CN103678939A - Degradation model consistency testing method catering to space distances and shapes and data distribution - Google Patents

Abstract

The invention discloses a degradation model consistency testing method catering to space distances and shapes and data distribution. The method includes six steps and includes the steps that firstly, data distribution consistence testing is carried out based on a likelihood ratio test so as to determine the fact that model data and verification test data have distribution consistency; then, the relational degree is determined according to improved grey correlation analysis; finally, model numerical value proximity and curve space shape similarity are determined. Data distribution uniformity, numerical value proximity and curve space similarity are considered in the method, and the defect that only degradation model consistency testing is considered in existing methods is overcome.

Description

A kind of space-oriented is apart from the degradation model consistency check method with shape and data distribution

Technical field

The present invention proposes a kind of space-oriented distance and the degradation model consistency check method that shape and data distribute, and belongs to the reliability engineering field of degeneration modeling.

Background technology

Raising along with product reliability and life level, no-failure product occurs in a large number, traditional reliability estimation method model based on fail data has been difficult to meet assessment of project requirement, in order to solve problems, degraded data is introduced in reliability assessment field, by degraded data, set up degradation model, become and solve important channel highly reliable, Long Life Products reliability assessment problem.

Yet, consistency check problem based on degradation model has become current problem demanding prompt solution, although, there is scholar to propose some consistency check methods, but, existing degradation model consistency check method is mainly considered from model numerical value consistance angle, almost not from mould shapes similarity, whether data obey with aspects such as distributions is considered model consistency check method, or even comprehensive several respects factor sophisticated model consistency check method of coming together, existing these methods have been investigated the error range of degradation model effectively, but fail to reflect the Changing Pattern of model, distribution character and spatial similarity.The present invention catches now methodical weak point exactly, find new breakthrough point, be mainly whether with distance and model space shape similarity three aspects: between distribution, model numerical value, to consider from data, propose a kind of new degradation model consistency check method.

Summary of the invention

For the modelling verification problem of setting up, from data, whether with distance and model space shape similarity three aspects: between distribution, model numerical value, study.The invention provides a kind of space-oriented apart from the degradation model consistency check method with shape and data distribution.

The object of the invention is to utilize the degradation model consistency check method of likelihood ratio test whether with distributing, to carry out decision analysis to model data, utilize improved Grey Incidence Analysis to compare research to the similarity of the distance between model numerical value and model space shape, the degradation model consistency check method that has finally proposed same distribution inspection based on likelihood ratio test and combined based on improved grey correlation numerical value proximity and spatial form similarity.

First the present invention carries out the data distribution consistency check based on likelihood ratio test, determines that model data and demonstration test data have distribution consistance; Then according to improved grey correlation analysis, determine the size of the degree of association, finally determine model numerical value proximity and space of curves shape similarity.

The present invention realizes by the following technical solutions, and a kind of space-oriented is apart from the degradation model consistency check method with shape and data distribution, and its step is as follows:

Step 1: choose the degradation model of product, this model characterizes the relation of performance parameter △ D and environmental stress S and time t, and its functional form can be expressed as △ D=f (S, t);

Step 2: analyzing and processing test figure, this test figure is the performance parameter of m product and the m of time data sequence { y under constant stress level _i(t), t=1,2 ..., N, i=1,2 ..., m};

A. calculate in test data sequence m sample in the average of each time point

formula is as follows:

$\hat{y}\left(t\right)=\frac{1}{m}\underset{i=1}{\overset{m}{\mathrm{\Σ}}}{y}_i\left(t\right);$

B. calculate in test data sequence m sample at the variance s of each time point ²(t), formula is as follows:

$s^2\left(t\right)=\frac{1}{m-1}\underset{i=1}{\overset{m}{\mathrm{\Σ}}}{[y_i\left(t\right)-\hat{y}\left(t\right)]}^2.$

Wherein, the m in m described product, a m data sequence, a m sample is same in this step, and the m occurring in subsequent step is all same;

Step 3: according to the stress level of test figure and selected degradation model, obtain the l-G simulation test data sequence of product degradation model under this stress level, i.e. the performance parameter of product and the data sequence of time relationship { x (t) under this stress level, t=1,2 ..., N};

Step 4: the data distribution consistency check based on likelihood ratio test, can realize as follows:

A. calculate that in test data sequence, m sample is at the standard deviation s of each time point (t), formula is as follows:

$s\left(t\right)=\sqrt{s^2\left(t\right)};$

B. calculate the test statistics T (t) under each time point, formula is as follows:

C. calculate critical value

under given confidence level 1-α and m sample, look into " t distribution table " and obtain t and divide and plant $t_{1-\frac{\mathrm{\α}}{2}}(m-1);$

D. the test statistics T (t) under each time point in step b is taken absolute value | T (t) |;

E. calculate the region of rejection W (t) under each time point, formula is as follows:

$W\left(t\right)=\{T\left(t\right):|T\left(t\right)|>t_{1-\frac{\mathrm{\α}}{2}}(m-1)\};$

If the test statistics T (t) f. under each time point all, not in region of rejection W (t), thinks by the data distribution consistency check based on likelihood ratio test, illustrate that these two time serieses have distribution consistance; Otherwise, think and do not pass through.

Step 5: the numerical value proximity based on improved grey correlation and the consistency check of space of curves shape similarity, can realize as follows:

A. the data of each time point of l-G simulation test data sequence are carried out to just value and process to obtain x'(t), formula is as follows: x'(t)=f (x (t))=x (t)/x (t ₀), wherein, t ₀for first time point;

B. the data of m in test data sequence each time point of sample are carried out to just value and process to obtain y ' _i(t), i=1,2 ..., m, formula is as follows:

Y ' _i(t)=f (y _i(t))=y _i(t)/y _i(t ₀), wherein, t ₀for first time point;

C. the average of m in test data sequence each time point of sample being carried out to first value processes

formula is as follows:

${\hat{y}}^{\′}\left(t\right)=f\left(\hat{y}\left(t\right)\right)=\hat{y}\left(t\right)/\hat{y}\left(t_0\right),$ Wherein, t ₀for first time point;

D. after first value is processed, calculate the absolute difference △ between the data of each each time point of sample in m sample of test data sequence and the data of corresponding each time point of l-G simulation test data sequence _i(t), formula is as follows:

△ _i(t)=|x'(t)-y′ _i(t)|,i=1,2,…,m；

E. after first value is processed, calculate the absolute difference △ between the average of m each time point of sample in test data sequence and the data of corresponding each time point of l-G simulation test data sequence _{on average}(t), formula is as follows:

F. all absolute difference △ that try to achieve in steps d and e _iand △ (t) _{on average}(t) in, find out maximal value and minimum value, try to achieve the two poles of the earth maximal value △ _i(t) _maxwith minimum value △ _i(t) _min, formula is as follows:

G. calculate the correlation coefficient γ of each sample and each time point of simulation sample in m sample _i(t), formula is as follows: wherein, ξ is resolution ratio, and ξ ∈ (0,1), and ξ often gets 0.5;

H. calculate the correlation coefficient γ of average and the emulated data of m each time point of sample _{on average}(t), formula is as follows:

Explain: γ _iand γ (t) _{on average}(t) embodied the similarity of two sequence curve spatial forms.

