CN101894215A - Likelihood ratio test error detection method - Google Patents

Abstract

The invention discloses a likelihood ratio test error detection method and relates to an application of an artificial intelligent technology in traditional Chinese medicine. The invention provides the likelihood ratio test error detection method based on probability integral transform. The invention provides an analysis tool with accurate and high-efficient prediction, the simulation experiments show that the method can be used for usual small samples; if the P value conclusion obtained by the likelihood ratio test is inconsistent with the appearance of a one-way ordinal contingency table, the P value error needs to be amended according to a 0.074 calibration parameter of the probability integral transform, thereby leading the P value conclusion to be consistent with the appearance of the one-way ordinal contingency table. The invention provides a tail prediction test method which is widely applicable to assess the prediction, and the method can evaluate the distribution of the whole prediction rather than a scalar or a range. The combination of information contents of the prediction distribution with after action knowledge is sufficient to establish a strong test, which can still meet the needs of a prediction program under the situation that the sample size is as small as 100.

Description

The detection method of likelihood ratio test error

Technical field

The present invention relates to of the application of a kind of artificial intelligence technology, especially relate to a kind of detection method of likelihood ratio test error in the traditional Chinese medical science.

Background technology

As far back as the Eastern Han Dynasty, Zhang Zhongjing is just attached great importance to the function of card type theory.Card type theory is meant the dialectical standard that ancient times, the doctor formulated and discusses and control rule." diagnosis and treatment " principle shows that dialectical and opinion is controlled and is used to Clinics and Practices.

Traditional Chinese medical science suggestion, the relation of dialectical theoretical description card type symptom, opinion is administered the relation that opinion is described card type prescription, and the relation of the theoretical contact of prescription card type Chinese medicine.

4 entities (i.e. card type, symptom, prescription and Chinese medicine) and 3 relations (i.e. the relation of the relation of card type symptom, card type prescription relation and card type Chinese medicine) are the marrow of the traditional Chinese medical science.The card type can have many symptoms, and a symptom can be comprised by many card types.One card type must comprise at least one symptom, but symptom not necessarily has the card type.The relation of card type symptom is meant and is used for representing the dialectical of one or more specific symptoms.

At present, the TCM Syndrome Type model only has ([1] Zhang such as Zhang Lianwen, N.L.Yuan, S., Chen, T.and Wang, Y.Latent tree models and diagnosis in traditional Chinese medicine.Artificial Intelligence in Medicine, 2008,42 (3): 229-245) use latent tree-model to analyze data set, find the natural aggregation of data set, well corresponding to TCM Syndrome Type.It provides statistics, and with the checking TCM Syndrome Type, and card pattern type is built in suggestion on dialectical basis.Yet this method thinks that single symptom belongs to specific card type, and uses the Bayesian network analysis to find the natural group in the data set.Those supposition are not consistent with reality, and the model that the result makes in this way and set up seldom produces quite positive performance.

In the actual life, comprise the traditional Chinese medical science etc., all need the prediction of science, and the order of accuarcy of prediction is the primary and foremost purpose of research.At present, the likelihood ratio test of small sample research is divided into interval prediction, density is predicted and tail is predicted three classes.The classic method of prediction lays particular emphasis on assessment interval prediction and density prediction.

Johansen ([2] Johansen, S.A small sample correction for tests of hypotheses on the cointegrating vectors[J] .Journal of Econometrics, 2002,111 (2): 195-221) carry out the whole inference that concerns of relevant association at whole to assisting (cointegrated) venture worth model, the asymptotic inferred results that draws small sample is not accurate enough, should obtain correction factor according to sample size and parameter.

([3] McSorley such as McSorley, E.O., Lu, J.C., and Li, C.S.Performance of Parameter-Estimates in Step-Stress Accelerated Life-Tests With Various Sample-Sizes[J] .IEEE Transactions on Reliability, 2002,51 (3): 271-277) adopt the emulation technology investigation to use the Gaussian approximation fiducial interval of large sample, and estimate at the required sample size of the ML of the limited sample situation of different model of fit.

([4] Wong such as Wong, Heung., Liu, F., Chen, M.and Cheung, W.Empirical likelihood based diagnostics for heteroscedasticity in partial linear models[J] .Computational Statistics and Data Analysis, 2009,53:3466-3477) the bootstrapping emulation of use experience likelihood overcomes the distortion of small sample.Because being one, likelihood ratio test has the asymptotic test that restriction card side distributes.Can overcome the distortion that the experience likelihood ratio test of small sample causes by experience likelihood bootstrapping critical value (EL bootstrap critical value).

In most of traditional Chinese medical science mechanism, not at the card type symptom modelling historical data of disease, in addition, and the time integral that the TCM Syndrome Type model need be longer than other medical science model, these facts show, need to be fit to the forecasting techniques of small sample.If traditional Chinese medical science model is estimated with small sample, the performance of model must have angle widely.But the density forecast assessment can be subjected to the influence of density inside, owing to be subjected to the many little interference of inner delineation, may significantly reduce the concern of traditional Chinese medical science managerial personnel to tail.The tail prediction measures has satisfied cover axiom reasonably directly perceived, and as monotonicity and subadditivity, the tail prediction causes than based on the lower loss of interval prediction.From statistical angle, obviously, the tail prediction measures comprises than interval more information.Tail forecast test power efficient relatively and require more parameter and basic assumption condition at present, has best detection effect.

