# Sequential Probabilistic Ratio Test for the Scale Parameter of theP-Norm Distribution

Authors: Huan Ren; Hongchang Hu; Zhen Zeng
Journal: Discrete Dynamics in Nature and Society
 (2021)

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## Abstract

We consider a series of independent observations from aP-norm distribution with the position parameter μ and the scale parameter σ. We test the simple hypothesis H0:σ=σ1 versus H1:σ=σ2. Firstly, we give the stop rule and decision rule of sequential probabilistic ratio test (SPRT). Secondly, we prove the existence of hσ which needs to satisfy the specific situation in SPRT method, and the approximate formula of the mean sample function is derived. Finally, a simulation example is given. The simulation shows that the ratio of sample size required by SPRT and the classic Neyman–Pearson N−P test is about 50.92% at most,38.30% at least.


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## Body

## 1. Introduction

P-norm distribution is a family of distributions, including normal distribution, Laplace distribution, uniform distribution, degenerate distribution and many unknown distributions. Let the density function of the random variable be given by(1)fx=pλ2σΓ1/pexp−λx−μσp.Then, it is said thatX follows the P-norm distribution, where λ=Γ3/p/Γ1/p1/2, the position parameter is μ, and the scale parameter is σ>0. Since it contains some important distributions, the P-norm distribution can better describe error distributions to some extent. There are many articles that studied the properties of the P-norm distribution. For example, Hu and Sun [1] systematically obtained unbiased estimators of parameters for the P-norm distribution. Sun and Hu [2] gave a density function of P-norm distribution and its sampling distribution. However, nobody investigated the sequential probabilistic ratio test for the scale parameter of P-norm distribution.Likelihood ratio method is a widely used test method, which can get a lot of in-depth results. For example, Self and Liang [3] gave the asymptotic distribution of maximum likelihood estimators and likelihood ratio statistics under nonstandard conditions. Fan and Zhang [4] proposed the sieve empirical likelihood ratio test for nonparametric functions. Ferrari and Cysneiros [5] used the Skovgaard’s modified likelihood ratio method to study the exponential family nonlinear model and obtained the approximate distribution of modified likelihood ratio statistics. Giampaoli and Singer [6] tested the variance parameter of the linear mixed model by the likelihood ratio method. Huang et al. [7] tested the shape parameter of the generalized extremum distribution with the Lq-likelihood ratio method. Qin and Priebe [8] proposed a robust Lq-likelihood ratio test for the general pollution distribution and obtained the asymptotic distribution of Lq-likelihood ratio test statistics.In order to meet the requirements of quality inspection of American munitions production during World War II, Wald [9] presented a sequential analysis method. Since then, many authors had studied sequence analysis methods and proposed various sequence tests to test different hypotheses. For example, Whitehead and Jones [10] and Jennison and Turnbull [11] provided extensive applications of sequence and group sequence based techniques in the formation and execution of clinical trials in their books. Darkhovsky [12] studied the sequence examination of two compound hypotheses and proposed a sequence process that minimizes the maximum Bayesian risk on a series of prior parameter distributions. Kachiashvili [13] proposed a sequential method to constrain the multiple test problem in the Bayesian hypothesis test task and proved the high quality of this method. Li et al. [14] extended the sequence probability ratio test and proved that the sequence test was asymptotically optimal when the error probability went to zero, so the sequence test could asymptotically obtain the minimum expected sample size. Wang et al. [15] proposed the weighted expected sample size (WESS) to evaluate the test problem of the composite hypothesis for the overall performance of three different regions. Nakamura et al. [16] proposed a sequential test procedure to determine the minimum dose with threshold effect. Li et al. [17] constructed a general sequence test to detect outliers in all collected observation sequences. Mudholkar et al. [18] deduced the sequential probabilistic ratio test method of M-Gaussian population model under the assumption that the discrete parameters were known. Zou et al. [19] proposed a nonparametric sequential test based on empirical likelihood to test the treatment effect. Tartakovsky et al. [20] discussed in detail recent advances in sequential hypothesis testing in their book.TheP-norm distribution describes the error distribution. In order to use the P-norm distribution, it is necessary to use the sequential likelihood ratio test method to study the P-norm distribution parameter. Therefore, this paper will apply the sequential probability ratio test method to study the parameter of P-norm distribution. The specific structure of this paper is as follows. Section 2 introduces the SPRT method. Section 3 gives some properties of SPRT. Section 4 performs a simulation study to confirm results. Proofs of theorems are contained in Section 5.