I. the x'(t obtaining according to preceding step), y ' _i(t) and calculate y ' _i(t)-x'(t) with

J. consider that the test data sequence curve that is symmetrical in emulated data sequence curve both sides has identical degree of closeness with emulated data sequence, note P _i(t), computing formula is as follows:

$P_i\left(t\right)=\left\{\begin{array}{c}\sqrt{|x^{\&prime;}\left(t\right){|}^{[1+\mathrm{sgn}({y_i}^{\&prime;}\left(t\right)-x^{\&prime;}\left(t\right)]}}\sqrt{{\left|{y_i}^{\&prime;}\left(t\right)\right|}^{[1-\mathrm{sgn}({y_i}^{\&prime;}\left(t\right)-x^{\&prime;}\left(t\right))]}}{y_i}^{\&prime;}\left(t\right)\&GreaterEqual;x^{\&prime;}\left(t\right)\\ \sqrt{{\left|x^{\&prime;}\left(t\right)\right|}^{[1+\mathrm{sgn}(x^{\&prime;}\left(t\right)-{y_i}^{\&prime;}\left(t\right)]}}\sqrt{{\left|{y_i}^{\&prime;}\left(t\right)\right|}^{[1-\mathrm{sgn}(x^{\&prime;}\left(t\right)-{y_i}^{\&prime;}\left(t\right))]}}{y_i}^{\&prime;}\left(t\right)<x^{\&prime;}\left(t\right)\end{array}\right.;$

K. consider that m the test figure average sequence curve that is symmetrical in emulated data sequence curve both sides has identical degree of closeness with emulated data, note P _{on average}(t), computing formula is as follows:

L. the degree of closeness of considering numerical value between two sequences not only with data between absolute difference relevant, be more subject to relative error domination and be designated as M _iand M (t) _{on average}(t), computing formula is as follows:

$M_i\left(t\right)=\frac{{\mathrm{\&Delta;}}_i\left(t\right)}{P_i\left(t\right)}+\frac{{\mathrm{\&Delta;}}_i\left(t\right)}{P_i\left(t\right)+{\mathrm{\&Delta;}}_i\left(t\right)},$

M. consider the proximity Q of two each time points of sequence numerical value _iand Q (t) _{on average}(t), computing formula is as follows:

Q _i(t)=exp (M _i(t)), Q _{on average}(t)=exp (M _{on average}(t));

N. the grey incidence coefficient γ of computed improved _i' (t) and γ _{on average}' (t), formula is as follows:

γ _i' (t)=[γ _i(t)] ^τ[Q _i(t)] ^β, γ _{on average}' (t)=[γ _{on average}(t)] ^τ[Q _{on average}(t)] ^β

Wherein, τ+β=1, τ, β ∈ (0,1), owing to considering that the spatial form similarity between two sequences is identical with the significance level of numerical value proximity in consistency check, so, τ=β=0.5 herein;

O. the degree of association γ (i) that calculates each sample and simulation sample in m sample, formula is as follows:

$\mathrm{\&gamma;}\left(i\right)=\frac{1}{N}\underset{t=1}{\overset{N}{\mathrm{\&Sigma;}}}{{\mathrm{\&gamma;}}_i}^{\&prime;}\left(t\right),i=\mathrm{1,2},...,m;$

P. the degree of association γ (on average) that calculates average and the simulation sample of m sample, formula is as follows:

Q. the degree of association size of obtaining according to step n and o is determined the similarity of model numerical value proximity and space of curves shape, if the degree of association of obtaining is all in 0.7 left and right, but allow due to particular product performance parameters present undulatory property Changing Pattern and change indivedual samples of obviously causing and the degree of association of simulation sample less than normal, but the average of m sample and the degree of association of simulation sample are in 0.7 left and right, think that model passes through numerical value proximity and the consistency check of spatial form similarity, otherwise, do not pass through.

Step 6: the degradation model simultaneously satisfied data based on likelihood ratio test, with distribution consistency check with based on improved grey correlation numerical value proximity and the consistency check of space of curves shape similarity, thinks that this model is by consistency check; Otherwise, think and do not pass through consistency check.

Wherein, in step 2, " the calculating in test data sequence m sample in the average of each time point " described in analyzing and processing test figure step a refers to the arithmetic mean value of each time point data of calculating; In " m sample is in the variance of each time point in calculating test data sequence " described in step b, refer to sample variance.

Wherein, at " stress level of test figure " described in step 3, refer to that obtaining test figure is under a given environmental stress (as temperature T=80 ℃, pressure F=50N etc.) condition; By by the numerical value substitution degradation model of given environmental stress, then the time value substitution degradation model of different time points is obtained to the l-G simulation test data sequence of each time point.

Wherein, in step 4, " likelihood ratio test " refers to m sequence { y of hypothesis _i(t), t=1,2 ..., N, i=1,2 ..., m} be from density function (or distributive law) for p (y (t): θ) (θ ∈ Θ) overall, consider that null hypothesis is as follows to the check problem of simple alternative hypothesis:

H ₀：θ∈θ ₀，H ₁：θ≠θ ₀(θ ₀∈Θ ₀)

Consider a method more directly perceived and natural, likelihood ratio test method is as follows:

$\mathrm{\&lambda;}\left(y\left(t\right)\right)=\frac{p(y_1\left(t\right),y_2\left(t\right),...,y_m\left(t\right);\mathrm{\&theta;})}{p(y_1\left(t\right),y_2\left(t\right),...,y_m\left(t\right);{\mathrm{\&theta;}}_0)}$

Due to

according to small probability event principles of inference, work as H ₀during establishment, λ (y (t)) has the trend of getting value less than normal, and when λ (y (t)) value is too large, has reason to suspect hypothesis H ₀correctness, refuse null hypothesis, otherwise, accept null hypothesis.

Wherein, in the step c in step 4, " confidence level " refers to that population parameter value drops on the probability in a certain district of sample statistics value; In steps d, " region of rejection " refers to as m sequence { y _i(t), t=1,2 ..., N, i=1,2 ..., m} is from normal population N (μ (t), σ ²(t) simple sample), μ (t) wherein, σ ²(t) the unknown, time provides check problem: H in level of significance α=0.05 ₀: μ (t)=μ ₀(t), H ₁: μ (t) ≠ μ ₀(t)

Likelihood ratio is: $\mathrm{\&lambda;}\left(y\left(t\right)\right)=\frac{p(y_1\left(t\right),y_2\left(t\right),...,y_m\left(t\right);\mathrm{\&mu;}\left(t\right))}{p(y_1\left(t\right),y_2\left(t\right),...,y_m\left(t\right);{\mathrm{\&mu;}}_0\left(t\right))}$

Owing to working as μ (t), σ ²(t) the unknown, has μ (t) and σ ²(t) maximum likelihood is estimated to be respectively:

