Average Sample Number Function for Pareto Heavy Tailed Distributions

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 case 1<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.

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 Section 1, 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 variable 𝑋 has a distribution infinitely divisible if and only if for all 𝑛,∃𝑋₁,…,𝑋_𝑛 independent with the same even law such as 𝑋=^𝑑𝑋₁+⋯+𝑋_𝑛, (1) where =^𝑑 means equality in distribution.

Remark 2. The random variables 𝑋_𝑖 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 distribution 𝒩(𝜇,𝜎²) is written as 𝜑_{𝒩(𝜇,𝜎2)}(𝑡)𝜎=exp𝑖𝜇𝑡−²𝑡²2=𝑖𝜇exp𝑛𝜎𝑡−²/𝑛2𝑡^2𝑛=𝜑_{𝒩(𝜇/𝑛,𝜎2/𝑛)}(𝑡)^𝑛, (2) as 𝑛th power of characteristic function of a normal distribution 𝒩(𝜇/𝑛,𝜎²/𝑛).

Example 4. The characteristic function of Cauchy distribution 𝐶(𝑐) is written as 𝜑_𝐶(𝑐)(𝑡){|𝑡=−𝑐=exp−𝑐|}exp𝑛|𝑡|^𝑛=𝜑_{𝒞(𝑐/𝑛)}(𝑡)^𝑛, (3) as 𝑛th power of the characteristic function of a Cauchy distribution 𝒞(𝑐/𝑛).

Example 5. The characteristic function of Poisson distribution 𝑃(𝜆) is written as 𝜑_𝒫(𝜆)(𝑡)=exp𝜆𝑒^𝑖𝑡=𝜆−1exp𝑛𝑒^𝑖𝑡−1^𝑛=𝜑_{𝒫(𝜆/𝑛)}(𝑡)^𝑛, (4) as 𝑛th power eme of the characteristic function of a Poisson law 𝒫(𝜆/𝑛).

Example 6. The characteristic function of Gamma distribution Γ(𝑟,𝜆) is written as 𝜑_{−(𝑟,𝜆)}(𝑡)=1(1−𝑖𝑡/𝜆)^𝑟=1(1−𝑖𝑡/𝜆)^{𝑟/𝑛𝑛}=𝜑_{−(𝑟/𝑛,𝜆)}(𝑡), (5) 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 variable 𝑋 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 variable 𝑋 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 𝑎𝑋₁+𝑏𝑋₂=𝑐𝑋+𝑑,indistribution (6) satisfies for some 𝑐∈ℝ⁺ and 𝑑∈ℝ.

Definition 10. A random variable 𝑋 is stable distribution if and only if, for all 𝑛 and any independent identique family 𝑋₁,𝑋₂,…,𝑋_𝑛 which have the same law that 𝑋, there is 𝑎_𝑛>0 and 𝑏_𝑛∈𝑅, as 𝑋₁+𝑋₂+⋯+𝑋=_𝑛^𝑑𝑎_𝑛𝑋+𝑏_𝑛. (7)

Proposition 11. If 𝑋 is stable, 𝑋 is divisible indefinitely. The inverse is false (see the example below, the Poisson distribution).

Proof. Just take the random variables 𝑌_𝑖=(𝑋_𝑖−𝑏_𝑛/𝑛)/𝑎_𝑛,𝑖=1,…,𝑛.
As 𝑋_𝑖 are independent, 𝑌_𝑖 are also independent, and by substitution, we have 𝑌₁+𝑌₂+⋯+𝑌=_𝑛^𝑑𝑋. (8)

Remark 12. This definition is coinside of the central limit theorem if 𝑋 is following a normal law.

Theorem 13. 𝑋 has domain of attraction; that is, there is a sequence of random variables 𝑖.𝑖.𝑑.(𝑋_𝑖)_𝑖∈𝑁, a sequence of positive real numbers (𝑎_𝑖)_𝑖∈ℕ, and a sequence of real numbers (𝑏_𝑖)_𝑖∈ℕ, such as 1𝑎_𝑛^𝑛_𝑖=1𝑋_𝑖−𝑏_𝑛⟶𝑋,𝑖𝑛𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛. (9)

Proof. The demonstration is detailed in Shiryayev [9].

