CN103413026B - A kind of random censorship type lifetime data distribution inspection method - Google Patents

Abstract

A kind of random censorship type lifetime data distribution inspection method, step has: 1. according to the feature of data bonded products unique characteristics, selected life-span distribution；2. disposal data, is ranked up lifetime data value；3. calculate the mean rank order of wherein complete data；4. utilize this mean rank order to all data demarcation intervals and the difference of its end points place complete data correspondence mean rank order respectively as this group random censorship type lifetime data drop on the observed frequency on each interval；5. obtain the Maximum-likelihood estimation of parameter in life-span distribution；6. estimate life of product and drop on the expecterd frequency on each interval；7. utilize observed frequency and expecterd frequency to calculate Pearson came χ²Inspection statistics value, given level of significance α, by χ²Distribution quantile obtains marginal value, the size of comparison test statistics value and marginal value, it is determined that whether this group random censorship type lifetime data obeys preliminary selected life-span distribution。The distribution inspection that the method is random censorship type lifetime data provides effective way。

Description

A kind of random censorship type lifetime data distribution inspection method

Technical field

The present invention relates to a kind of random censorship type lifetime data distribution inspection method, by Mean rank order and Pearson came χ²Inspection combines, it is possible to effectively solve the distribution inspection problem of random censorship type lifetime data, it is adaptable to the correlative technology field such as data analysis, reliability assessment。

Background technology

In engineering reality, at the lifetime data that test or onsite application etc. obtain, its life-span distribution the unknown often。For effectively carrying out the work such as reliability assessment, typically via exploratory data analysis, the life-span distribution of preliminary selected data, therefore, it is desirable to use distribution inspection to come whether further verification and validation data obey the life-span distribution of primary election。

Distribution inspection is to utilize given life of product data, infers whether life of product is obeyed edit and analyzed selected distribution, and the foundation of deduction is test of goodness of fit。The goodness of fit is to observe the tolerance of matching degree between the distribution of data and selected theoretical distribution。

At present, the goodness-of-fit test method for complete data and fixed number, the right censored data of timing is comparatively perfect, for instance conventional Pearson came χ²Method of inspection, kolmogorov test method etc.；But, for random censorship data, still there is no effective distribution inspection method；For this, the present invention provides a kind of random censorship type lifetime data distribution inspection method。

Summary of the invention

(1) purpose of the present invention:

The present invention is directed to random censorship type lifetime data and still there is no the problem of distribution inspection method, utilize mean rank order to be grouped and calculate the actual frequency of each group, to traditional Pearson came χ²Method of inspection improves, and provides a kind of effective distribution inspection method for random censorship type lifetime data, namely provides a kind of random censorship type lifetime data distribution inspection method。

(2) technical scheme:

The present invention be directed to the distribution inspection method of random censorship type life of product data, therefore, first censored data and random censorship type lifetime data are briefly described。L deadline of in advance regulation test or observation in engineering, some individualities life-span when the cut-off of test or observation does not terminate, and at this moment claims life-span of this individuality at L by truncation, and title L is censored data。And random censorship type lifetime data generally refers to, n individual life-span is observed (or investigation, lower same), it was observed that data are to (t₁,δ₁)、(t₂,δ₂)、…、(t_n,δ_n), wherein t_i, i=1,2 ..., n is lifetime data value, δ_i, i=1,2 ..., n is the Boolean variable making truncation mark, works as t_iWhen being complete data (being again data of dying of old age), make δ_i=0, work as t_iWhen being censored data, make δ_i=1。Recordable it is: (t_i,δ_i),i=1,2,...,n。

Below the distribution inspection problem discussed in the present invention is briefly described: life of product refers to from (the t=0 that starts working, t express time) to occur first lost efficacy working time, it be one [0 ,+∞) continuous random variables of upper value, conventional T represents。Its distribution was distributed also known as the life-span, its distribution function F (t)=F (t;θ)=P (T≤t) is also called cumulative distribution function, wherein θ=(θ₁,θ₂,...,θ_m) it is unknown parameter vector in distribution function, θ₁,θ₂,...,θ_mIt is m unknown parameter of distribution function。It is now to, by the one of product group of lifetime data, infer whether life of product obeys preliminary selected distribution, then sets up null hypothesis H₀: product life distribution function F (t;θ)=F₀(t;θ)；Wherein, F₀(t, θ) is preliminary selected life distribution function, such as common exponential, Weibull distribution, normal distribution, logarithm normal distribution etc., θ=(θ₁,θ₂,...,θ_m) it is unknown parameter vector in distribution function, θ₁,θ₂,...,θ_mIt is m unknown parameter of distribution function。

The present invention is being applicable to observe completely traditional Pearson came χ of data²On the basis of method of inspection, carry out being grouped and calculate each group of actual frequency according to the mean rank order of complete data, then use Pearson came χ²Method of inspection carries out distribution inspection。A kind of random censorship type lifetime data distribution inspection method of the present invention, it is as follows that it is embodied as involved theoretical foundation:

1. Pearson came χ²Method of inspection

Utilize Pearson came χ²Method of inspection, it is necessary first to the lifetime data of product is divided into k mutual exclusive interval, drops on the difference of the expecterd frequency on each interval and observed frequency again through the comparative product life-span, use statisticCheck H₀, as follows:

${\widehat{\mathrm{\χ}}}^2=\underset{i=1}{\overset{k}{\mathrm{\Σ}}}\frac{{(n_i-{n\hat{p}}_i)}^2}{n{\hat{p}}_i}---\left(1\right)$