$\widehat{\mathrm{\&mu;}}\left(t\right)=\hat{y}\left(t\right)=\frac{1}{m}\underset{i=1}{\overset{m}{\mathrm{\&Sigma;}}}{y}_i\left(t\right),{\widehat{\mathrm{\&sigma;}}}^2\left(t\right)=\frac{1}{m}\underset{i=1}{\overset{m}{\mathrm{\&Sigma;}}}{[y_i\left(t\right)-\hat{y}\left(t\right)]}^2$

Again as μ (t)=μ ₀(t)=x (t) (x (t), t=1,2 ..., N is emulated data sequence) and when known, σ ²(t) maximum likelihood is estimated as: ${{\widehat{\mathrm{\&sigma;}}}_0}^2\left(t\right)=\frac{1}{m}\underset{i=1}{\overset{m}{\mathrm{\&Sigma;}}}{[y_i\left(t\right)-{\mathrm{\&mu;}}_0\left(t\right)]}^2$

Obtaining likelihood ratio is:

$\begin{array}{c}\mathrm{\&lambda;}\left(y\left(t\right)\right)=\frac{{\left(\frac{1}{\sqrt{2\mathrm{\&pi;}}\widehat{\mathrm{\&sigma;}}\left(t\right)}\right)}^m\exp \{-\frac{1}{{2\widehat{\mathrm{\&sigma;}}}^2\left(t\right)}{\mathrm{\&Sigma;}}_{i=1}^m{(y_i\left(t\right)-\hat{y}\left(t\right))}^2\}}{{\left(\frac{1}{\sqrt{2\mathrm{\&pi;}}{\widehat{\mathrm{\&sigma;}}}_0\left(t\right)}\right)}^m\exp \{-\frac{1}{{2\widehat{\mathrm{\&sigma;}}}_0^2\left(t\right)}{\mathrm{\&Sigma;}}_{i=1}^m{(y_i\left(t\right)-{\mathrm{\&mu;}}_0\left(t\right))}^2\}}\\ ={\left(\frac{{\widehat{\mathrm{\&sigma;}}}_0^2\left(t\right)}{{\widehat{\mathrm{\&sigma;}}}^2\left(t\right)}\right)}^{\frac{m}{2}}={(1+\frac{m{(\hat{y}\left(t\right)-x\left(t\right))}^2}{(m-1)s^2\left(t\right)})}^{\frac{m}{2}}\end{array}$

Order $T\left(t\right)=\frac{\hat{y}\left(t\right)-x\left(t\right)}{s\left(t\right)/\sqrt{m}},$ Have $\mathrm{\&lambda;}\left(y\left(t\right)\right)={(1+\frac{T^2\left(t\right)}{m-1})}^{\frac{m}{2}}$

Because λ (y (t)) is | T (t) | monotonic increasing function, so inequality λ (y (t)) >c and | T (t) | >c ₁equivalence, so there is P{ λ (y (t)) >c|H ₀set up=P{|T (t) | >c ₁| H ₀set up=α

Again because work as H ₀during establishment, have $T\left(t\right)=\frac{\hat{y}\left(t\right)-x\left(t\right)}{s\left(t\right)/\sqrt{m}}~t(m-1)$

Obtain critical value $c_1=t_{1-\frac{\mathrm{\&alpha;}}{2}}(m-1),$ So test statistics $T\left(t\right)=\frac{\hat{y}\left(t\right)-x\left(t\right)}{s\left(t\right)/\sqrt{m}},$ Under level of significance α, region of rejection is $W\left(t\right)=\{T\left(t\right):|T\left(t\right)|>t_{1-\frac{\mathrm{\&alpha;}}{2}}(m-1)\}.$

Wherein, in step 5, " the numerical value proximity of improved grey correlation and spatial form similarity " refers to and considers numerical value proximity between two sequences, see that Grey Incidence Analysis improves, by improved grey correlation, obtain grey relational grade, with it, analyze and definite model numerical value proximity and spatial form similarity degree, two models to be changed to the quantitative and qualitative comparison and analysis of situation, if variation situation consistance is higher relatively in change procedure for two models, both grey relational grades are just larger.

Wherein, in a in step 5, b, c " initial value dataization processing " refer to data nondimensionalization, be about to the data of first time point as denominator, then the data using first time point to last time point are removed denominator as molecule successively, obtain the initial value data result of each time point; In steps d and e, " absolute difference " refers to the absolute value of the difference of data in two data sequences; In step g and h, " correlation coefficient " refers to that two sequences are in the correlation degree value of each time point; In step j and k, " sgn " refers to sign function, i.e. sgn=1 during x>0, sgn=0 during x=0, sgn=-1 during x<0; In step o and p, " degree of association " refers to the quantitaes of the correlation coefficient of each time point being got to correlation degree between two sequences that arithmetic mean obtains.

A kind of space-oriented of the present invention is apart from the degradation model consistency check method with shape and data distribution, and its advantage is:

1. the present invention is from the consistance of distance between the whether same distribution of data, model numerical value and model space shape similarity three aspects: testing model and test figure, the method has been considered data distribution consistance, numerical value proximity and space of curves shape similarity, has made up the limitation of just unilaterally considering degradation model consistency check in existing method;

2. the present invention can verify for degradation model, and method is easy to implement, applied widely, workable.

Accompanying drawing explanation

Fig. 1 is the inventive method process flow diagram.

Embodiment

Below in conjunction with drawings and Examples, the present invention is described in further detail.

Following examples are to implement according to the flow process shown in accompanying drawing 1, the present embodiment has been chosen accelerometer constant multiplier degradation model for implementing example, scale factor is as one of accelerometer critical performance parameters, and this method provides effective way for the consistency check of accelerometer scale factor degradation model.

The degradation model consistency check method that a kind of space-oriented of the present invention distance and shape and data distribute, take certain accelerometer scale factor is example, its concrete implementation step is as follows:

Step 1: the degradation model of choosing accelerometer constant multiplier is

$K_1=({0.152e}^{-\frac{2704.701}{T}}+{0.081}^{-\frac{2987.713}{T}}+{0.06e}^{-\frac{3053.967}{T}})t^{0.48},$ Wherein environmental stress is temperature T, and unit is K, and the time is t, and unit is the moon.