Corollary 14 (Levy-Khinchin). The characteristic function of a stable random variable 𝑋 admits the following form: 𝜓_𝑋(𝑡)=exp−𝜎^𝛼|𝑡|^𝛼1−𝑖𝛽⋅sgn(𝑡)⋅𝑡𝑔𝛼𝜋2,|𝑡|2+𝑖𝜇𝑡𝛼≠1exp−𝜎1+𝜋(𝑡)|𝑡|,(𝑡)=𝑖𝛽⋅sgn⋅ln+𝑖𝜇𝑡𝛼=1,𝑤ℎ𝑒𝑟𝑒sgn1,𝑡>00,𝑡=0−1,𝑡<0. (10)

Proof. The demonstration is detailed in Gnedenko and Kolmogorov [10].

The random variable 𝑋 is a stable law with parameters 𝛼, 𝛽, 𝜎, and 𝜇; we note that 𝑋≡𝑆_𝛼𝛽,𝜎,𝜇,where0<𝛼≤2,−1≤𝛽≤1,𝜎≥0,𝜇∈ℝ. (11)

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 where 𝛼=1 and 𝛽=0; otherwise, there are other parameterization of the FC more adapted to different problems. Consider 𝜓_𝑋(𝑡)=exp−𝜎^𝛼|𝑡|^𝛼1+𝑖𝛽⋅sgn(𝑡)⋅tan𝛼𝜋2|𝜎𝑡|^1−𝛼−1+𝑖𝜇₀𝑡,|𝑡|2𝛼≠1exp−𝜎1+𝜋𝑖𝛽⋅sgn(𝑡)⋅ln|𝑡|+ln𝜎+𝑖𝜇₀𝑡,𝛼=1. (12)

This representation of Zolotarev’s 𝑀 parameterization notes 𝑋∼𝑆⁰_𝛼(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 𝜇𝜇=⁰−𝛽𝜎tan𝛼𝜋2𝜇,𝛼≠1₀−2𝜋𝛽𝜎ln𝜎,𝛼=1. (13)

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 parameterization 𝑆¹ is given by the following: 𝜓_𝑋(𝑡)=exp−𝜎^𝛼₂|𝑡|^𝛼exp−𝑖𝛽₂(𝑡)⋅𝜋⋅sgn2𝐾(𝛼)+𝑖𝜇𝑡,𝛼≠1exp−𝜎₂|𝑡|𝜋2+𝑖𝛽₂⋅sgn(𝑡)⋅ln|𝑡|+𝑖𝜇𝑡,𝛼=1, (14) where 𝐾(𝛼)=𝛼−1+sgn(1−𝛼)=𝛼,𝛼≤1𝛼−2,𝛼≥1, (15) where parameters 𝛼 and 𝜇 are the same as for standard parameterization; other parameters meet the following relationships: 𝜎₂=𝜎1+𝛽²tan²𝛼𝜋2^1/2𝛼,tan𝛽₂𝜋𝐾(𝛼)2=𝛽tan𝛼𝜋2𝜎,𝛼≠1,₂=2𝜎𝜋,𝛽₂=𝛽,𝛼=1. (16)

Remark 15. The main disadvantage is that the densities of stable laws are unknown except in three cases.(1)The Gaussian distribution 𝑆₂(0,𝜎,𝜇) where 𝑓(𝑥,2,0,𝜎,𝜇)=(1/2𝜎√𝜋)exp(−(𝑥−𝜇)²/4𝜎²), 𝑥∈ℝ.(2)The Cauchy distribution 𝑆₁(0,𝜎,𝜇) where 𝑓(𝑥,1,0,𝜎,𝜇)=(2𝜎/𝜋((𝑥−𝜇)²+4𝜎²)),𝑥∈ℝ.(3)The Lévy distribution 𝑆_1/2(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 density 𝑓(𝑥)=(1/2𝜋)∫^+∞_−∞𝑒^{−𝑖𝑡𝑥}𝜓_𝑋(𝑡)𝑑𝑡.

