 3.
Generate aSα(β,σ,μ) law using these transformations:
(26)Y=σX+μ,ifα≠1,Y=σX+2πβσlnσ+μ,ifα=1.
For more details see Samorodnitsky and Taqqu, [12, page 43].


## 2.1. Interpretation of the Parameters of the Characteristic Function of a Stable Law

(1)
The parameterα known as an exponent characteristic or a stability index describes the form of distribution or the degree of thickness of the tail distribution (0<α≤2). If α is smaller, then the tails of the distribution are thick. In other words, the more small α is, the more we see the existence of very large fluctuations (see Figure 1). A Gaussian distribution that has the maximum value of α is α=2.
(2)
The parameterβ gives an idea of asymmetric distribution. This is the parameter of asymmetry. If β is equal to -1 (resp., 1), distribution is totally asymmetrical left or right. When β=0, then the distribution is symmetric around μ (see Figure 2).
(3)
The parameterσ is called a scale factor. σ is more grand, if more data are volatile (see Figure 3). The parameter σ has the possibility to bend more or less the body of the distribution.
(4)
The parameterμ location corresponds for α>1 to the mean of the distribution. If β=0, then μ is the median. In other cases, μ cannot be interpreted.Figure 1
pdf plot for different values of alpha.Figure 2
pdf plot for different values of beta.Figure 3
pdf plot for different values of sigma.


## 2.2. Arithmetic Properties

Stable random variablesSα(β,σ,μ) have the following properties.(i)
X1,X2 are two independent random variables that follow, respectively, stable laws Sα(β1,σ1,μ1), Sα(β2,σ2,μ2); then, X1+X2 follows a stable law Sα(β,σ,μ) with
(17)σ=(σ1α+σ2α)1/α,β=β1σ1α+β2σ2ασ1α+σ2α,μ=μ1+μ2.
Note that, if β1=β2, then β=β1=β2.(ii)
IfX~Sα(β,σ,μ), a∈ℝ, then X+a~Sα(β,σ,μ+a).(iii)
IfX1,X2 follows a stable law Sα(β,σ,μ),a∈ℝ+,b∈ℝ+,c∈ℝ, then
(18)aX1+bX2+c=dSα(β,σ(aα+bα)1/α,μ(aα+bα)1/α+c),
where “=d” means “has the same distribution.”(iv)
X is a symmetric random variable around μ if and only if the law of X is Sα(0,σ,μ). We note that SαS is a symmetric stable random variable with μ=0.(v)
LetX be a random variable law Sα(β,σ,μ) with α∈(0,2], then
(19)E(|X|p)<∞ifp∈]0;α[,E(|X|p)=∞ifp∈]α;∞[.(vi)
LetX be a random variable law Sα(β,σ,μ) with α∈]1,2[, then
(20)E(X)=μ.In this section, it is assumed that the samplesX1,X2,…,Xn are independent and equally distributed stable random variables, such that Xi⇝Sα(β,σ,μ), i=1,n.¯


## 2.3. The Asymptotic Behavior

Theorem 16 (see [11]).
For0<α<2, -1≤β≤1, σ≥0, μ∈ℝ, If X~Sα(β,σ,μ), then
(21)limx→∞xαP(X>x)=Cα(1+β)σα,whereCα=(2∫0+∞x-αsin(x)dx)-1=1πΓ(α)sin(απ2),
by derivation, for large data,
(22)f(x,α,β,σ,μ)~ασα(1+β)sin(απ2)Γ(α)πx-(α+1),
or Γ denotes the Gamma function defined by Γ(α)=∫0+∞e-ttα-1dt.Asx→∞,P(X>x)=Cα(1+β)σαx-α⇒F(x)=P(x≤x)=1-P(X>x)=1-Cα(1+β)σαx-α, then by derivation f(x)=(F(x))′=αCα(1+β)σαx-α-1=αCα(1+β)σαx-(α+1).