Step 2: analyzing and processing test figure, this test figure is 3 data sequence { y of 80 ℃ of lower 3 accelerometer constant multipliers of stress level and time relationship _i(t), t=1,2 ..., 18, i=1,2 ..., 3};

A. calculate in test figure 3 accelerometer constant multipliers in the average of each time point

by formula, obtained: the average of first time point is

$\hat{y}\left(1\right)=\frac{1}{3}\underset{i=1}{\overset{3}{\mathrm{\&Sigma;}}}{y}_i\left(1\right)=\frac{1}{3}\underset{i=1}{\overset{3}{\mathrm{\&Sigma;}}}(58.29738+72.51166+32.69466)=54.50123$

In like manner, calculate the average of other each time point

similar with above-mentioned computation process, as shown in table 1 by calculating the average of each time point:

In table 1 test figure, 3 accelerometer constant multipliers are in the average of each time point

B. calculate in test figure 3 accelerometer constant multipliers at the variance s of each time point ²(t), by formula, obtained: the variance of first time point is

$s^2\left(1\right)=\frac{1}{3-1}\underset{i=1}{\overset{3}{\mathrm{\&Sigma;}}}{[y_i\left(1\right)-\hat{y}\left(1\right)]}^2=407.1565$

In like manner, calculate the variance of other each time point

similar with above-mentioned computation process, as shown in table 2 by calculating the variance of each time point:

In table 2 test figure, 3 accelerometer constant multipliers are in the variance of each time point

Step 3: according to the model in 80 ℃ of the stress levels of test figure and step 1, obtain the l-G simulation test data sequence of product degradation model at 80 ℃, i.e. data sequence { the x (t) of accelerometer constant multiplier and time relationship at 80 ℃, t=1,2,, 18}, is obtained by formula: the accelerometer constant multiplier of first time point is

$x\left(1\right)=K_1=({0.152e}^{-\frac{2704.701}{273+80}}+{0.081}^{-\frac{2987.713}{273+80}}+{0.06e}^{-\frac{3053.967}{273+80}}){\left(\frac{1}{6}\right)}^{0.48}=43.7962$

In like manner, accelerometer constant multiplier and the above-mentioned computation process of calculating other each time point are similar, as shown in table 3 by calculating the accelerometer constant multiplier of each time point:

The accelerometer constant multiplier of each time point of table 3

Step 4: the data distribution consistency check based on likelihood ratio test, can realize as follows:

A. calculate 3 accelerometer constant multipliers of test figure at the standard deviation s of each time point (t), by formula, obtained: the standard deviation of first time point is

$s\left(1\right)\sqrt{s^2\left(1\right)}=\sqrt{\frac{1}{3-1}\underset{i=1}{\overset{3}{\mathrm{\&Sigma;}}}{[y_i\left(1\right)-\hat{y}\left(1\right)]}^2}=\sqrt{407.1565}=20.1781$

In like manner, standard deviation and the above-mentioned computation process of calculating other each time point are similar, as shown in table 4 by calculating the standard deviation of each time point:

The standard deviation of table 43 each time point of accelerometer constant multiplier

B. calculate the test statistics T (t) under each time point, by formula, obtained: the test statistics under first time point is $T\left(1\right)=\frac{\hat{y}\left(1\right)-x\left(1\right)}{s\left(1\right)/\sqrt{m}}=\frac{54.50123-43.7961}{20.1781/\sqrt{3}}=0.9189$

In like manner, test statistics and the above-mentioned computation process calculated under other each time point are similar, as shown in table 5 by the test statistics calculating under each time point:

Test statistics under each time point of table 5

Time/h	120	240	360	480	600	720	840	960	1080
										T(t)	0.9189	0.3462	0.3386	-1.7533	-1.2705	0.1222	-0.4295	-0.1790	-0.1131
Time/h	1200	1320	1440	1560	1680	1800	1920	2040	2160
										T(t)	-0.7623	-0.2892	-0.3811	-0.3844	0.2604	-0.1468	-0.2236	0.1246	0.3004

C. calculate critical value

given confidence level 1-α=0.95, looks into " t distribution table " and obtains t and divide and plant $t_{1-\frac{\mathrm{\&alpha;}}{2}}(m-1)=t_{1-\frac{0.5}{2}}(3-1)=4.3027;$

D. the test statistics T (t) under each time point in step b is taken absolute value | T (t) |, result is as shown in table 6:

The absolute value of the test statistics under each time point of table 6

Time/h	120	240	360	480	600	720	840	960	1080
										|T(t)|	0.9189	0.3462	0.3386	1.7533	1.2705	0.1222	0.4295	0.1790	0.1131
Time/h	1200	1320	1440	1560	1680	1800	1920	2040	2160
										|T(t)|	0.7623	0.2892	0.3811	0.3844	0.2604	0.1468	0.2236	0.1246	0.3004

E. calculate the region of rejection W (t) under each time point, by step c obtain W (t)=T (t): | T (t) | >4.3027};

F. through contrasting, the test statistics T (t) under each time point all not in region of rejection W (t), thinks by the data based on likelihood ratio test with distribution consistency check, illustrates that these two time serieses have distribution consistance.

Step 5: the numerical value proximity based on improved grey correlation and the consistency check of space of curves shape similarity, can realize as follows:

A. the data of each time point of l-G simulation test data sequence are carried out to just value and process to obtain x'(t), by formula, obtained: the initial value data result of first time point is

x'(1)=f(x(1))=x(1)/x(t ₀)=x(1)/x(1)=1

T wherein ₀for first time point, in like manner, the data of other each time point carry out just value process with above-mentioned computation process similar, as shown in table 6 by calculating the initial value data result of each time point;

B. the data of 3 each time points of accelerometer constant multiplier in test data sequence are carried out to just value and process to obtain y _i' (t), i=1,2 ..., m, is obtained by formula: the initial value data result of first time point is

y _i'(1)=f(y _i(1))=y _i(1)/y _i(t ₀)=y _i(1)/y _i(1)=1

T wherein ₀for first time point, in like manner, the data of other each time point carry out just value process with above-mentioned computation process similar, as shown in table 6 by calculating the initial value data result of each time point;

C. the just value of the average of 3 each time points of accelerometer constant multiplier in test data sequence is processed

by formula, obtained: the average of first time point just value result is

${\hat{y}}^{\&prime;}\left(1\right)=f\left(\hat{y}\left(1\right)\right)=\hat{y}\left(1\right)/\hat{y}\left(t_0\right)=\hat{y}\left(1\right)/\hat{y}\left(1\right)=1$

T wherein ₀for first time point, in like manner, it is similar with above-mentioned computation process that the average of other each time point is carried out just value processing, and by calculating the average of each time point, just value result is as shown in table 7;

Table 7 accelerometer constant multiplier initial value data result

D. after first value is processed, calculate the absolute difference △ of accelerometer constant multiplier between the data of each time point and the data of corresponding each time point of l-G simulation test data sequence in test data sequence sample 1 ₁(t), by formula, obtained: both are at the absolute difference between the data of first time point

△ ₁(1)=|x'(1)-y ₁'(1)|=|1-1|=0

In like manner, calculate both absolute difference and above-mentioned computation processes between the data of other times point similar, for sample 2 and sample 3, calculate the same, so as shown in table 8 by the absolute difference calculating between the data of each time point;

E. after first value is processed, calculate in test data sequence the absolute difference △ between the data of the average of each time point of accelerometer constant multiplier in 3 samples and each time point of l-G simulation test data sequence accordingly _{on average}(t), by formula, obtained: both are at the absolute difference between the data of first time point

In like manner, calculate both absolute difference and above-mentioned computation processes between the data of other each time points similar, as shown in table 8 by calculating both absolute differences between the data of each time point;

Absolute difference between data under each time point of table 8

F. all absolute difference △ that try to achieve in steps d and e _iand △ (t) _{on average}(t) in, find out maximal value and minimum value, try to achieve the two poles of the earth maximal value △ _i(t) _maxwith minimum value △ _i(t) _min, by formula, obtained:

G. calculate the correlation coefficient γ of accelerometer constant multiplier and each time point of emulated data in sample 1 ₁(t), by formula, obtained: the correlation coefficient of first time point is

${\mathrm{\&gamma;}}_1\left(1\right)=\frac{0+0.5\&times;3.21}{{\mathrm{\&Delta;}}_1\left(1\right)+0.5\&times;3.21}=\frac{1.605}{0+1.605}=1$

In like manner, correlation coefficient and the above-mentioned computation process of calculating other each time points are similar, the same for sample 2 and sample 3, so as shown in table 9 by calculating the correlation coefficient of each time point;

H. calculate the correlation coefficient γ of accelerometer constant multiplier and each time point of emulated data in 3 samples _{on average}(t), by formula, obtained: the correlation coefficient of first time point is

In like manner, correlation coefficient and the above-mentioned computation process of calculating other each time points are similar, as shown in table 9 by calculating the correlation coefficient of each time point;

The correlation coefficient of each time point of table 9

I. the x'(t obtaining according to preceding step), y _i' (t) and

further process and calculate $y_i^{\&prime;}\left(t\right)-x^{\&prime;}\left(t\right)$ With ${\hat{y}}^{\&prime;}\left(t\right)-x^{\&prime;}\left(t\right),$

For sample 1, under first time point, there is y ₁' (1)-x'(1)=1-1=0, in like manner, calculating and above-mentioned computation process under other each time point are similar, and the same for sample 2,3 and the further processing procedure of 3 sample average, final process result is as shown in table 10:

Table 10 sample 1,2,3 and the further result of 3 sample average

J. consider that the test data sequence that is symmetrical in emulated data sequence curve both sides has identical degree of closeness with emulated data sequence, note P _i(t), by step I, then obtain according to formula: at the P of first time point _i(t) be: y ₁' (1)=x'(1) time,

$P_1\left(1\right)=\sqrt{{\left|x^{\&prime;}\left(1\right)\right|}^{[1+\mathrm{sgn}({y_1}^{\&prime;}\left(1\right)-x^{\&prime;}\left(1\right)]}}\sqrt{{\left|{y_1}^{\&prime;}\left(1\right)\right|}^{[1-\mathrm{sgn}({y_1}^{\&prime;}\left(1\right)-x^{\&prime;}\left(1\right))]}}=\sqrt{{\left|x^{\&prime;}\left(1\right)\right|}^{[1+0]}}\sqrt{{\left|{y_1}^{\&prime;}\left(1\right)\right|}^{[1-0]}}=1$

In like manner, at the P of other each time point _i(t) calculating is similar with said process, and can similarly provide for sample 2,3, and result is as shown in table 11;

K. consider that 3 sampling test data mean value sequences that are symmetrical in emulated data sequence curve both sides have identical degree of closeness P with emulated data sequence _{on average}(t), by step I, then obtain according to formula: at the P of first time point _{on average}(t) be: ${\hat{y}}^{\&prime;}\left(1\right)=x^{\&prime;}\left(1\right)$ Time,

In like manner, at the P of other each time point _{on average}(t) calculating is similar with said process, and result is as shown in table 11:

The closeness value of each time point of table 11

Time/h	120	240	360	480	600	720	840	960	1080
										P 1(t)	1	1.39	1.69	1.95	2.17	2.36	2.54	2.71	2.87
P 2(t)	1	1.39	1.69	1.95	2.17	2.36	2.54	2.71	2.87
										P 3(t)	1	1.39	1.69	1.95	2.17	2.36	2.54	2.71	2.87
P On average(t)	1	1.39	1.69	1.95	2.17	2.36	2.54	2.71	2.87
										Time/h	1200	1320	1440	1560	1680	1800	1920	2040	2160
P 1(t)	3.02	3.16	3.30	3.43	3.55	3.67	3.78	3.90	4.00
										P 2(t)	3.02	3.16	3.30	3.43	3.55	3.67	3.78	3.90	4.00
P 3(t)	3.02	3.16	3.30	3.43	3.55	3.67	3.78	3.90	4.00
										P On average(t)	3.02	3.16	3.30	3.43	3.55	3.67	3.78	3.90	4.00

L. the degree of closeness of considering two sequence numerical value is not only relevant with the absolute difference of data, is more subject to relative error domination and is designated as M _iand M (t) _{on average}(t), by formula, obtained: sample 1 test data sequence and emulated data sequence have in the degree of closeness of first time point:

$M_1\left(1\right)=\frac{{\mathrm{\&Delta;}}_1\left(1\right)}{P_1\left(1\right)}+\frac{{\mathrm{\&Delta;}}_1\left(1\right)}{P_1\left(1\right)+{\mathrm{\&Delta;}}_1\left(1\right)}=\frac{0}{1}+\frac{0}{1+0}=0$

In like manner, the degree of closeness of other each time point is calculated with said process similar.Above-mentioned result of calculation is as shown in table 12:

The degree of closeness of table 12 liang sequence under each time point

Time/h

120

240

360

480

600

720

840

960

1080

M 1(t)	0	0.071	0.024	0.732	0.501	0.075	0.107	0.101	0.141
										M 2(t)	0	0.523	0.457	0.740	0.742	0.519	0.622	0.541	0.607
M 3(t)	0	0.192	0.214	1.018	0.999	0.491	0.984	0.678	0.801
										M On average(t)	0	0.257	0.266	0.792	0.715	0.322	0.530	0.421	0.405
Time/h	1200	1320	1440	1560	1680	1800	1920	2040	2160
										M 1(t)	0.370	0.111	0.054	0.058	0.086	0.091	0.057	0.200	0.337
M 2(t)	0.626	0.537	0.583	0.541	0.030	0.046	0.444	0.237	0.088
										M 3(t)	0.988	0.888	1.108	1.258	1.041	1.249	1.306	1.095	1.071
M On average(t)	0.616	0.471	0.526	0.545	0.204	0.438	0.482	0.292	0.173

M. consider the proximity Q of two each time points of sequence numerical value _iand Q (t) _{on average}(t), by formula, obtained:

Sample 1 test data sequence and emulated data sequence at the numerical value proximity of first time point are:

Q ₁(1)=exp (M ₁(1))=exp (0)=1, in like manner, the numerical value proximity of other each time point calculates with said process similar, and calculates the same for the numerical value proximity of sample 2,3 test data sequences and each time point of emulated data sequence; For 3 sampling test data sequences and emulated data sequence, at the numerical value proximity of first time point, be: Q _{on average}(1)=exp (M _{on average}(1))=exp (0)=1, in like manner, the numerical value proximity of other each time point calculates with said process similar.Above-mentioned result of calculation is as shown in table 13:

The numerical value proximity of table 13 liang each time point of sequence

Time/h	120	240	360	480	600	720	840	960	1080
										Q 1(t)	1	0.932	0.976	0.481	0.606	0.927	0.899	0.904	0.868
Q 2(t)	1	0.593	0.633	0.477	0.476	0.595	0.537	0.582	0.545
										Q 3(t)	1	0.825	0.807	0.361	0.368	0.612	0.374	0.508	0.449
Q On average(t)	1	0.773	0.766	0.453	0.489	0.725	0.589	0.656	0.667
										Time/h	1200	1320	1440	1560	1680	1800	1920	2040	2160