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

(1)The parameter 𝛼 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 parameter 𝛽 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 parameter 𝜎 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 parameter 𝜇 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 variables 𝑆_𝛼(𝛽,𝜎,𝜇) have the following properties.(i)𝑋₁,𝑋₂ are two independent random variables that follow, respectively, stable laws 𝑆_𝛼(𝛽₁,𝜎₁,𝜇₁), 𝑆_𝛼(𝛽₂,𝜎₂,𝜇₂); then, 𝑋₁+𝑋₂ follows a stable law 𝑆_𝛼(𝛽,𝜎,𝜇) with 𝜎=𝜎^𝛼₁+𝜎^𝛼₂^1/𝛼,𝛽𝛽=¹𝜎^𝛼₁+𝛽₂𝜎^𝛼₂𝜎^𝛼₁+𝜎^𝛼₂,𝜇=𝜇₁+𝜇₂. (17) Note that, if 𝛽₁=𝛽₂, then 𝛽=𝛽₁=𝛽₂.(ii)If 𝑋∼𝑆_𝛼(𝛽,𝜎,𝜇), 𝑎∈ℝ, then 𝑋+𝑎∼𝑆_𝛼(𝛽,𝜎,𝜇+𝑎).(iii)If 𝑋₁,𝑋₂ follows a stable law 𝑆_𝛼(𝛽,𝜎,𝜇),𝑎∈ℝ⁺,𝑏∈ℝ⁺,𝑐∈ℝ, then 𝑎𝑋₁+𝑏𝑋₂+𝑐=^𝑑𝑆_𝛼𝛽,𝜎𝑎^𝛼+𝑏^𝛼1/𝛼,𝜇𝑎^𝛼+𝑏^𝛼1/𝛼+𝑐, (18) 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 𝑆_𝛼(0,𝜎,𝜇). We note that 𝑆𝛼𝑆 is a symmetric stable random variable with 𝜇=0.(v)Let 𝑋 be a random variable law 𝑆_𝛼(𝛽,𝜎,𝜇) with 𝛼∈(0,2], then 𝐸|𝑋|^𝑝][,𝐸|𝑋|<∞if𝑝∈0;𝛼_𝑝=∞if𝑝∈]𝛼;∞[. (19)(vi)Let 𝑋 be a random variable law 𝑆_𝛼(𝛽,𝜎,𝜇) with 𝛼∈]1,2[, then 𝐸(𝑋)=𝜇. (20)

In this section, it is assumed that the samples 𝑋₁,𝑋₂,…,𝑋_𝑛 are independent and equally distributed stable random variables, such that 𝑋_𝑖⇝𝑆_𝛼(𝛽,𝜎,𝜇), 𝑖=1,𝑛.

2.3. The Asymptotic Behavior

Theorem 16 (see [11]). For 0<𝛼<2, −1≤𝛽≤1, 𝜎≥0, 𝜇∈ℝ, If 𝑋∼𝑆_𝛼(𝛽,𝜎,𝜇), then lim_𝑥→∞𝑥^𝛼𝑃(𝑋>𝑥)=𝐶_𝛼1+𝛽𝜎^𝛼,𝑤ℎ𝑒𝑟𝑒𝐶_𝛼=2₀^+∞𝑥^−𝛼sin(𝑥)𝑑𝑥⁻¹=1𝜋Γ(𝛼)sin𝛼𝜋2, (21) by derivation, for large data, 𝑓𝑥,𝛼,𝛽,𝜎,𝜇∼𝛼𝜎^𝛼1+𝛽sin𝛼𝜋2Γ(𝛼)𝜋𝑥^−(𝛼+1), (22) or Γ denotes the Gamma function defined by Γ(𝛼)=∫₀^+∞𝑒^−𝑡𝑡^𝛼−1𝑑𝑡.

As 𝑥→∞,𝑃(𝑋>𝑥)=𝐶_𝛼(1+𝛽)𝜎^𝛼𝑥^−𝛼⇒𝐹(𝑥)=𝑃(𝑥≤𝑥)=1−𝑃(𝑋>𝑥)=1−𝐶_𝛼(1+𝛽)𝜎^𝛼𝑥^−𝛼, then by derivation 𝑓(𝑥)=(𝐹(𝑥))^′=𝛼𝐶_𝛼(1+𝛽)𝜎^𝛼𝑥^−𝛼−1=𝛼𝐶_𝛼(1+𝛽)𝜎^𝛼𝑥^−(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 of 𝑋, the tails of stable laws come in a power of alpha; see (22); then, we consider a linear regression as ln𝑃(𝑋>𝑥)∼ln𝐶_𝛼1+𝛽𝜎^𝛼−𝛼ln𝑥, (23) where the slope of the line will be an estimator of −𝛼.