## 2.4. Parameter Estimation

### 2.4.1. Tail Exponent Estimation

There are many methods to give estimators of the index exponent of stable distributions, but we describe here the regression method, which is based on the effect that, for large enough values ofX, the tails of stable laws come in a power of alpha; see (22); then, we consider a linear regression as
(23)lnP(X>x)~lnCα(1+β)σα-αlnx,
where the slope of the line will be an estimator of -α.


## 2.4.1. Tail Exponent Estimation

There are many methods to give estimators of the index exponent of stable distributions, but we describe here the regression method, which is based on the effect that, for large enough values ofX, the tails of stable laws come in a power of alpha; see (22); then, we consider a linear regression as
(23)lnP(X>x)~lnCα(1+β)σα-αlnx,
where the slope of the line will be an estimator of -α.


## 2.5. Simulation of anα-Stable Distribution

To simulate the stable laws, there is an algorithm developed by Chambers et al. [5]. It can generate a law Sα(β,1,0); then to obtain a Sα(β,σ,μ) stable law, just make a change of variables.Step 1.
Generate two random variables as follows.
LetU be uniform on ]0,1[. Then, W=πU-(π/2) is uniform on ]-π/2,π/2[, and Z=-ln(1-U) is exponentially distributed with mean 1.Step 2.
GenerateX a Sα(β,1,0) law.(i)
Forα≠1(24)X=ζ+sinαW-ζcosαW(cosW)1/α×(cos(1-α)W-ζsin(1-α)WZ)((1-α)/α).(ii)
Forα=1(25)X=2π[(π2+βW)tanY-βln(π/2)cosW(π/2)+βW],Step 3.
Generate aSα(β,σ,μ) law using these transformations:
(26)Y=σX+μ,ifα≠1,Y=σX+2πβσlnσ+μ,ifα=1.
For more details see Samorodnitsky and Taqqu, [12, page 43].