## 2. SPRT Method

Suppose thatX1,X2,… is an independent and identically distributed (i.i.d.) random sample sequence from the P-norm distribution. Let x1,x2,… be the sequence of their observed values. The following is a simple hypothesis test problem (let us assume that parameter μ is known. Without loss of generality, let parameter μ=0):(2)H0:σ=σ1↔H1:σ=σ2.Here,σ1,σ2∈Θ are known numbers and 0<σ1<σ2.Let us give the stopping rule and decision rule of SPRT in turn. First, consider the likelihood ratio statistic:(3)λn=∏i=1nfXi,σ2∏i=1nfXi,σ1=σ1σ2nexpλpσ1−p−σ2−p∑i=1nXip.LetSnp=∑i=1nXip. Taking the logarithm of equation (3), we can get(4)lnλn=nlnσ1σ2+λpσ1−p−σ2−pSnp.Given the test levelα>0 and β>0, α+β<1. Taking A=β/1−α and B=1−β/α, we can get(5)c=−lnσ1/σ2λpσ1−p−σ2−p,d1=lnBλpσ1−p−σ2−p,d2=lnAλpσ1−p−σ2−p.We easily knowc>0,d1>0,andd2>0. It is given by λn≥B that(6)Snp≥−lnσ1/σ2λpσ1−p−σ2−p+lnBλpσ1−p−σ2−p=cn+d1=Rn.Similarly, it is given byλn≥A that(7)Snp≤−lnσ1/σ2λpσ1−p−σ2−p+lnAλpσ1−p−σ2−p=cn+d2=An.It is not difficult to see that the sufficient and necessary condition forλn≥B is Snp≥Rn, the sufficient and necessary condition of λn≤A is Snp≤An. Then, the stopping rule of the SPRT for the P-norm distribution is(8)τ∗=infn:n≥1andSnp∉An,Rn.The decision rule of theP-norm distribution SPRT is as follows:(i)
WhenSnp≥Rn, we should stop the experiment and accept H0(ii)
WhenSnp≤An, we should stop the experiment and reject H0