Q 1(t)	0.691	0.895	0.947	0.947	0.918	0.913	0.945	0.819	0.714
										Q 2(t)	0.535	0.585	0.558	0.582	0.970	0.955	0.642	0.789	0.916
Q 3(t)	0.372	0.411	0.330	0.353	0.353	0.289	0.271	0.335	0.343
										Q On average(t)	0.540	0.624	0.591	0.580	0.816	0.645	0.618	0.747	0.841

N. the grey incidence coefficient γ of computed improved _i' (t) and γ _{on average}' (t), by formula, obtained:

The improved grey incidence coefficient of sample 1 test data sequence and first time point of emulated data sequence is:

in like manner, the improved grey incidence coefficient of other each time point calculates with said process similar, and calculates the same for the improved grey incidence coefficient of sample 2,3 test data sequences and each time point of emulated data sequence; For the equal value sequence of 3 samples and the improved grey incidence coefficient of first time point of emulated data sequence, be:

in like manner, the improved grey incidence coefficient of other each time point calculates with said process similar.Above-mentioned result of calculation is as shown in table 14:

The improved grey incidence coefficient of each time point of table 14 liang sequence

Time/h	120	240	360	480	600	720	840	960	1080
										γ 1'(t)	1	0.9470	0.9810	0.5616	0.6620	0.9381	0.9102	0.9114	0.8760
γ 2'(t)	1	0.6871	0.7062	0.5588	0.5444	0.6449	0.5859	0.6196	0.5792
										γ 3'(t)	1	0.8699	0.8484	0.4520	0.4476	0.6602	0.4366	0.5520	0.4921
γ On average'(t)	1	0.8309	0.8168	0.5363	0.5586	0.7607	0.6324	0.6861	0.6900
										Time/h	1200	1320	1440	1560	1680	1800	1920	2040	2160
γ 1'(t)	0.7073	0.8967	0.9508	0.9446	0.9124	0.9091	0.9394	0.8077	0.5919
										γ 2'(t)	0.5646	0.6045	0.5755	0.5933	0.9678	0.7957	0.6373	0.7782	0.9073
γ 3'(t)	0.4162	0.4470	0.3685	0.3593	0.3809	0.3190	0.3004	0.3542	0.3583
										γ On average'(t)	0.5695	0.6411	0.6058	0.5914	0.8132	0.6438	0.6153	0.7365	0.8292

O. the degree of association γ (i) that calculates each sample and simulation sample in m sample, is obtained by formula:

$\mathrm{\&gamma;}\left(1\right)=\frac{1}{N}\underset{t=1}{\overset{N}{\mathrm{\&Sigma;}}}{{\mathrm{\&gamma;}}_1}^{\&prime;}\left(t\right)=\frac{1}{18}(1+0.9470+...+0.5919)=0.8582$

$\mathrm{\&gamma;}\left(2\right)=\frac{1}{N}\underset{t=1}{\overset{N}{\mathrm{\&Sigma;}}}{{\mathrm{\&gamma;}}_2}^{\&prime;}\left(t\right)=\frac{1}{18}(1+0.6871+...+0.9073)=0.6861$

$\mathrm{\&gamma;}\left(3\right)=\frac{1}{N}\underset{t=1}{\overset{N}{\mathrm{\&Sigma;}}}{{\mathrm{\&gamma;}}_3}^{\&prime;}\left(t\right)=\frac{1}{18}(1+0.8699+...+0.3583)=0.5035$

P. the degree of association γ (on average) that calculates average and the simulation sample of m sample, formula is as follows:

Q. the degree of association size of obtaining according to step o and p, the degree of association of sample 1 and 2 test data sequences and emulated data sequence is more than 0.68, the degree of association of sample 3 test data sequences and emulated data sequence has just reached 0.5, this may obviously cause because accelerometer constant multiplier presents undulatory property Changing Pattern and changes, and the degree of association of the average of 3 sampling test data sequences and emulated data sequence almost reaches 0.7, therefore, think that this model is by numerical value proximity and the consistency check of space of curves shape similarity.

Step 6: because model meets data distribution consistency check based on likelihood ratio test and based on improved grey correlation numerical value proximity and the consistency check of space of curves shape similarity, thinks that this model is by consistency check simultaneously

In the present invention, applying alphabetical physical significance illustrates as following table:

Claims (7)

1. the degradation model consistency check method that space-oriented distance and shape and data distribute, is characterized in that: the method concrete steps are as follows:

Step 1: choose the degradation model of product, this model characterizes the relation of performance parameter △ D and environmental stress S and time t, and its functional form is expressed as △ D=f (S, t);

Step 2: analyzing and processing test figure, this test figure is the performance parameter of m product and the m of time data sequence { y under constant stress level _i(t), t=1,2 ..., N, i=1,2 ..., m};

A. calculate in test data sequence m sample in the average of each time point

formula is as follows:

$\hat{y}\left(t\right)=\frac{1}{m}\underset{i=1}{\overset{m}{\mathrm{\&Sigma;}}}{y}_i\left(t\right);$

B. calculate in test data sequence m sample at the variance s of each time point ²(t), formula is as follows:

$s^2\left(t\right)=\frac{1}{m-1}\underset{i=1}{\overset{m}{\mathrm{\&Sigma;}}}{[y_i\left(t\right)-\hat{y}\left(t\right)]}^2;$

Wherein, described " m product ", " m data sequence ", the m in " m sample " are same in this step, and the m occurring in subsequent step is all same;

Step 3: according to the stress level of test figure and selected degradation model, obtain the l-G simulation test data sequence of product degradation model under this stress level, i.e. the performance parameter of product and the data sequence of time relationship { x (t) under this stress level, t=1,2 ..., N};

Step 4: the data distribution consistency check based on likelihood ratio test, realizes as follows:

A. calculate that in test data sequence, m sample is at the standard deviation s of each time point (t), formula is as follows:

$s\left(t\right)=\sqrt{s^2\left(t\right)};$

B. calculate the test statistics T (t) under each time point, formula is as follows:

C. calculate critical value under given confidence level 1-α and m sample, look into " t distribution table " and obtain t and divide and plant $t_{1-\frac{\mathrm{\&alpha;}}{2}}(m-1);$

D. the test statistics T (t) under each time point in step b is taken absolute value | T (t) |;

E. calculate the region of rejection W (t) under each time point, formula is as follows:

$W\left(t\right)=\{T\left(t\right):|T\left(t\right)|>t_{1-\frac{\mathrm{\&alpha;}}{2}}(m-1)\};$

If the test statistics T (t) f. under each time point all, not in region of rejection W (t), thinks by the data distribution consistency check based on likelihood ratio test, illustrate that these two time serieses have distribution consistance; Otherwise, think and do not pass through;

Step 5: the numerical value proximity based on improved grey correlation and the consistency check of space of curves shape similarity, realize as follows:

A. the data of each time point of l-G simulation test data sequence are carried out to just value and process to obtain x'(t), formula is as follows: x'(t)=f (x (t))=x (t)/x (t ₀), wherein, t ₀for first time point;

B. the data of m in test data sequence each time point of sample are carried out to just value and process to obtain y ' _i(t), i=1,2 ..., m, formula is as follows: y ' _i(t)=f (y _i(t))=y _i(t)/y _i(t ₀), wherein, t ₀for first time point;

C. the average of m in test data sequence each time point of sample being carried out to first value processes

formula is as follows: ${\hat{y}}^{\&prime;}\left(t\right)=f\left(\hat{y}\left(t\right)\right)=\hat{y}\left(t\right)/\hat{y}\left(t_0\right),$ Wherein, t ₀for first time point;

D. after first value is processed, calculate the absolute difference △ between the data of each each time point of sample in m sample of test data sequence and the data of corresponding each time point of l-G simulation test data sequence _i(t), formula is as follows:

△ _i(t)=|x'(t)-y′ _i(t)|,i=1,2,…,m；

E. after first value is processed, calculate the absolute difference △ between the average of m each time point of sample in test data sequence and the data of corresponding each time point of l-G simulation test data sequence _{on average}(t), formula is as follows:

F. all absolute difference △ that try to achieve in steps d and e _iand △ (t) _{on average}(t) in, find out maximal value and minimum value, try to achieve the two poles of the earth maximal value △ _i(t) _maxwith minimum value △ _i(t) _min, formula is as follows:

G. calculate the correlation coefficient γ of each sample and each time point of simulation sample in m sample _i(t), formula is as follows:

wherein, ξ is resolution ratio, and ξ ∈ (0,1), and ξ often gets 0.5;

H. calculate the correlation coefficient γ of average and the emulated data of m each time point of sample _{on average}(t), formula is as follows:

Explain: γ _iand γ (t) _{on average}(t) embodied the similarity of two sequence curve spatial forms;

I. the x'(t obtaining according to preceding step), y _i' (t) and

calculate

with

J. consider that the test data sequence curve that is symmetrical in emulated data sequence curve both sides has identical degree of closeness with emulated data sequence, note P _i(t), computing formula is as follows:

$P_i\left(t\right)=\left\{\begin{array}{c}\sqrt{|x^{\&prime;}\left(t\right){|}^{[1+\mathrm{sgn}({y_i}^{\&prime;}\left(t\right)-x^{\&prime;}\left(t\right)]}}\sqrt{{\left|{y_i}^{\&prime;}\left(t\right)\right|}^{[1-\mathrm{sgn}({y_i}^{\&prime;}\left(t\right)-x^{\&prime;}\left(t\right))]}}{y_i}^{\&prime;}\left(t\right)\&GreaterEqual;x^{\&prime;}\left(t\right)\\ \sqrt{{\left|x^{\&prime;}\left(t\right)\right|}^{[1+\mathrm{sgn}(x^{\&prime;}\left(t\right)-{y_i}^{\&prime;}\left(t\right)]}}\sqrt{{\left|{y_i}^{\&prime;}\left(t\right)\right|}^{[1-\mathrm{sgn}(x^{\&prime;}\left(t\right)-{y_i}^{\&prime;}\left(t\right))]}}{y_i}^{\&prime;}\left(t\right)<x^{\&prime;}\left(t\right)\end{array}\right.;$

K. consider that m the test figure average sequence curve that is symmetrical in emulated data sequence curve both sides has identical degree of closeness with emulated data, note P _{on average}(t), computing formula is as follows:

L. the degree of closeness of considering numerical value between two sequences not only with data between absolute difference relevant, be more subject to relative error domination and be designated as M _iand M (t) _{on average}(t), computing formula is as follows:

$M_i\left(t\right)=\frac{{\mathrm{\&Delta;}}_i\left(t\right)}{P_i\left(t\right)}+\frac{{\mathrm{\&Delta;}}_i\left(t\right)}{P_i\left(t\right)+{\mathrm{\&Delta;}}_i\left(t\right)},$

M. consider the proximity Q of two each time points of sequence numerical value _iand Q (t) _{on average}(t), computing formula is as follows:

Q _i(t)=exp (M _i(t)), Q _{on average}(t)=exp (M _{on average}(t));

N. the grey incidence coefficient γ ' of computed improved _iand γ (t) _{on average}' (t), formula is as follows:

γ ' _i(t)=[γ _i(t)] ^τ[Q _i(t)] ^β, γ _{on average}' (t)=[γ _{on average}(t)] ^τ[Q _{on average}(t)] ^β

Wherein, τ+β=1, τ, β ∈ (0,1), owing to considering that the spatial form similarity between two sequences is identical with the significance level of numerical value proximity in consistency check, so, τ=β=0.5 herein;

O. the degree of association γ (i) that calculates each sample and simulation sample in m sample, formula is as follows:

$\mathrm{\&gamma;}\left(i\right)=\frac{1}{N}\underset{t=1}{\overset{N}{\mathrm{\&Sigma;}}}{{\mathrm{\&gamma;}}_i}^{\&prime;}\left(t\right),i=\mathrm{1,2},...,m;$

P. the degree of association γ (on average) that calculates average and the simulation sample of m sample, formula is as follows:

Q. the degree of association size of obtaining according to step n and o is determined the similarity of model numerical value proximity and space of curves shape, if the degree of association of obtaining is all in 0.7 left and right, but allow due to particular product performance parameters present undulatory property Changing Pattern and change indivedual samples of obviously causing and the degree of association of simulation sample less than normal, but the average of m sample and the degree of association of simulation sample are in 0.7 left and right, think that model passes through numerical value proximity and the consistency check of spatial form similarity, otherwise, do not pass through;

Step 6: the degradation model simultaneously satisfied data based on likelihood ratio test, with distribution consistency check with based on improved grey correlation numerical value proximity and the consistency check of space of curves shape similarity, thinks that this model is by consistency check; Otherwise, think and do not pass through consistency check.

2. the degradation model consistency check method that a kind of space-oriented according to claim 1 distance and shape and data distribute, is characterized in that: " the calculating in test data sequence m sample in the average of each time point " in step 2 described in analyzing and processing test figure step a refers to the arithmetic mean value of each time point data of calculating; In " m sample is in the variance of each time point in calculating test data sequence " described in step b, refer to sample variance.

3. a kind of space-oriented according to claim 1 distance and the degradation model consistency check method that shape and data distribute, is characterized in that: at " stress level of test figure " described in step 3, refer to that obtaining test figure is under a given environmental stress conditions; By by the numerical value substitution degradation model of given environmental stress, then the time value substitution degradation model of different time points is obtained to the l-G simulation test data sequence of each time point.