Wherein, k is the interval number divided, and n is the number observing life of product,It is the life of product expected frequency that drops on that i-th is interval,It is the life of product expecterd frequency that drops on that i-th is interval, n_iIt it is the life of product observed frequency that drops on that i-th is interval。Given level of significance α, by χ²Distribution quantile obtains marginal valueWherein m is preliminary selected distribution F₀(t;The number of unknown parameter in θ)。When inspection statistics value is more than marginal value, namelyTime, refuse null hypothesis, it is believed that life of product disobeys preliminary selected life-span distribution, otherwise accepts null hypothesis, it is believed that life of product obeys preliminary selected life-span distribution。

2. Mean rank order

For one group of complete lifetime data, it is possible to be arranged in one group of order statistic by the size of life of product value, each of which life value has a serial number, and this order is called rank。For one group of truncation lifetime data, due to wherein censored data corresponding product life value cannot it is expected that therefore their life-span rank determine with regard to bad, but but it is estimated that the possible rank of all of which, then obtain mean rank order。Mean rank order specifically available following formula solves:

$A_j=A_{j-1}+\frac{n+1-A_{j-1}}{n-i+2}---\left(2\right)$

Wherein: j is the serial number being sized by all complete datas；I is the serial number being sized in all data by jth complete data；；A_jIt it is jth complete data mean rank order in all data。

3. Maximum-likelihood estimation

If (t_i,δ_i), i=1,2 ..., n is the random censorship type lifetime data of above-mentioned introduction, wherein t_i, i=1,2 ..., n is lifetime data, δ_i, i=1,2 ..., n is the Boolean variable making truncation mark, and its overall probability density function is f (t;θ), distribution function is F (t;θ), wherein θ=(θ₁,θ₂,...,θ_m) it is unknown parameter vector, θ in population distribution function₁,θ₂,...,θ_mIt is m unknown parameter of population distribution function。The likelihood function of random censorship type lifetime data is defined as:

$L\left(\mathrm{\θ}\right)=\underset{i=1}{\overset{n}{\mathrm{\Π}}}{\left[f(t_i;\mathrm{\θ})\right]}^{1-{\mathrm{\δ}}_i}{[1-F(t_i;\mathrm{\θ})]}^{{\mathrm{\δ}}_i}---\left(3\right)$

If there is a statisticMake

$L\left(\widehat{\mathrm{\θ}}\right)=\underset{\mathrm{\θ}}{\max }\left\{L\left(\mathrm{\θ}\right)\right\}---\left(4\right)$

Then claimIt is the Maximum-likelihood estimation (MLE) of θ, namely makes likelihood function L (θ) obtain the parameter vector of maximum。It can be provided by definition, it is also possible to by likelihood function L (θ) and log-likelihood function l (θ)=lnL (θ) derivation is drawn。

Four kinds of distribution patterns relatively conventional in preliminary selected life-span distribution mainly reality in the present invention: exponential, Weibull distribution, normal distribution and logarithm normal distribution, are briefly described below the relevant nature of the common life-span distribution of these four。

1) exponential

Probability density function is:

$f(t;\mathrm{\&lambda;})=\left\{\begin{array}{cc}{\mathrm{\&lambda;e}}^{-\mathrm{\&lambda;t}},& t\&GreaterEqual;0,\\ 0,& t<0.\end{array}\right.---\left(5\right)$

Cumulative Distribution Function is:

$F(t;\mathrm{\&lambda;})=\left\{\begin{array}{cc}{1-e}^{-\mathrm{\&lambda;t}},& t\&GreaterEqual;0,\\ 0,& t<0.\end{array}\right.---\left(6\right)$

Wherein λ is unknown parameter θ in exponential, and crash rate λ > 0；

2) Weibull distribution

Probability density function is:

$f(t;\mathrm{\&eta;},m)=\left\{\begin{array}{cc}\frac{m}{\mathrm{\&eta;}}{\left(\frac{t}{\mathrm{\&eta;}}\right)}^{m-1}{e}^{{-(t/\mathrm{\&eta;})}^m},& t\&GreaterEqual;0;\\ 0,& t<0.\end{array}\right.---\left(7\right)$

Cumulative distribution function is:

$F(t;\mathrm{\&eta;},m)=1-e^{{-(t_i/\mathrm{\&eta;})}^m}---\left(8\right)$

Wherein (η, m) is unknown parameter θ in Weibull distribution, and scale parameter η > 0, form parameter m > 0；

3) normal distribution

Probability density function is:

$f(t;{\mathrm{\&mu;}}_1,{\mathrm{\&sigma;}}_1)=\frac{1}{{\mathrm{\&sigma;}}_1\sqrt{2\mathrm{\&pi;}}}{e}^{{-(t-{\mathrm{\&mu;}}_1)}^2/2{{\mathrm{\&sigma;}}_1}^2},-\&infin;<t<+\&infin;---\left(9\right)$

Cumulative Distribution Function is:

$F(t;{\mathrm{\&mu;}}_1,{\mathrm{\&sigma;}}_1)={\&Integral;}_0^t\frac{1}{{\mathrm{\&sigma;}}_1\sqrt{2\mathrm{\&pi;}}}{e}^{{-(t-{\mathrm{\&mu;}}_1)}^2/2{{\mathrm{\&sigma;}}_1}^2}=\mathrm{\&Phi;}\left(\frac{t-u_1}{{\mathrm{\&sigma;}}_1}\right)---\left(10\right)$

Wherein (μ₁,σ₁) it is unknown parameter θ in normal distribution, and average is μ₁, standard deviation sigma₁> 0；

4) logarithm normal distribution

Probability density function is:

$f(t;{\mathrm{\&mu;}}_2,{\mathrm{\&sigma;}}_2)=\left\{\begin{array}{cc}\frac{1}{{\mathrm{t\&sigma;}}_2\sqrt{2\mathrm{\&pi;}}}{e}^{{-(\ln t-{\mathrm{\&mu;}}_2)}^2/2{{\mathrm{\&sigma;}}_2}^2,}& t>0;\\ 0,& t\&le;0.\end{array}\right.---\left(11\right)$

Cumulative Distribution Function is:

$F(t;{\mathrm{\&mu;}}_2,{\mathrm{\&sigma;}}_2)={\&Integral;}_0^t\frac{1}{{\mathrm{t\&sigma;}}_2\sqrt{2}\mathrm{\&pi;}}{e}^{{-(\ln t-{\mathrm{\&mu;}}_2)}^2/{{2\mathrm{\&sigma;}}_2}^2}=\mathrm{\&Phi;}\left(\frac{\ln t-u_2}{{\mathrm{\&sigma;}}_2}\right)---\left(12\right)$

Wherein (μ₂,σ₂) it is unknown parameter θ in logarithm normal distribution, and logarithmic average is μ₂, logarithm standard deviation σ₂> 0；

One random censorship type lifetime data distribution inspection method of the present invention, the method specifically comprises the following steps that

Step one: for one group of given random censorship type lifetime data (t_i,δ_i), i=1,2 ..., n, wherein t_i, i=1,2 ..., n is lifetime data value, δ_iI=1,2, ..., n is the Boolean variable making truncation mark, first according to the feature of these group data bonded products own characteristic, and the life-span distribution of preliminary selected product, four kinds of life-span distributions common in reality are mainly discussed here, including exponential, Weibull distribution, normal distribution and logarithm normal distribution。Thus can set up null hypothesis H₀: product life distribution function F (t;θ)=F₀(t;θ)；Wherein, F₀(t, θ) is preliminary selected life distribution function, θ=(θ₁,θ₂,...,θ_m) it is unknown parameter vector in distribution function, θ₁,θ₂,...,θ_mIt is m unknown parameter of distribution function；

Step 2: disposal data。

By all of lifetime data value t_i, i=1,2 ..., n is ranked up from small to large, obtains t₍₁₎≤t₍₂₎≤...≤t_(n), work as t_(i)When being complete data, remember δ_(i)=0；Work as t_(i)When being right censored data, remember δ_(i)=1, reduced data can be expressed as (t_(i),δ_(i)), i=1,2 ..., n, wherein total y of complete data；

Step 3: for the random censorship type lifetime data (t after arranging_(i),δ_(i)), i=1,2 ..., n, calculate the mean rank order that y complete data is corresponding, its computational methods are as follows:

$A_j=A_{j-1}+\frac{n+1-A_{j-1}}{n-i+2}---\left(13\right)$

Wherein: j is the serial number being sized by all complete datas；I is the serial number that jth complete data is sized in all data；A_jIt it is jth complete data mean rank order in all data；

Step 4: according to the result calculated in step 3, organizes this y tactic complete data in random censorship type lifetime data and is divided into k mutual exclusive interval according to the mean rank order of its correspondence, and wherein k is for being not more thanMaximum integer, y is the number of all complete datas in these group data, and makes the difference (△ A) of mean rank order corresponding to upper two the end points place complete datas in each interval after dividing_i, i=1,2 ..., k is roughly equal, (△ A)_i, i=1,2 ..., k drops on i-th interval [t as this group random censorship type lifetime data_(i1),t_(i2)), i=1,2 ..., the observed frequency n on k_i, i=1,2 ..., k, wherein t_(i1),t_(i2)Represent the complete data at interval upper two the end points places of i-th respectively；

Step 5: according to this group random censorship type lifetime data, calculates the Maximum-likelihood estimation of parameter θ in preliminary selected life-span distributionThe Maximum-likelihood estimation of parameter θ in preliminary selected life-span distributionMethod for solving as follows:

1) likelihood function of exponential is:

$L\left(\mathrm{\&lambda;}\right)=\underset{i=1}{\overset{n}{\mathrm{\&Pi;}}}{\left(\mathrm{\&lambda;exp}\right\{-{\mathrm{\&lambda;t}}_i\left\}\right)}^{1-{\mathrm{\&delta;}}_i}{\left(\exp \right\{-{\mathrm{\&lambda;t}}_i\left\}\right)}^{{\mathrm{\&delta;}}_i}---\left(14\right)$

Wherein, (t_i,δ_i), i=1,2 ..., n is the random censorship type lifetime data of above-mentioned introduction, t_i, i=1,2 ..., n is lifetime data value, δ_i, i=1,2 ..., n is the Boolean variable making truncation mark, and λ is unknown parameter θ in exponential。Find out oneL (λ) in formula (14) is made to take maximum, thenIt it is the Maximum-likelihood estimation of index distributed constant θ=λ；

2) likelihood function of Weibull distribution is:

$L(\mathrm{\&eta;},m)=\underset{i=1}{\overset{n}{\mathrm{\&Pi;}}}{\left[\frac{m}{\mathrm{\&eta;}}{\left(\frac{t_i}{\mathrm{\&eta;}}\right)}^{m-1}{e}^{{-(t_i/\mathrm{\&eta;})}^m}\right]}^{1-{\mathrm{\&delta;}}_i}{\left[e^{{-(t_i/\mathrm{\&eta;})}^m}\right]}^{{\mathrm{\&delta;}}_i}---\left(15\right)$

Wherein, (t_i,δ_i), i=1,2 ..., n is the random censorship type lifetime data of above-mentioned introduction, t_i, i=1,2 ..., n is lifetime data value, δ_i, i=1,2 ..., n is the Boolean variable making truncation mark, and (η m) is unknown parameter θ in Weibull distribution。Find out one(η m) takes maximum, then to make L in formula (15)It is Weibull distribution parameters θ=(η, Maximum-likelihood estimation m)；