## 3. Sample Size Required for SPRT

From the above analysis, we know that the SPRT method depends on the constantsA and B that have been selected beforehand. For the selection of A and B, we need to start from some properties of SPRT. The study of SPRT properties can be turned into the study of random walks. Then, this section will study the random walks of the SPRT method for the P-norm distribution and give some properties of SPRT.Givenα and β (which satisfy α+β<1), we can get constants A and B. Let zi=lnfXi,σ2/fXi,σ1i=1,2,…. Then, the random walk sequence is Zn=∑i=1nzi,n≥1. We have Lσ=PσacceptH1=PσZn≤lnA for ∀σ∈Θ.By (4), we can get(9)Z1=lnfX1,σ2fX1,σ1=lnσ1σ2expλX1pσ1−p−σ2−p=lnσ1σ2−λX1pσ2−p−σ1−p,(10)EZ1=Elnσ1σ2−λX1pσ2−p−σ1−p=lnσ1σ2−λpEX1pσ2−p−σ1−p.Note thatEXip=1/pσ/λp and EXi2p=1+p/p2σ/λ2p (Hu and Sun [1]). By (10), we have(11)EZ1=lnσ1σ2−σppσ2−p−σ1−p.IfEZ1≥0, then(12)σp≥plnσ1/σ2σ2−p−σ1−p.Letσ∗p=plnσ1/σ2/σ2−p−σ1−p we know that(i)
Whenσp=σ∗p, EZ1=0 can be given(ii)
Whenσp>σ∗p, EZ1>0 can be given(iii)
Whenσp<σ∗p, EZ1<0 can be givenSimilarly, we have(13)Z12=lnσ1σ2−λX1pσ2−p−σ1−p2=lnσ1σ22+λ2pσ2−p−σ1−p2X12p−2lnσ1σ2λpX1pσ2−p−σ1−p,EZ12=lnσ1σ22+λ2pσ2−p−σ1−p2X12p=lnσ1σ22+p+1pσ2−p−σ1−p2σ2p−2plnσ1σ2σ2−p−σ1−pσp.For simplicity,h=hσ is denoted as a function of σ.(14)exphZ1=exphlnσ1σ2−λX1pσ2−p−σ1−p,EexphZ1=exphlnσ1σ2pλ2σΓ1/p∫−∞+∞exp−λphσ2−p−σ1−p+σ−pX1pdX1.Leth∗=σ−p/σ1−p−σ2−p, h∗>0. We have(15)EexphZ1=1σexphlnσ1σ2hσ2−p−σ1−p+σ−p−1/p,h<h∗,∞,h≥h∗.By studying random walks, we can derive some properties of SPRT. Next, we will give some properties of SPRT method forP-norm distributions. At the same time, we will also give the average sample size required for the SPRT method.Theorem 1.
Ifh<σ−p/σ1−p−σ2−p exists, then there must be some h≠0 that satisfies EexphZ1=1.In order to find out the average sample size in the SPRT method, we need to verify whether conditionh<σ−p/σ1−p−σ2−p in Theorem 1 is true. It is given by Theorem 2.Theorem 2.
Forσ1 and σ2 in (2), h0σi<σi−p/σ1−p−σ2−pi=1,2.The operational characteristic (OC) of Wald’s SPRT is the probability of accepting the null hypothesis. The average sample number (ASN) is the average number of observations that we would have to collect in order to make a decision regarding the statistical hypotheses put forth. According to Mudholkar et al. [18], we can give the OC function and ANS of SPRT for the P-norm distribution. These are given by Theorems 3 and 4.Theorem 3.
Ifh<σ−p/σ1−p−σ2−p and EexphZ1=1, then the OC function of SPRT is given by(16)Lσ=PσacceptH1≈BhσBhσ−Ahσ.By Theorem3, we can get the approximate formulas α≈1−L1 and β≈L2, where L1=Lσ1 and L2=Lσ2. Thus, we can give Theorem 4.Theorem 4.
The ANS of SPRT for theP-norm distribution is that(1)
Ifσp=plnσ1/σ2/σ2−p−σ1−p, then Eστ∗≈LσlnA2+1−LσlnB2/EσZ12(2)
Ifσp≠plnσ1/σ2/σ2−p−σ1−p, then Eστ∗≈LσlnA+1−LσlnB/EσZ1For the given probabilityα and β which are the type 1 and the type 2 errors, A=β/1−α and B=1−β/α are taken, respectively. According to Theorem 4, it can be obtained that the average sample size of the SPRT method is(17)Eτ∗=12Eσ1τ∗+Eσ2τ∗.Eσ1τ∗ and Eσ2τ∗ here are determined in four cases as follows:(1)
Ifσip=plnσ1/σ2/σ2−p−σ1−p, i=1,2, then(18)Eσ1τ∗≈1−αlnβ/1−α2+αln1−β/α2Eσ1Z12,Eσ2τ∗≈βlnβ/1−α2+1−βln1−β/α2Eσ2Z12,where(19)Eσ1Z12=lnσ1σ22+p+1pσ2−p−σ1−p2σ12p−2plnσ1σ2σ1σ2p−1,Eσ2Z12=lnσ1σ22+p+1pσ2−p−σ1−p2σ22p−2plnσ1σ21−σ2σ1p.(2)
Ifσip≠plnσ1/σ2/σ2−p−σ1−p, i=1,2, then(20)Eσ1τ∗≈1−αlnβ/1−α+αln1−β/αEσ1Z1,Eσ2τ∗≈βlnβ/1−α+1−βln1−β/αEσ2Z1,where(21)Eσ1Z1=lnσ1σ2−σ1/σ2p−1p,Eσ2Z1=lnσ1σ2−1−σ2/σ1pp.(3)
Ifσ1p=plnσ1/σ2/σ2−p−σ1−p and σ2p≠plnσ1/σ2/σ2−p−σ1−p, then(22)Eσ1τ∗≈1−αlnβ/1−α2+αln1−β/α2Eσ1Z12,Eσ2τ∗≈βlnβ/1−α+1−βln1−β/αEσ2Z1.(4)
Ifσ1p≠plnσ1/σ2/σ2−p−σ1−p and σ2p=plnσ1/σ2/σ2−p−σ1−p, then(23)Eσ1τ∗≈1−αlnβ/1−α+αln1−β/αEσ1Z1,Eσ2τ∗≈βlnβ/1−α2+1−βln1−β/α2Eσ2Z12.To compare the methods of the simulation example in Section4, we briefly introduce the Neyman–Pearson test method to calculate the average sample size. The principle of the N−P test is to control the probability of making the first type of error within a given range and to find the test to make the probability of making the second type of error as small as possible, that is, to maximize the effectiveness of the test. Before proceeding to the results, we introduce these notions: Xipi=1,2,…,n is an independent sequence, and we take Snp=∑i=1nXip and X¯p=Snp/n. Likewise, we know EXip=1/pσ/λp and DXip=1/pσ/λ2p (Hu and Sun [1]).According to the Neyman–Pearson theory, the optimal fixed quantity is as follows. We should look forn and c that satisfy(24)Pσ1X¯p≥c=α,Pσ2X¯p<c=β.By the central limit theorem, we know thatX¯p approximately follows a normal distribution ANEXip,DXip/n, namely,(25)nX¯p−EXipDXipisAN0,1.We can get(26)Pσ1nX¯p−Eσ1XipDσ1Xip≥nc−Eσ1XipDσ1Xip=α,Pσ2nX¯p−Eσ2XipDσ2Xip<nc−Eσ2XipDσ2Xip=β.That is,(27)Kα=nc−Eσ1XipDσ1Xip,K1−β=nc−Eσ2XipDσ2Xip,where Kα and K1−β represent quantiles of the standard normal distribution.Then, we have(28)n=KαDσ1Xipc−Eσ1Xip,c=KαDσ1XipEσ1Xip−K1−βDσ2XipEσ2XipKαDσ1Xip−K1−βDσ2Xip.And the ratio of sample size required by SPRT andN−P test is that(29)r=Eσ1τ∗+Eσ2τ∗2n.