2.5. Simulation of an 𝛼-Stable Distribution

To simulate the stable laws, there is an algorithm developed by Chambers et al. [5]. It can generate a law 𝑆_𝛼(𝛽,1,0); then to obtain a 𝑆_𝛼(𝛽,𝜎,𝜇) stable law, just make a change of variables.

Step 1. Generate two random variables as follows.
Let 𝑈 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. Generate 𝑋 a 𝑆_𝛼(𝛽,1,0) law.(i)For 𝛼≠1𝑋=𝜁+sin𝛼𝑊−𝜁cos𝛼𝑊(cos𝑊)^1/𝛼×cos(1−𝛼)𝑊−𝜁sin(1−𝛼)𝑊𝑍^{((1−𝛼)/𝛼)}. (24)(ii)For 𝛼=12𝑋=𝜋𝜋2()+𝛽𝑊tan𝑌−𝛽ln𝜋/2cos𝑊(),𝜋/2+𝛽𝑊 (25)

Step 3. Generate a 𝑆_𝛼(𝛽,𝜎,𝜇) law using these transformations: 2𝑌=𝜎𝑋+𝜇,if𝛼≠1,𝑌=𝜎𝑋+𝜋𝛽𝜎ln𝜎+𝜇,if𝛼=1. (26) For more details see Samorodnitsky and Taqqu, [12, page 43].

3. Probability Ratio Test Applying to Stable Laws in the Case 𝑥→+∞

We will use 𝜃 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 𝑋_𝑖≡𝑆_𝜃(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 of 𝑋₁,𝑋₂,…,𝑋_𝑛 independent random variables which follow the same law 𝑋≡𝑆_𝜃(0,1,0) with unknown 𝜃.