3) likelihood function of normal distribution is:

$L({\mathrm{\&mu;}}_1,{\mathrm{\&sigma;}}_1)=\underset{i=1}{\overset{n}{\mathrm{\&Pi;}}}{\left[\frac{1}{{\mathrm{\&sigma;}}_1\sqrt{2\mathrm{\&pi;}}}{e}^{{-(t_i-{\mathrm{\&mu;}}_1)}^2/{{2\mathrm{\&sigma;}}_1}^2}\right]}^{1-{\mathrm{\&delta;}}_i}{[1-\mathrm{\&Phi;}\left(\frac{t_i-{\mathrm{\&mu;}}_1}{{\mathrm{\&sigma;}}_1}\right)]}^{{\mathrm{\&delta;}}_i}---\left(16\right)$

Wherein, (t_i,δ_i), i=1,2 ..., n is the random censorship type lifetime data of above-mentioned introduction, t_i, i=1,2 ..., n is lifetime data value, δ_i, i=1,2 ..., n is the Boolean variable making truncation mark, (μ₁,σ₁) it is unknown parameter θ in normal distribution。Find out oneMake L (μ in formula (16)₁,σ₁) take maximum, thenIt is Parameters of Normal Distribution θ=(μ₁,σ₁) Maximum-likelihood estimation；

4) likelihood function of logarithm normal distribution is:

$L({\mathrm{\&mu;}}_2,{\mathrm{\&sigma;}}_2)=\underset{i=1}{\overset{n}{\mathrm{\&Pi;}}}{\left[\frac{1}{{t_i\mathrm{\&sigma;}}_2\sqrt{2\mathrm{\&pi;}}}{e}^{{-(t_i-{\mathrm{\&mu;}}_2)}^2/{{2\mathrm{\&sigma;}}_2}^2}\right]}^{1-{\mathrm{\&delta;}}_i}{[1-\mathrm{\&Phi;}\left(\frac{\ln t_i-{\mathrm{\&mu;}}_2}{{\mathrm{\&sigma;}}_2}\right)]}^{{\mathrm{\&delta;}}_i}---\left(17\right)$

Wherein, (t_i,δ_i), i=1,2 ..., n is the random censorship type lifetime data of above-mentioned introduction, t_i, i=1,2 ..., n is lifetime data value, δ_i, i=1,2 ..., n is the Boolean variable making truncation mark, (μ₂,σ₂) it is unknown parameter θ in logarithm normal distribution。Find out oneMake L (μ in formula (17)₂,σ₂) take maximum, thenIt is lognormal distribution parameter θ=(μ₂,σ₂) Maximum-likelihood estimation；

Step 6: the Maximum-likelihood estimation of unknown parameter θ in distribution of calculated primary election life-span in step 5Substitute into its corresponding Cumulative Distribution Function F₀(t;In θ), and then estimate life of product T and drop on each interval [t_(i1),t_(i2)), i=1,2 ..., the expected frequency on kIts solution formula is as follows:

${\hat{p}}_i=P\{t_{\left(i2\right)}\&le;T<t_{\left(i1\right)}\}=F_0(t_{\left(i2\right)};\widehat{\mathrm{\&theta;}})-F_0(t_{\left(i1\right)};\widehat{\mathrm{\&theta;}}),i=\mathrm{1,2},...,k---\left(18\right)$

Specifically, for different primary election life-span distributions, it is utilized respectively formula (18) and estimates that life of product T drops on each interval [t_(i1),t_(i2)), i=1,2 ..., the expected frequency on kIts method for solving is as follows:

1) exponential:

${\hat{p}}_i=e^{-\widehat{\mathrm{\&lambda;}}{t}_{\left(i1\right)}-e^{-\widehat{\mathrm{\&lambda;}}{t}_{\left(i2\right)}},i=\mathrm{1,2},...,k---\left(19\right)}$

Wherein,It it is the Maximum-likelihood estimation of index distributed constant λ；T_(i1),t_(i2)Represent the complete data at interval upper two the end points places of i-th respectively；

2) Weibull distribution:

${\hat{p}}_i=e^{-{(t_{\left(i1\right)}/\widehat{\mathrm{\&eta;}})}^{\hat{m}}}-e^{-{(t_{\left(i2\right)}/\widehat{\mathrm{\&eta;}})}^{\hat{m}}},i=\mathrm{1,2},...,k---\left(20\right)$

Wherein,It is Weibull distribution parameters (η, Maximum-likelihood estimation m)；T_(i1),t_(i2)Represent the complete data at interval upper two the end points places of i-th respectively；

3) normal distribution:

${\hat{p}}_i=\mathrm{\&Phi;}\left(\frac{t_{\left(i2\right)}-{\hat{u}}_1}{{\widehat{\mathrm{\&sigma;}}}_1}\right)-\mathrm{\&Phi;}\left(\frac{t_{\left(i1\right)}-{\hat{u}}_1}{{\widehat{\mathrm{\&sigma;}}}_1}\right),i=\mathrm{1,2},...,k---\left(21\right)$

Wherein,It is Parameters of Normal Distribution (μ₁,σ₁) Maximum-likelihood estimation；T_(i1),t_(i2)Represent the complete data at interval upper two the end points places of i-th respectively；

4) logarithm normal distribution:

${\hat{p}}_i=\mathrm{\&Phi;}\left(\frac{{\ln t}_{\left(i2\right)}-{\hat{u}}_2}{{\widehat{\mathrm{\&sigma;}}}_2}\right)-\mathrm{\&Phi;}\left(\frac{\ln t_{\left(i1\right)}-{\hat{u}}_2}{{\widehat{\mathrm{\&sigma;}}}_2}\right),i=\mathrm{1,2},...,k---\left(22\right)$