4. the degradation model consistency check method that a kind of space-oriented according to claim 1 distance and shape and data distribute, is characterized in that: in step 4 " likelihood ratio test " refer to m sequence { y of hypothesis _i(t), t=1,2 ..., N, i=1,2 ..., m} be from density function or distributive law be p (y (t): θ) (θ ∈ Θ) overall, consider that null hypothesis is as follows to the check problem of simple alternative hypothesis:

H ₀：θ∈θ ₀，H ₁：θ≠θ ₀(θ ₀∈Θ ₀)

Consider a method more directly perceived and natural, likelihood ratio test method is as follows:

$\mathrm{\&lambda;}\left(y\left(t\right)\right)=\frac{p(y_1\left(t\right),y_2\left(t\right),...,y_m\left(t\right);\mathrm{\&theta;})}{p(y_1\left(t\right),y_2\left(t\right),...,y_m\left(t\right);{\mathrm{\&theta;}}_0)}$

Due to

according to small probability event principles of inference, work as H ₀during establishment, λ (y (t)) has the trend of getting value less than normal, and when λ (y (t)) value is too large, has reason to suspect hypothesis H ₀correctness, refuse null hypothesis, otherwise, accept null hypothesis.

5. a kind of space-oriented according to claim 1 distance and the degradation model consistency check method that shape and data distribute, is characterized in that: in the step c in step 4, " confidence level " refers to that population parameter value drops on the probability in a certain district of sample statistics value; In steps d, " region of rejection " refers to as m sequence { y _i(t), t=1,2 ..., N, i=1,2 ..., m} is from normal population N (μ (t), σ ²(t) simple sample), μ (t) wherein, σ ²(t) the unknown, time provides check problem: H in level of significance α=0.05 ₀: μ (t)=μ ₀(t), H ₁: μ (t) ≠ μ ₀(t)

Likelihood ratio is: $\mathrm{\&lambda;}\left(y\left(t\right)\right)=\frac{p(y_1\left(t\right),y_2\left(t\right),...,y_m\left(t\right);\mathrm{\&mu;}\left(t\right))}{p(y_1\left(t\right),y_2\left(t\right),...,y_m\left(t\right);{\mathrm{\&mu;}}_0\left(t\right))}$

Owing to working as μ (t), σ ²(t) the unknown, has μ (t) and σ ²(t) maximum likelihood is estimated to be respectively:

$\widehat{\mathrm{\&mu;}}\left(t\right)=\hat{y}\left(t\right)=\frac{1}{m}\underset{i=1}{\overset{m}{\mathrm{\&Sigma;}}}{y}_i\left(t\right),{\widehat{\mathrm{\&sigma;}}}^2\left(t\right)=\frac{1}{m}\underset{i=1}{\overset{m}{\mathrm{\&Sigma;}}}{[y_i\left(t\right)-\hat{y}\left(t\right)]}^2$

Again as μ (t)=μ ₀(t)=x (t) (x (t), t=1,2 ..., N is emulated data sequence) and when known, σ ²(t) maximum likelihood is estimated as: ${{\widehat{\mathrm{\&sigma;}}}_0}^2\left(t\right)=\frac{1}{m}\underset{i=1}{\overset{m}{\mathrm{\&Sigma;}}}{[y_i\left(t\right)-{\mathrm{\&mu;}}_0\left(t\right)]}^2$

Obtaining likelihood ratio is:

$\begin{array}{c}\mathrm{\&lambda;}\left(y\left(t\right)\right)=\frac{{\left(\frac{1}{\sqrt{2\mathrm{\&pi;}}\widehat{\mathrm{\&sigma;}}\left(t\right)}\right)}^m\exp \{-\frac{1}{{2\widehat{\mathrm{\&sigma;}}}^2\left(t\right)}{\mathrm{\&Sigma;}}_{i=1}^m{(y_i\left(t\right)-\hat{y}\left(t\right))}^2\}}{{\left(\frac{1}{\sqrt{2\mathrm{\&pi;}}{\widehat{\mathrm{\&sigma;}}}_0\left(t\right)}\right)}^m\exp \{-\frac{1}{{2\widehat{\mathrm{\&sigma;}}}_0^2\left(t\right)}{\mathrm{\&Sigma;}}_{i=1}^m{(y_i\left(t\right)-{\mathrm{\&mu;}}_0\left(t\right))}^2\}}\\ ={\left(\frac{{\widehat{\mathrm{\&sigma;}}}_0^2\left(t\right)}{{\widehat{\mathrm{\&sigma;}}}^2\left(t\right)}\right)}^{\frac{m}{2}}={(1+\frac{m{(\hat{y}\left(t\right)-x\left(t\right))}^2}{(m-1)s^2\left(t\right)})}^{\frac{m}{2}}\end{array}$

Order $T\left(t\right)=\frac{\hat{y}\left(t\right)-x\left(t\right)}{s\left(t\right)/\sqrt{m}},$ Have $\mathrm{\&lambda;}\left(y\left(t\right)\right)={(1+\frac{T^2\left(t\right)}{m-1})}^{\frac{m}{2}}$

Because λ (y (t)) is | T (t) | monotonic increasing function, so inequality λ (y (t)) >c and | T (t) | >c ₁equivalence, so there is P{ λ (y (t)) >c|H ₀set up=P{|T (t) | >c ₁| H ₀set up=α

Again because work as H ₀during establishment, have $T\left(t\right)=\frac{\hat{y}\left(t\right)-x\left(t\right)}{s\left(t\right)/\sqrt{m}}~t(m-1)$

Obtain critical value $c_1=t_{1-\frac{\mathrm{\&alpha;}}{2}}(m-1),$ So test statistics $T\left(t\right)=\frac{\hat{y}\left(t\right)-x\left(t\right)}{s\left(t\right)/\sqrt{m}},$ Under level of significance α, region of rejection is $W\left(t\right)=\{T\left(t\right):|T\left(t\right)|>t_{1-\frac{\mathrm{\&alpha;}}{2}}(m-1)\}.$

6. a kind of space-oriented according to claim 1 is apart from the degradation model consistency check method with shape and data distribution, it is characterized in that: in step 5, " the numerical value proximity of improved grey correlation and spatial form similarity " refers to and consider numerical value proximity between two sequences, see that Grey Incidence Analysis improves, by improved grey correlation, obtain grey relational grade, with it, analyze and definite model numerical value proximity and spatial form similarity degree, two models to be changed to the quantitative and qualitative comparison and analysis of situation, if variation situation consistance is higher relatively in change procedure for two models, both grey relational grades are just larger.

7. a kind of space-oriented according to claim 1 is apart from the degradation model consistency check method with shape and data distribution, it is characterized in that: in a in step 5, b, c " initial value dataization processing " refer to data nondimensionalization, be about to the data of first time point as denominator, then the data using first time point to last time point are removed denominator as molecule successively, obtain the initial value data result of each time point; In steps d and e, " absolute difference " refers to the absolute value of the difference of data in two data sequences; In step g and h, " correlation coefficient " refers to that two sequences are in the correlation degree value of each time point; In step j and k, " sgn " refers to sign function, i.e. sgn=1 during x>0, sgn=0 during x=0, sgn=-1 during x<0; In step o and p, " degree of association " refers to the quantitaes of the correlation coefficient of each time point being got to correlation degree between two sequences that arithmetic mean obtains.

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