We want to use the sequential probability ratio test SPRT, for the hypothesis 𝐻₀: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 for 𝛼 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)accept 𝐻₀ if ∑^𝑚_𝑖=1𝑍_𝑖≤ln𝐵;(ii)accept 𝐻₁ if ∑^𝑚_𝑖=1𝑍_𝑖≥ln𝐴;(iii)continue the test if ln𝐴<∑^𝑚_𝑖=1𝑍_𝑖<ln𝐵, with 𝑍_𝑖 defined by the relationship in general by 𝑍_𝑖𝑓𝑥=ln^𝑖,𝜃₁,𝛿,𝜎,𝜇𝑓𝑥_𝑖,𝜃₀𝜃,𝛿,𝜎,𝜇=ln¹𝜎^𝜃1(1+𝛿)sin(𝛼𝜋/2)(Γ(𝛼)/𝜋)𝑥^−(𝜃1+1)𝜃₀𝜎^𝜃0(1+𝛿)sin(𝛼𝜋/2)(Γ(𝛼)/𝜋)𝑥^−(𝜃0+1). (27)In our case of symmetric standard alpha stable random variables, we have 𝑍_𝑖𝑓𝑥=ln^𝑖,𝜃₁,0,1,0𝑓𝑥_𝑖,𝜃₀𝜃,0,1,0∼ln¹sin𝜃₁𝜋/2Γ𝜃₁/𝜋𝑥^−(𝜃1_𝑖⁺¹⁾𝜃₀sin𝜃₀𝜋/2Γ𝜃₀/𝜋𝑥^−(𝜃0_𝑖⁺¹⁾, (28) so, 𝑍_𝑖𝜃=ln¹Γ𝜃₁sin(1/2)𝜋𝜃₁𝜃₀Γ𝜃₀sin(1/2)𝜋𝜃₀𝑥^{𝜃0−𝜃1}_𝑖𝑍_𝑖,𝜃=ln¹𝜃₀Γ𝜃+ln¹Γ𝜃₀()+lnsin1/2𝜋𝜃₁sin(1/2)𝜋𝜃₀+𝜃₀−𝜃₁ln𝑥_𝑖. (29)(30) In (30), we note that for 1<𝜃<2,sin((1/2)𝜋𝜃)>0, then ^𝑚_𝑖=1𝑍_𝑖𝜃∼𝑚ln¹𝜃₀𝜃+𝑚lnsin₁𝜋/2sin𝜃₀Γ𝜃𝜋/2+𝑚ln¹Γ𝜃₀+𝜃₀−𝜃₁^𝑚_𝑖=1ln𝑥_𝑖. (31) To summarize, we accept 𝐻₀ if 𝜃𝑚ln¹𝜃₀𝜃+𝑚lnsin₁𝜋/2sin𝜃₀Γ𝜃𝜋/2+𝑚ln¹Γ𝜃₀+𝜃₀−𝜃₁^𝑚_𝑖=1ln𝑥_𝑖𝛽≤ln𝐵=ln⟺1−𝛼_𝑚𝑖=1ln𝑥_𝑖≤1𝜃₁−𝜃₀×𝜃𝑚ln¹𝜃₀𝜃+𝑚lnsin₁𝜋/2sin𝜃₀Γ𝜃𝜋/2+𝑚ln¹Γ𝜃₀𝛽−ln1−𝛼^def=𝑎_𝑚. (32)(i)Accept 𝐻₁, if ^𝑚_𝑖=1𝑍_𝑖𝜃∼𝑚ln¹𝜃₀𝜃+𝑚lnsin₁𝜋/2sin𝜃₀Γ𝜃𝜋/2+𝑚ln¹Γ𝜃₀+𝜃₀−𝜃₁^𝑚_𝑖=1ln𝑥_𝑖≥ln𝐴=ln1−𝛽𝛼⟺^𝑚_𝑖=1ln𝑥_𝑖≥1𝜃₁−𝜃₀×𝜃𝑚ln¹𝜃₀𝜃+𝑚lnsin₁𝜋/2sin𝜃₀Γ𝜃𝜋/2+𝑚ln¹Γ𝜃₀−ln1−𝛽𝛼^def=𝑟_𝑚. (33)(ii)We take an additional observation if 𝑎_𝑚<^𝑚_𝑖=1ln𝑥_𝑖<𝑟_𝑚. (34)In practice, we trace the following two parallel lines: 𝐿₀1:𝑦=𝜃₁−𝜃₀×𝜃𝑚ln¹𝜃₀𝜃+𝑚lnsin₁𝜋/2sin𝜃₀Γ𝜃𝜋/2+𝑚ln¹Γ𝜃₀𝛽−ln,𝐿1−𝛼₁1:𝑦=𝜃₁−𝜃₀×𝜃𝑚ln¹𝜃₀𝜃+𝑚lnsin₁𝜋/2sin𝜃₀Γ𝜃𝜋/2+𝑚ln¹Γ𝜃₀−ln1−𝛽𝛼. (35) Then we place on the same graph the points of coordinates (𝑚,∑^𝑚_𝑖=1ln(𝑥_𝑖)) 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 of 𝑋₁,𝑋₂,…,𝑋_𝑛 independent random variables which follow the same law 𝑋≡𝑆_𝜃(0,1,0) with unknown 𝜃, we want to use the sequential probability ratio test SPRT, for the hypothesis 𝐻₀:𝜃=𝜃₀=1.3 against the hypothesis,  𝐻₁:𝜃=𝜃₁=1.5, where 𝜃₁>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 

Accepting 𝐻0.

Example 18. We assume that we have a sample of 𝑋₁,𝑋₂,…,𝑋_𝑛 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=𝜃₀ against the hypothesis, 𝐻₁:𝜃=𝜃₁,where 𝜃₁>𝜃₀>0; if we follow the same steps, we obtain 𝐿₀𝜃:𝑦=¹+𝜃₀2𝜎𝑚+²𝜃₁−𝜃₀𝛽ln,𝐿1−𝛼₁𝜃:𝑦=¹+𝜃₀2𝜎𝑚+²𝜃₁−𝜃₀ln1−𝛽𝛼. (36)

3.1. Efficiency Function and Average Sample Number Function

Definition 19. The efficiency function of the probability ratio test is 𝜃⟶ℙ_𝜃Accepts𝐻₀. (37)

It is possible to obtain additional information on the sequential probability test, especially on the number of observations 𝑁 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 following average sample number (ASN) function: 𝜃⟶𝔼_𝜃(𝑁). (38)

Theorem 21. Let a fixed 𝜃∈Θ. If 𝔼_𝜃(|𝑍₁|)<+∞, then 𝔼_𝜃(𝑁)𝔼_𝜃𝑍₁=𝔼_𝜃𝑆_𝑁, (39) where 𝑆_𝑁=∑^𝑁_𝑖=1𝑍_𝑖 and 𝑍_𝑖 is defined by (29).