Wherein,It is lognormal distribution parameter (μ₂,σ₂) Maximum-likelihood estimation；T_(i1),t_(i2)Represent the complete data at interval upper two the end points places of i-th respectively；

Step 7: calculate Pearson came chi-square test statisticIts method for solving is as follows:

${\widehat{\mathrm{\&chi;}}}^2=\underset{i=1}{\overset{k}{\mathrm{\&Sigma;}}}\frac{{(n_i-n\widehat{p_i})}^2}{n\widehat{p_i}}---\left(23\right)$

Wherein, k is the interval number divided, and n is the number of this group random censorship type lifetime data, n_iIt is that this group random censorship type lifetime data drops on the observed frequency that i-th is interval,It is the life of product expected frequency that drops on i-th interval,It it is the life of product expecterd frequency that drops on i-th interval。Given level of significance α, by χ²Distribution quantile obtains marginal valueWherein m is preliminary selected life-span distribution F₀(t;The number of unknown parameter in θ)。When inspection statistics value is more than marginal value, namelyTime, refuse null hypothesis, it is believed that life of product disobeys preliminary selected life-span distribution, otherwise accepts null hypothesis, it is believed that life of product obeys preliminary selected life-span distribution。

(3) advantage

The present invention be directed to a kind of distribution inspection method of random censorship type lifetime data, by traditional Pearson came χ²Method of inspection improves, it is proposed to a kind of Pearson came χ in conjunction with mean rank order²Method of inspection。Its advantage is as follows:

1. the present invention is by means of Mean rank order, efficiently solves the determination problem of complete data life-span rank in random censorship type lifetime data。

2. the Pearson came χ in conjunction with mean rank order that the present invention proposes²Method of inspection efficiently solves the distribution inspection problem of random censorship type life of product data, provides technical support for work such as follow-up reliability assessments。

Accompanying drawing explanation

Fig. 1 the inventive method implementing procedure figure。

Detailed description of the invention

Generating the exponential random number of one group of right truncation, sample size n=80 with Matlab, delete mistake ratio p=0.15, density function is f (t)=0.05e^-0.05t, now it can be used as the one of certain product group random censorship type lifetime data, be expressed as (t_i,δ_i), i=1,2 ..., 80, wherein t_i, i=1,2 ..., 80 is lifetime data value, δ_i, i=1,2 ..., 80 is the Boolean variable making truncation mark, and as shown in table 1, examination judges the life-span distribution of product。

1 one groups of random censorship type lifetime datas of table

t(i)	1.59	3.94	10.82	1.26	2.65	31.43	5.01	1.46	3.38	0.90
											δ(i)	0	0	1	0	0	0	0	0	0	0
t(i)	4.03	12.07	9.48	4.72	40.50	2	5.08	8.64	62.92	7.51
											δ(i)	0	0	0	1	0	0	0	0	0	0 6 -->
t(i)	8.10	35.56	8.69	23.19	9.38	4.59	9.56	14.40	10.65	1.19
											δ(i)	1	0	0	1	0	0	1	0	0	0
t(i)	11.06	16.87	4.18	12.59	37.42	13.44	13.55	10.39	16.11	16.47
											δ(i)	0	0	0	0	1	0	0	0	0	0
t(i)	16.61	29.35	83.46	12.05	17.15	32.72	18.13	30.12	18.93	24.30
											δ(i)	0	0	0	1	0	0	0	0	1	0
t(i)	19.46	16.76	33.77	28.40	8.79	23.35	43.94	24	19.43	26.77
											δ(i)	1	0	0	0	0	0	0	0	0	0
t(i)	27.09	22.87	16.62	29.60	18.85	0.41	17.89	33.32	20.82	33.91
											δ(i)	0	0	0	0	0	0	0	0	0	0
t(i)	35.36	6.25	12.74	4.96	42.84	23.80	50.24	56.34	6.82	19.97

δ(i)

0

0

1

0

0

1

0

0

0

0

Use the random censorship type lifetime data distribution inspection method that provides of the present invention, as it is shown in figure 1, its to be embodied as step as follows:

Step one: first according to this group random censorship type lifetime data (t_i,δ_i), i=1,2 ..., 80, wherein t_i, i=1,2 ..., 80 is lifetime data value, δ_i, i=1,2 ..., 80 is the Boolean variable making truncation mark, and the life-span of preliminary selected product is distributed as exponential, sets up null hypothesis H₀: product life distribution function F (t;θ)=F₀(t;λ), wherein, F₀(t, λ) is the exponential distribution function of primary election, and λ is unknown parameter θ in exponential；

Step 2: by all of lifetime data value t_i, i=1,2 ..., 80 arrange from small to large, obtain t₍₁₎≤t₍₂₎≤...≤t₍₈₀₎, work as t_(i)When being complete data, remember δ_(i)=0；Work as t_(i)When being right censored data, remember δ_(i)=1, the random censorship type lifetime data after arrangement is as shown in table 2:

Random censorship type lifetime data after table 2 arrangement

t(i)	0.41	0.90	1.19	1.26	1.46	1.59	2	2.65	3.38	3.94
											δ(i)	0	0	0	0	0	0	0	0	0	0
t(i)	4.03	4.18	4.59	4.72	4.96	5.01	5.08	6.25	6.82	7.51
											δ(i)	0	0	0	1	0	0	0	0	0	0
t(i)	8.10	8.64	8.69	8.79	9.38	9.48	9.56	10.39	10.65	10.82
											δ(i)	1	0	0	0	0	0	1	0	0	1
t(i)	11.06	12.05	12.07	12.59	12.74	13.44	13.55	14.40	16.11	16.47
											δ(i)	0	1	0	0	1	0	0	0	0	0
t(i)	16.61	16.62	16.76	16.87	17.15	17.89	18.13	18.85	18.93	19.43
											δ(i)	0	0	1	0	0	0	0	0	1	0

t(i)	19.46	19.97	20.82	22.87	23.19	23.35	23.80	24	24.30	26.77
											δ(i)	1	0	0	0	1	0	1	0	0	0
t(i)	27.09	28.40	29.35	29.60	30.12	31.43	32.72	33.32	33.77	33.91
											δ(i)	0	0	0	0	0	0	0	0	0	0 7 -->
t(i)	35.36	35.56	37.42	40.50	42.84	43.94	50.24	56.34	62.92	83.46
											δ(i)	0	0	1	0	0	0	0	0	0	0

Step 3: utilize formula (13) to calculate all 68 tactic complete data t_(i), mean rank order A corresponding to 1≤i≤80_j, j=1,2 ..., 68, result is as shown in table 3:

The mean rank order computer chart that table 3 complete data is corresponding

t(i)	0.41	0.90	1.19	1.26	1.46	1.59	2	2.65	3.38	3.94
											Aj	1	2	3	4	5	6	7	8	9	10
t(i)	4.03	4.18	4.59	4.96	4.96	5.08	6.25	6.82	7.51	8.64
											Aj	11	12	13	14.01	15.03	16.04	17.06	18.07	19.09	20.12
t(i)	8.69	8.79	9.38	9.48	10.39	10.65	11.06	12.07	12.59	13.44
											Aj	21.15	22.19	23.22	24.25	25.30	26.35	27.42	28.52	29.61	30.73
t(i)	13.55	14.40	16.11	16.47	16.61	16.62	16.87	17.15	17.89	18.13
											Aj	31.84	32.96	34.08	35.20	36.31	37.43	38.58	39.72	40.87	42.02
t(i)	18.85	19.43	19.97	20.82	22.87	23.35	24	24.30	26.77	27.09
											Aj	43.16	44.34	45.57	46.79	48.01	49.28	50.60	51.92	53.24	54.57
t(i)	28.40	29.35	29.60	30.12	31.43	32.72	33.32	33.77	33.91	35.36
											Aj	55.89	57.21	58.53	59.85	61.17	62.50	63.82	65.14	66.46	67.78

t(i)	35.56	40.50	42.84	43.94	50.24	56.34	62.92	83.46
											Aj	69.10	70.59	72.08	73.57	75.05	76.54	78.03	79.51

Step 4: according to the result calculated in step 3, organizes this all 68 tactic complete datas in random censorship type lifetime data and is divided into k mutual exclusive interval according to the mean rank order of its correspondence, and wherein k is for being not more thanMaximum integer, i.e. k=8, and make the difference (△ A) of mean rank order corresponding to upper two the end points place complete datas in each interval after dividing_i, i=1,2 ..., 8 is roughly equal, (△ A)_i, i=1,2 ..., 8 drop on i-th interval [t as this group random censorship type lifetime data_(i1),t_(i2)), i=1,2 ..., the observed frequency n on 8_i, i=1,2 ..., 8, wherein t_(i1),t_(i2)Representing the complete data at interval upper two the end points places of i-th respectively, the observed frequency result of calculation that the product data so tried to achieve drop on each interval is as shown in table 4:

Table 4 life of product drops on the observed frequency computer chart on each interval

(t(i1),t(i2)]	ni
		(0.41,3.94]	9
(4.03,8.64]	9.12
		(8.69,12.59]	8.46
(13.44,17.15]	9.00
		(17.89,23.35]	8.41
(24,30.12]	9.25
		(31.43,35.56]	7.93
(40.5,83.46]	8.92

Step 5: according to this group random censorship type lifetime data, utilizes formula (14) to calculate the Maximum-likelihood estimation of parameter lambda in the exponential of primary electionObtain

Step 6: the Maximum-likelihood estimation of parameter lambda in the exponential of primary electionSubstitute in formula (19), estimate life of product T and drop on each interval [t_(i1),t_(i2)), i=1,2 ..., the expected frequency on 8Its method for solving is as follows:

${\hat{p}}_i=e^{{-0.045t}_{\left(i1\right)}}-e^{{-0.045t}_{\left(i2\right)}},i=\mathrm{1,2},...,8---\left(24\right)$

Wherein t_(i1),t_(i2)Represent the complete data at interval upper two the end points places of i-th respectively；

Step 7: by observed frequency n calculated in step 4, step 6_iAnd expected frequencySubstitute in formula (23), calculate Pearson came χ²Inspection statistics value, its method for solving is as follows:

${\widehat{\mathrm{\&chi;}}}^2=\underset{i=1}{\overset{k}{\mathrm{\&Sigma;}}}\frac{{(n_i-n\widehat{p_i})}^2}{n{\hat{p}}_i}=\underset{i=1}{\overset{8}{\mathrm{\&Sigma;}}}\frac{{(n_i-80{\hat{p}}_i)}^2}{80{\hat{p}}_i}=10.38---\left(25\right)$

Given level of significance α=0.05, by χ²Distribution quantile obtains marginal valueThe wherein number m=1 of unknown parameter in the exponential of primary election。It can be seen that Pearson came χ²Inspection statistics value is less than marginal value, namelySo accepting null hypothesis, it is believed that life of product obeys preliminary selected exponential。

Claims (1)

1. a random censorship type lifetime data distribution inspection method, it is characterised in that: the method specifically comprises the following steps that