## 3. Probability Ratio Test Applying to Stable Laws in the Casex→+∞

We will useθ for the exponent characteristic and δ for the asymmetry parameter to avoid confusion with the symbols α and β which were used in type I error and type II error in Wald’s sequential probability ratio test, and for simplifying the process, let us consider symmetric standard alpha stable random variables Xi≡Sθ(0,1,0),i=1,2,…,m, in other terms (β=0,σ=1,μ=0) where θ is unknown.We assume that we have a sample ofX1,X2,…,Xn independent random variables which follow the same law X≡Sθ(0,1,0) with unknown θ.We want to use the sequential probability ratio testSPRT, for the hypothesisH0:θ=θ0,against the hypothesis,H1:θ=θ1,where θ1>θ0>0, if we want to get a test as P(RejectsH0/H0istrue)=α and P(AcceptsH0/H0isfalse)=β.We are willing to choose values forα and β as the Wald’s [7] approximation A=(1-β)/α,B=β/(1-α). Where α,β∈[0.01;0.05], at the end of the mth observation:(i)
acceptH0 if ∑i=1mZi≤lnB;(ii)
acceptH1 if ∑i=1mZi≥lnA;(iii)
continue the test iflnA<∑i=1mZi<lnB, with Zi defined by the relationship in general by
(27)Zi=lnf(xi,θ1,δ,σ,μ)f(xi,θ0,δ,σ,μ)=lnθ1σθ1(1+δ)sin(απ/2)(Γ(α)/π)x-(θ1+1)θ0σθ0(1+δ)sin(απ/2)(Γ(α)/π)x-(θ0+1).In our case of symmetric standard alpha stable random variables, we have
(28)Zi=lnf(xi,θ1,0,1,0)f(xi,θ0,0,1,0)~lnθ1sin(θ1π/2)(Γ(θ1)/π)xi-(θ1+1)θ0sin(θ0π/2)(Γ(θ0)/π)xi-(θ0+1),
so,
(29)Zi=ln(θ1Γ(θ1)sin((1/2)πθ1)θ0Γ(θ0)sin((1/2)πθ0)xiθ0-θ1),(30)Zi=lnθ1θ0+lnΓ(θ1)Γ(θ0)+lnsin((1/2)πθ1)sin((1/2)πθ0)+(θ0-θ1)lnxi.
In (30), we note that for 1<θ<2,sin((1/2)πθ)>0, then
(31)∑i=1mZi~mlnθ1θ0+mlnsin(θ1π/2)sin(θ0π/2)+mlnΓ(θ1)Γ(θ0)+(θ0-θ1)∑i=1mln(xi).
To summarize, we accept H0 if
(32)mlnθ1θ0+mlnsin(θ1π/2)sin(θ0π/2)+mlnΓ(θ1)Γ(θ0)+(θ0-θ1)∑i=1mln(xi)≤lnB=lnβ1-α⟺∑i=1mln(xi)≤1(θ1-θ0)×(mlnθ1θ0+mlnsin(θ1π/2)sin(θ0π/2)+mlnΓ(θ1)Γ(θ0)-lnβ1-α)=defam.(i)
AcceptH1, if
(33)∑i=1mZi~mlnθ1θ0+mlnsin(θ1π/2)sin(θ0π/2)+mlnΓ(θ1)Γ(θ0)+(θ0-θ1)∑i=1mln(xi)≥lnA=ln1-βα⟺∑i=1mln(xi)≥1(θ1-θ0)×(mlnθ1θ0+mlnsin(θ1π/2)sin(θ0π/2)+mlnΓ(θ1)Γ(θ0)-ln1-βα)=defrm.(ii)
We take an additional observation if(34)am<∑i=1mln(xi)<rm.In practice, we trace the following two parallel lines:
(35)L0:y=1(θ1-θ0)×(mlnθ1θ0+mlnsin(θ1π/2)sin(θ0π/2)+mlnΓ(θ1)Γ(θ0)-lnβ1-α),L1:y=1(θ1-θ0)×(mlnθ1θ0+mlnsin(θ1π/2)sin(θ0π/2)+mlnΓ(θ1)Γ(θ0)-ln1-βα).
Then we place on the same graph the points of coordinates (m,∑i=1mln(xi)) as long as they are in the band plan defined by two straight, we decides H0 whether L0 is crossed  the band formed by the two straights, and we decides H1 if L1 is crossed. We continue the sampling if the points are in the band.Example 17.
For a sample ofX1,X2,…,Xn independent random variables which follow the same law X≡Sθ(0,1,0) with unknown θ, we want to use the sequential probability ratio test SPRT, for the hypothesisH0:θ=θ0=1.3against the hypothesis,H1:θ=θ1=1.5,whereθ1>θ0>0. Let α=0.01 and β=0.05, then A=(1-β)/α=(1-0.05)/0.01=95.0, B=β/(1-α)=0.05/(1-0.01)=5.0505×10-2.lnA=ln(95.0)=4.5539 and lnB=ln(5.0505×10-2)=-2.9857.L0:y=-0.50360m+14.929 with thick black line (straight).L1:y=-0.50360m-22.770 with red line (straight) (see Figure 4).Figure 4
AcceptingH0.Example 18.
We assume that we have a sample ofX1,X2,…,Xn independent random variables which follow the same normal distribution N(θ,σ2) with unknown θ. We want to use the sequential probability ratio test SPRT, for the hypothesis H0:θ=θ0against the hypothesis,H1:θ=θ1,where θ1>θ0>0; if we follow the same steps, we obtain
(36)L0:y=θ1+θ02m+σ2θ1-θ0ln(β1-α),L1:y=θ1+θ02m+σ2θ1-θ0ln(1-βα).