## 4. A Simulation Example

In this section, we will calculate the average sample size obtained by the SPRT method and the Neyman–Pearson test method, respectively, and compare them to verify the superiority of the SPRT method.Suppose thatσ1=1,σ2=2,p=1,μ=0,andα=β=0.05. The approximate values of the average sample size required by the SPRT method at the given test level are shown in Tables 1–3.Table 1
Eσ1τ∗ approximation.
βα0.010.030.050.10.0123.315022.431521.625519.78060.0317.684916.917316.227214.67200.0515.067714.355213.720112.30260.111.517710.882510.32489.1007Table 2
Eσ2τ∗ approximation.
βα0.010.030.050.10.0114.675511.13179.48427.24980.0314.119410.64859.03586.84990.0513.612110.21418.63606.49890.112.45089.23537.74385.7284Table 3
Eτ∗ approximation.
βα0.010.030.050.10.0118.995216.781615.554913.51520.0315.902213.782912.631510.76100.0514.339912.284611.17819.40070.111.984310.05899.03437.4146In order to illustrate the advantages of the SPRT method, the sample size required by the Neyman–Pearson test at the given test level is presented here (see Table4). Meanwhile, Table 5 shows the ratio of the sample size required by the SPRT method and Neyman–Pearson test method at the given test level.Table 4
Sample size required for theN−P test method.
βα0.010.030.050.10.0148.860142.771639.816135.28360.0337.088131.809629.268125.40160.0531.696926.832424.502520.97640.123.912119.713617.745614.7456Table 5
Ratio of sample size required for the SPRT andN−P test (%).
βα0.010.030.050.10.010.38880.39240.39070.38300.030.42880.43330.43160.42360.050.45240.45780.45620.44820.10.50120.51030.50920.5028From the above example, we can see that the ratio of sample size required by SPRT andN−P test is about 50.92% at most and 38.30% at least. It can be seen that the sample size required by the SPRT method is much less than the optimal sample size required by the Neyman–Pearson test method, which also shows the advantages of the SPRT method.