Step one: for one group of given random censorship type lifetime data (t_i,δ_i), wherein t_iIt is lifetime data value, δ_iIt is the Boolean variable making truncation mark, first according to the feature of these group data bonded products own characteristic, the life-span distribution of preliminary selected product；Wherein, i=1,2 ..., n；

The distribution of this life-span includes exponential, Weibull distribution, normal distribution and logarithm normal distribution；Set up null hypothesis H₀: product life distribution function F (t；θ)=F₀(t；θ)；Wherein, F₀(t, θ) is preliminary selected life distribution function, θ=(θ₁,θ₂,…,θ_m) it is unknown parameter vector in distribution function, θ₁,θ₂,…,θ_mIt is m unknown parameter of distribution function；

Step 2: disposal data

By all of lifetime data value t_iIt is ranked up from small to large, obtains t₍₁₎≤t₍₂₎≤…≤t_(n), work as t_(i)When being complete data, remember δ_(i)=0；Work as t_(i)When being right censored data, remember δ_(i)=1, reduced data is expressed as (t_(i),δ_(i)), wherein total y of complete data；Wherein, i=1,2 ..., n；

Step 3: for the random censorship type lifetime data (t after arranging_(i),δ_(i)), calculating the mean rank order that y complete data is corresponding, its computational methods are as follows: wherein, i=1, and 2 ..., n；

$A_j=A_{j-1}+\frac{n+1-A_{j-1}}{n-i+2}---\left(13\right)$

Wherein: j is the serial number being sized by all complete datas；I is the serial number that jth complete data is sized in all data；A_jIt it is jth complete data mean rank order in all data；

Step 4: according to the result calculated in step 3, organizes this y tactic complete data in random censorship type lifetime data and is divided into k mutual exclusive interval according to the mean rank order of its correspondence, and wherein k is for being not more thanMaximum integer, y is the number of all complete datas in these group data, and makes the difference (Δ A) of mean rank order corresponding to upper two the end points place complete datas in each interval after dividing_iEqual, (Δ A)_iI-th interval [t is dropped on as this group random censorship type lifetime data_(i1),t_(i2)) on observed frequency n_i, wherein t_(i1),t_(i2)Represent the complete data at interval upper two the end points places of i-th respectively；Wherein, i=1,2 ..., k；

Step 5: according to this group random censorship type lifetime data, calculates the Maximum-likelihood estimation of parameter θ in preliminary selected life-span distributionThe Maximum-likelihood estimation of parameter θ in preliminary selected life-span distributionMethod for solving as follows:

1) likelihood function of exponential is:

$L\left(\mathrm{\&lambda;}\right)=\underset{i=1}{\overset{n}{\mathrm{\&Pi;}}}{\left(\mathrm{\&lambda;}\exp \right\{-{\mathrm{\&lambda;t}}_i\left\}\right)}^{1-{\mathrm{\&delta;}}_i}{\left(\exp \right\{-{\mathrm{\&lambda;t}}_i\left\}\right)}^{{\mathrm{\&delta;}}_i}---\left(14\right)$

Wherein, (t_i,δ_i) it is the random censorship type lifetime data of above-mentioned introduction, t_iIt is lifetime data value, δ_iBeing the Boolean variable making truncation mark, λ is unknown parameter θ in exponential；Find out oneL (λ) in formula (14) is made to take maximum, thenIt it is the Maximum-likelihood estimation of index distributed constant θ=λ；Wherein, i=1,2 ..., n；

2) likelihood function of Weibull distribution is:

$L(\mathrm{\&eta;},m)=\underset{i=1}{\overset{n}{\mathrm{\&Pi;}}}{\&lsqb;\frac{m}{\mathrm{\&eta;}}{\left(\frac{t_i}{\mathrm{\&eta;}}\right)}^{m-1}{e}^{-{(t_i/\mathrm{\&eta;})}^m}\&rsqb;}^{1-{\mathrm{\&delta;}}_i}{\&lsqb;e^{-{(t_i/\mathrm{\&eta;})}^m}\&rsqb;}^{{\mathrm{\&delta;}}_i}---\left(15\right)$

Wherein, (t_i,δ_i) it is the random censorship type lifetime data of above-mentioned introduction, t_iIt is lifetime data value, δ_iBeing the Boolean variable making truncation mark, (η m) is unknown parameter θ in Weibull distribution；Find out one(η m) takes maximum, then to make L in formula (15)It is Weibull distribution parameters θ=(η, Maximum-likelihood estimation m)；Wherein, i=1,2 ..., n；

3) likelihood function of normal distribution is:

$L({\mathrm{\&mu;}}_1,{\mathrm{\&sigma;}}_1)=\underset{i=1}{\overset{n}{\mathrm{\&Pi;}}}{\&lsqb;\frac{1}{{\mathrm{\&sigma;}}_1\sqrt{2\mathrm{\&pi;}}}{e}^{-{(t_i-{\mathrm{\&mu;}}_1)}^2/2{{\mathrm{\&sigma;}}_1}^2}\&rsqb;}^{1-{\mathrm{\&delta;}}_i}{\&lsqb;1-\mathrm{\&Phi;}\left(\frac{t_i-{\mathrm{\&mu;}}_1}{{\mathrm{\&sigma;}}_1}\right)\&rsqb;}^{{\mathrm{\&delta;}}_i}---\left(16\right)$

Wherein, (t_i,δ_i) it is the random censorship type lifetime data of above-mentioned introduction, t_iIt is lifetime data value, δ_iIt is the Boolean variable making truncation mark, (μ₁,σ₁) it is unknown parameter θ in normal distribution；Find out oneMake L (μ in formula (16)₁,σ₁) take maximum, thenIt is Parameters of Normal Distribution θ=(μ₁,σ₁) Maximum-likelihood estimation；Wherein, i=1,2 ..., n；