### 3.1. Efficiency Function and Average Sample Number Function

Definition 19.
The efficiency function of the probability ratio test is(37)θ⟶ℙθ(AcceptsH0).It is possible to obtain additional information on the sequential probability test, especially on the number of observationsN needed to take decision, which is a discrete random variable that admits moments of all kinds; consider the function average sample number (ASN).Definition 20.
One calls the followingaverage sample number (ASN) function:
(38)θ⟶𝔼θ(N).Theorem 21.
Let a fixedθ∈Θ. If 𝔼θ(|Z1|)<+∞, then
(39)𝔼θ(N)𝔼θ(Z1)=𝔼θ(SN),
where SN=∑i=1NZi and Zi is defined by (29).Theorem 22.
Let a fixedθ∈Θ. If 𝔼θ(|Z1|)<+∞ and 𝔼θ(Z1)≠0, then
(40)𝔼θ(N)~(1-L(θ))lnA+L(θ)lnB𝔼θ(Z1).In the case of stable densities with1<α<2,𝔼(X)=μ<+∞, see (v) in arithmetic properties,
(41)𝔼θ(|Z1|)=𝔼θ(|lnθ1θ0+lnΓ(θ1)Γ(θ0)+lnsin((1/2)πθ1)sin((1/2)πθ0)+(θ0-θ1)lnx1sin((1/2)πθ1)sin((1/2)πθ0)|)=lnθ1θ0+lnΓ(θ1)Γ(θ0)+lnsin((1/2)πθ1)sin((1/2)πθ0)+(θ0-θ1)μ<∞,
and with a simple calculation, and for symmetric standard alpha stable random variable (β=0,σ=1,μ=0),
(42)𝔼θ(Z1)=lnθ1θ0+lnΓ(θ1)Γ(θ0)+lnsin((1/2)πθ1)sin((1/2)πθ0).
Then,
(43)𝔼θ(N)~(1-L(θ))lnA+L(θ)lnB𝔼θ(Z1),
where L(θ) is the function of efficiency. With substitution in (29)
(44)𝔼θ(Z1)=𝔼θ0(Z1)=lnθ1θ0+lnΓ(θ1)Γ(θ0)+lnsin((1/2)πθ1)sin((1/2)πθ0),
finally, under H0(45)𝔼θ0(N)~αln((1-β)/α)+(1-α)ln(β/(1-α))ln(θ1/θ0)+ln(Γ(θ1)/Γ(θ0))+ln(sin((1/2)πθ1)/sin((1/2)πθ0)),becauseL(θ0)=1-α, and under H1(46)𝔼θ1(N)~(1-β)ln((1-β)/α)+βln(β/(1-α))ln(θ1/θ0)+ln(Γ(θ1)/Γ(θ0))+ln(sin((1/2)πθ1)/sin((1/2)πθ0)),becauseL(θ1)=β.Example 23.
Letα=0.01 and β=0.05; then, for the index parameter of symmetric stable distributions θ1=1.5, θ0=1.3,
(47)𝔼θ0(N)=28.907.In the case ofnormaldistribution α=2, that is, 𝒩(θ,1), we use the same steps, but here we have the analytic form of the density function, and the average sample number function is under H0(48)𝔼θ(N)~αln((1-β)/α)+(1-α)ln(β/(1-α))-(1/2)(θ1-θ0)2,
because L(θ0)=1-α, and under H1(49)𝔼θ(N)~(1-β)ln((1-β)/α)+βln(β/(1-α))(1/2)(θ1-θ0)2,
because L(θ1)=β, where
(50)𝔼θ(Z1)=(θ1-θ0)θ-12(θ12-θ02).