## 5. Proof of SPRT Properties

Proof.
of Theorem1. Suppose that there exists hσ≠0 such that EexphZ1=1. Hence,(30)1σexphlnσ1σ2hσ2−p−σ1−p+σ−p−1/p=1,exphlnσ1σ2−σhσ2−p−σ1−p+σ−p1/p=0.
Here, letgh=exphlnσ1/σ2−σhσ2−p−σ1−p+σ−p1/p. By calculation, we know g0=0, g−∞>0, and gh∗=exph∗lnσ1/σ2>0. To get gh=0, there exists h such that gh<0, namely,(31)exphlnσ1σ2−σhσ2−p−σ1−p+σ−p1/p<0,(32)σ1σ2hp<hσpσ2−p−σ1−p+1.
Letfh=σ1/σ2hp−hσpσ2−p−σ1−p−1; we have(33)f′h=pσ1σ2hplnσ1σ2−σpσ2−p−σ1−p,f″h=plnσ1σ22σ1σ2hp>0.
So, we know thatfh is convex. Let us set f′h=0 here, and we get(34)h0=lnσpσ2−p−σ1−p/plnσ1/σ2plnσ1/σ2.
By calculation, we can knowh0≠0. We substitute h0 into fh and talk about the following two cases:(1)
Ifσ=σ1, then(35)fh0=σ1pσ2−p−1plnσ1/σ21−lnσ1pσ2−p−1plnσ1/σ2−1.Here, we study the properties of the functionVt=t1−lnt−1, where t>0. After calculation, we can know that the function value of Vt in its domain is always less than 0. And fh0<0 is always true if σ1pσ2−p−1/plnσ1/σ2 is regarded as a value of the independent variable in (35), where σ1pσ2−p−1/plnσ1/σ2>0. Thus, we can find h0 to make the inequality (32) true, and then there is h to make gh<0.(2)
Ifσ=σ2, then(36)fh0=1−σ1−pσ2pplnσ1/σ21−ln1−σ1−pσ2pplnσ1/σ2−1.
And by the same token, we getfh0<0 here, we have the situation where h makes gh<0.
To sum up, whenhσ<σ−p/σ1−p−σ2−p, it can be proved that hσ≠0 satisfies EexphZ1=1 by the intermediate value theorem. In the end, the proposition has been proved.Proof.
of Theorem2. When σ=σ1, we can get(37)h0σ1<σ1−pσ1−p−σ2−p,that is,(38)lnσ1/σ2p−1/plnσ1/σ2plnσ1/σ2<11−σ1/σ2p.
By observing the right part of (38), we know that its range is 1,+∞. Then, formula (38) can also be written as(39)lnσ1/σ2p−1/plnσ1/σ2plnσ1/σ2≤1.
By simplifying (39), it can be obtained that(40)lnσ1/σ2p−1plnσ1/σ2≥plnσ1σ2,σ1σ2p−1≤pσ1σ2plnσ1σ2.
Letx=σ1/σ2,x∈0,1,andTx=xp−pxplnx−1; then,(41)T′x=pxp−1−pxp−1−p2xp−1lnx=−p2xp−1lnx>0.
From the above, we can see that the functionTx increases monotonically on 0,1, and T1=0; hence, Tx≤0 holds on (0,1). And we can deduce that σ1/σ2p−1≤pσ1/σ2plnσ1/σ2; therefore, h0σ1<σ1−p/σ1−p−σ2−p holds.
Similarly, whenσ=σ2, the inequality h0σ2<σ2−p/σ1−p−σ2−p also holds. To sum up, the proposition is proved.



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