4) likelihood function of logarithm normal distribution is:

$L({\mathrm{\&mu;}}_2,{\mathrm{\&sigma;}}_2)=\underset{i=1}{\overset{n}{\mathrm{\&Pi;}}}{\&lsqb;\frac{1}{t_i{\mathrm{\&sigma;}}_2\sqrt{2\mathrm{\&pi;}}}{e}^{-{(t_i-{\mathrm{\&mu;}}_2)}^2/2{{\mathrm{\&sigma;}}_2}^2}\&rsqb;}^{1-{\mathrm{\&delta;}}_i}{\&lsqb;1-\mathrm{\&Phi;}\left(\frac{\ln t_i-{\mathrm{\&mu;}}_2}{{\mathrm{\&sigma;}}_2}\right)\&rsqb;}^{{\mathrm{\&delta;}}_i}---\left(17\right)$

Wherein, (t_i,δ_i) it is above-mentioned random censorship type lifetime data, t_iIt is lifetime data value, δ_iIt is the Boolean variable making truncation mark, (μ₂,σ₂) it is unknown parameter θ in logarithm normal distribution；Find out oneMake L (μ in formula (17)₂,σ₂) take maximum, thenIt is lognormal distribution parameter θ=(μ₂,σ₂) Maximum-likelihood estimation；Wherein, i=1,2 ..., n；

Step 6: the Maximum-likelihood estimation of unknown parameter θ in distribution of calculated primary election life-span in step 5Substitute into its corresponding Cumulative Distribution Function F₀(t；In θ), and then estimate life of product T and drop on each interval [t_(i1),t_(i2)) on expected frequencyIts solution formula is as follows: wherein, i=1, and 2 ..., k；

${\hat{p}}_i=P\{t_{\left(i2\right)}\&le;T<t_{\left(i1\right)}\}=F_0(t_{\left(i2\right)};\widehat{\mathrm{\&theta;}})-F_0(t_{\left(i1\right)};\widehat{\mathrm{\&theta;}})---\left(18\right)$

Specifically, for different primary election life-span distributions, it is utilized respectively formula (18) and estimates that life of product T drops on each interval [t_(i1),t_(i2)) on expected frequencyIts method for solving is as follows:

1) exponential:

${\hat{p}}_i=e^{-\widehat{\mathrm{\&lambda;}}{t}_{\left(i1\right)}}-e^{-\widehat{\mathrm{\&lambda;}}{t}_{\left(i2\right)}},i=1,2,...,k---\left(19\right)$

Wherein,It it is the Maximum-likelihood estimation of index distributed constant λ；T_(i1),t_(i2)Represent the complete data at interval upper two the end points places of i-th respectively；

2) Weibull distribution:

${\hat{p}}_i=e^{-{(t_{\left(i1\right)}/\widehat{\mathrm{\&eta;}})}^{\hat{m}}}-e^{-{(t_{\left(i2\right)}/\widehat{\mathrm{\&eta;}})}^{\hat{m}}},i=1,2,...,k---\left(20\right)$

Wherein,It is Weibull distribution parameters (η, Maximum-likelihood estimation m)；T_(i1),t_(i2)Represent the complete data at interval upper two the end points places of i-th respectively；

3) normal distribution:

${\hat{p}}_i=\mathrm{\&Phi;}\left(\frac{t_{\left(i2\right)}-{\widehat{\mathrm{\&mu;}}}_1}{{\widehat{\mathrm{\&sigma;}}}_1}\right)-\mathrm{\&Phi;}\left(\frac{t_{\left(i1\right)}-{\widehat{\mathrm{\&mu;}}}_1}{{\widehat{\mathrm{\&sigma;}}}_1}\right),i=1,2,...,k---\left(21\right)$

Wherein,It is Parameters of Normal Distribution (μ₁,σ₁) Maximum-likelihood estimation；T_(i1),t_(i2)Represent the complete data at interval upper two the end points places of i-th respectively；

4) logarithm normal distribution:

${\hat{p}}_i=\mathrm{\&Phi;}\left(\frac{\ln t_{\left(i2\right)}-{\widehat{\mathrm{\&mu;}}}_2}{{\widehat{\mathrm{\&sigma;}}}_2}\right)-\mathrm{\&Phi;}\left(\frac{\ln t_{\left(i1\right)}-{\widehat{\mathrm{\&mu;}}}_2}{{\widehat{\mathrm{\&sigma;}}}_2}\right),i=1,2,...,k---\left(22\right)$

Wherein,It is lognormal distribution parameter (μ₂,σ₂) Maximum-likelihood estimation；T_(i1),t_(i2)Represent the complete data at interval upper two the end points places of i-th respectively；

Step 7: calculate Pearson came chi-square test statisticIts method for solving is as follows:

${\widehat{\mathrm{\&chi;}}}^2=\underset{i=1}{\overset{k}{\mathrm{\&Sigma;}}}\frac{{(n_i-n{\hat{p}}_i)}^2}{{\mathrm{np}}_i}---\left(23\right)$

Wherein, k is the interval number divided, and n is the number of this group random censorship type lifetime data, n_iIt is that this group random censorship type lifetime data drops on the observed frequency that i-th is interval,It is the life of product expected frequency that drops on i-th interval,It it is the life of product expecterd frequency that drops on i-th interval；Given level of significance α, by χ²Distribution quantile obtains marginal valueWherein m is preliminary selected life-span distribution F₀(t；The number of unknown parameter in θ)；When inspection statistics value is more than marginal value, namelyTime, refuse null hypothesis, it is believed that life of product disobeys preliminary selected life-span distribution, otherwise accepts null hypothesis, it is believed that life of product obeys preliminary selected life-span distribution。