## 3.1. Efficiency Function and Average Sample Number Function

Definition 19.
The efficiency function of the probability ratio test is(37)θ⟶ℙθ(AcceptsH0).It is possible to obtain additional information on the sequential probability test, especially on the number of observationsN needed to take decision, which is a discrete random variable that admits moments of all kinds; consider the function average sample number (ASN).Definition 20.
One calls the followingaverage sample number (ASN) function:
(38)θ⟶𝔼θ(N).Theorem 21.
Let a fixedθ∈Θ. If 𝔼θ(|Z1|)<+∞, then
(39)𝔼θ(N)𝔼θ(Z1)=𝔼θ(SN),
where SN=∑i=1NZi and Zi is defined by (29).Theorem 22.
Let a fixedθ∈Θ. If 𝔼θ(|Z1|)<+∞ and 𝔼θ(Z1)≠0, then
(40)𝔼θ(N)~(1-L(θ))lnA+L(θ)lnB𝔼θ(Z1).In the case of stable densities with1<α<2,𝔼(X)=μ<+∞, see (v) in arithmetic properties,
(41)𝔼θ(|Z1|)=𝔼θ(|lnθ1θ0+lnΓ(θ1)Γ(θ0)+lnsin((1/2)πθ1)sin((1/2)πθ0)+(θ0-θ1)lnx1sin((1/2)πθ1)sin((1/2)πθ0)|)=lnθ1θ0+lnΓ(θ1)Γ(θ0)+lnsin((1/2)πθ1)sin((1/2)πθ0)+(θ0-θ1)μ<∞,
and with a simple calculation, and for symmetric standard alpha stable random variable (β=0,σ=1,μ=0),
(42)𝔼θ(Z1)=lnθ1θ0+lnΓ(θ1)Γ(θ0)+lnsin((1/2)πθ1)sin((1/2)πθ0).
Then,
(43)𝔼θ(N)~(1-L(θ))lnA+L(θ)lnB𝔼θ(Z1),
where L(θ) is the function of efficiency. With substitution in (29)
(44)𝔼θ(Z1)=𝔼θ0(Z1)=lnθ1θ0+lnΓ(θ1)Γ(θ0)+lnsin((1/2)πθ1)sin((1/2)πθ0),
finally, under H0(45)𝔼θ0(N)~αln((1-β)/α)+(1-α)ln(β/(1-α))ln(θ1/θ0)+ln(Γ(θ1)/Γ(θ0))+ln(sin((1/2)πθ1)/sin((1/2)πθ0)),becauseL(θ0)=1-α, and under H1(46)𝔼θ1(N)~(1-β)ln((1-β)/α)+βln(β/(1-α))ln(θ1/θ0)+ln(Γ(θ1)/Γ(θ0))+ln(sin((1/2)πθ1)/sin((1/2)πθ0)),becauseL(θ1)=β.Example 23.
Letα=0.01 and β=0.05; then, for the index parameter of symmetric stable distributions θ1=1.5, θ0=1.3,
(47)𝔼θ0(N)=28.907.In the case ofnormaldistribution α=2, that is, 𝒩(θ,1), we use the same steps, but here we have the analytic form of the density function, and the average sample number function is under H0(48)𝔼θ(N)~αln((1-β)/α)+(1-α)ln(β/(1-α))-(1/2)(θ1-θ0)2,
because L(θ0)=1-α, and under H1(49)𝔼θ(N)~(1-β)ln((1-β)/α)+βln(β/(1-α))(1/2)(θ1-θ0)2,
because L(θ1)=β, where
(50)𝔼θ(Z1)=(θ1-θ0)θ-12(θ12-θ02).


## 4. Conclusion

We use the fact that the tails of the Lévy-stable distributions are asymptotically equivalent to a Pareto law when1<α<2 for large data as x→∞,P(X>x)=1-F(x)→σαCα(1+β)x-α, where Cα=(2∫0∞x-αsinxdx)-1=(1/π)Γ(α)sin(απ/2) to obtain the probability ratio, and we compare the fit of two Pareto stable models using the sequential probability ratio test which is not possible in the other cases because of the lack of the analytical formulas of stable densities; then, we apply the theorem giving the average sample number function.Can one give a bound of the average sample number function of stable distributions in our case, and in other cases?Can one say that this amount of data is sufficient to make the decision that this population follows this stable law with this alpha parameter or the other laws with the second alpha parameter?One can find many applications of the Pareto stable distributions in many fields when the data are very large, such as the flow of profits or revenues in public and private companies with huge capital, such as banks, insurance companies, and oil companies. Actuaries give speculation of instant transactions reliability and test hypotheses and advice to guide managers; however, it would be prudent in their decisions in cases where the data are incomplete in the sense that the mean of the number of data required to take decision is not achieved. As in the likelihood, we use the log data, which simplifies the calculus.



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