Theorem 22. Let a fixed 𝜃∈Θ. If 𝔼_𝜃(|𝑍₁|)<+∞ and 𝔼_𝜃(𝑍₁)≠0, then 𝔼_𝜃(𝑁)∼(1−𝐿(𝜃))ln𝐴+𝐿(𝜃)ln𝐵𝔼_𝜃𝑍₁. (40)

In the case of stable densities with 1<𝛼<2,𝔼(𝑋)=𝜇<+∞, see (v) in arithmetic properties, 𝔼_𝜃𝑍₁=𝔼_𝜃𝜃ln¹𝜃₀Γ𝜃+ln¹Γ𝜃₀()+lnsin1/2𝜋𝜃₁sin(1/2)𝜋𝜃₀+𝜃₀−𝜃₁ln𝑥₁𝜃=ln¹𝜃₀Γ𝜃+ln¹Γ𝜃₀)+lnsin(1/2𝜋𝜃₁sin(1/2)𝜋𝜃₀+𝜃₀−𝜃₁𝜇<∞, (41) and with a simple calculation, and for symmetric standard alpha stable random variable (𝛽=0,𝜎=1,𝜇=0), 𝔼_𝜃𝑍₁𝜃=ln¹𝜃₀Γ𝜃+ln¹Γ𝜃₀()+lnsin1/2𝜋𝜃₁sin(1/2)𝜋𝜃₀. (42) Then, 𝔼_𝜃(𝑁)∼(1−𝐿(𝜃))ln𝐴+𝐿(𝜃)ln𝐵𝔼_𝜃𝑍₁, (43) where 𝐿(𝜃) is the function of efficiency. With substitution in (29) 𝔼_𝜃𝑍₁=𝔼_𝜃0𝑍₁𝜃=ln¹𝜃₀Γ𝜃+ln¹Γ𝜃₀()+lnsin1/2𝜋𝜃₁sin(1/2)𝜋𝜃₀, (44) finally, under 𝐻₀𝔼_𝜃0(𝑁)∼𝛼ln1−𝛽/𝛼+(1−𝛼)ln𝛽/(1−𝛼)ln𝜃₁/𝜃₀+lnΓ𝜃₁/Γ𝜃₀+lnsin(1/2)𝜋𝜃₁/sin(1/2)𝜋𝜃₀, (45)

because 𝐿(𝜃₀)=1−𝛼, and under 𝐻₁𝔼_𝜃1(𝑁)∼1−𝛽ln1−𝛽/𝛼+𝛽ln𝛽/(1−𝛼)ln𝜃₁/𝜃₀+lnΓ𝜃₁/Γ𝜃₀+lnsin(1/2)𝜋𝜃₁/sin(1/2)𝜋𝜃₀, (46)

because 𝐿(𝜃₁)=𝛽.

Example 23. Let 𝛼=0.01 and 𝛽=0.05; then, for the index parameter of symmetric stable distributions 𝜃₁=1.5, 𝜃₀=1.3, 𝔼_𝜃0(𝑁)=28.907. (47)

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 𝐻₀𝔼_𝜃(𝑁)∼𝛼ln1−𝛽/𝛼+(1−𝛼)ln𝛽/(1−𝛼)−(1/2)𝜃₁−𝜃₀², (48) because 𝐿(𝜃₀)=1−𝛼, and under 𝐻₁𝔼_𝜃(𝑁)∼1−𝛽ln1−𝛽/𝛼+𝛽ln𝛽/(1−𝛼)(1/2)𝜃₁−𝜃₀², (49) because 𝐿(𝜃₁)=𝛽, where 𝔼_𝜃𝑍₁=𝜃₁−𝜃₀1𝜃−2𝜃²₁−𝜃²₀. (50)

4. Conclusion

We use the fact that the tails of the Lévy-stable distributions are asymptotically equivalent to a Pareto law when 1<𝛼<2 for large data as 𝑥→∞,𝑃(𝑋>𝑥)=1−𝐹(𝑥)→𝜎^𝛼𝐶_𝛼(1+𝛽)𝑥^−𝛼, where 𝐶_𝛼=(2∫^∞₀𝑥^−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.