Summary of the invention

The object of the present invention is to provide a kind of detection method of the likelihood ratio test error based on probability integral transformation.

The detection method of a kind of likelihood ratio test error based on probability integral transformation of the present invention may further comprise the steps:

Step 1) is set mathematical model: establish the stochastic variable X value 0,1 of parent, then the probability function of X is:

$f(X=x;p)={\left(\frac{1}{2}\right)}^x{\left(\frac{1}{2}\right)}^{1-x},x=\mathrm{0,1}---\left(12\right)$

Take out n sample X thus at random ₁, X ₂..., X _n(X _i=0,1 i=1,2 ..., n), in the formula (12), it is 0 or 1 that x represents possible outcome, it is 0 probability (each test is all identical) that p represents the x possible outcome.

Step 2) establishes stochastic variable Y=X ₁+ X ₂+ ... + X _n, because of the distribution function of Y is

$h(Y=y;n,p)=\left(\begin{array}{c}n\\ y\end{array}\right){\left(\frac{1}{2}\right)}^y{\left(\frac{1}{2}\right)}^{n-y},y=\mathrm{1,2},...,n---\left(13\right)$

In the formula (13), h represents in the binomial experiment, tests that to have be for y time the probability distribution of 0 stochastic variable X for n time; It is 0 number of times that y represents the result, and n represents test number (TN), and p represents each 0 probability that occurs.

So can be by following its right value of asking

$P(Y\≥S)=\underset{y=S}{\overset{n}{\mathrm{\Σ}}}\left(\begin{array}{c}n\\ y\end{array}\right){\left(\frac{1}{2}\right)}^y{\left(\frac{1}{2}\right)}^{n-y}---\left(14\right)$

In the formula (14), P represent n test to have S time at least (S＜n) is 0 probability, and S represents that to have S time at least be 0, and n represents test number (TN), and it is 0 number of times that y represents the result.

Step 3) can utilize following its approximate value of asking of central limit theorem to be for related with Gauss's likelihood:

$\stackrel{\‾}{X}=\frac{1}{n}\underset{k=1}{\overset{n}{\mathrm{\Σ}}}{X}_k,$

Wherein, n representative sample size.

Step 4) knows that by central limit theorem its distribution is similar to

Wherein, N represents normal distribution, n representative sample size,

Cause

P is 0 probability, and q is 1 probability), so

$t=\frac{\stackrel{\‾}{X}-\frac{1}{2}}{\sqrt{\frac{1}{4n}}}---\left(15\right)$

Distribution be similar to N (0,1), wherein on behalf of t, t distribute,

The representative sample size is the average of the stochastic variable X of n, and n representative sample size is n,

Cause

$Y\≥S\≅\stackrel{\‾}{X}\≥\frac{S}{n}\≅t\≥\frac{2S-n}{\sqrt{n}}---\left(16\right)$

In the formula (16), it is 0 number of times that y represents the result, S represent the result be 0 have at least S time (S＜n), n representative sample size,

So $P\{Y\&GreaterEqual;S\}=P(t\&GreaterEqual;\frac{2S-n}{\sqrt{n}})~\frac{1}{\sqrt{2\mathrm{\&pi;}}}{\&Integral;}_{\frac{2s-n}{\sqrt{n}}}^{\&infin;}{e}^{-\frac{{\mathrm{<figure-callout\; id="2"\; label="t"\; filenames="HSA00000183714800012.png"\; state="\{\{state\}\}">t</figure-callout>}}^2}{2}\mathrm{dt}}$

$=0.5-\frac{1}{\sqrt{2\mathrm{\&pi;}}}{\&Integral;}_0^{\frac{2s-n}{\sqrt{n}}}{e}^{-\frac{{\mathrm{<figure-callout\; id="2"\; label="t"\; filenames="HSA00000183714800012.png"\; state="\{\{state\}\}">t</figure-callout>}}^2}{2}}\mathrm{dt}=0.5-\mathrm{\&Phi;}\left(\frac{2S-n}{\sqrt{n}}\right)---\left(17\right)$

Wherein, the accurate single argument normal distribution of bidding $\mathrm{\&Phi;}\left(x\right)=\frac{1}{\sqrt{2\mathrm{\&pi;}}}{\&Integral;}_0^x{e}^{-\frac{{\mathrm{<figure-callout\; id="2"\; label="t"\; filenames="HSA00000183714800012.png"\; state="\{\{state\}\}">t</figure-callout>}}^2}{2}}\mathrm{dt}).$

Step 5) when n=∞ (n=1000), by

Can look into gaussian distribution table (the most important continuous probability distribution table of statistics) gets

$\frac{2S-1000}{31.6}=2.33---\left(20\right)$

Compare with n again after solving S, can draw calibration parameter.

Step 6) is when the looks of likelihood ratio test resulting P value conclusion and one-dimensional order contingency table are inconsistent, and the calibration parameter correction P value error according to based on probability integral transformation makes P value conclusion consistent with the looks of one-dimensional order contingency table.

Outstanding advantage of the present invention is as follows:

The present invention proposes the detection method of relevant likelihood ratio test error, is used for the prediction of one-dimensional order contingency table and small sample.Its objective is provides prediction accuracy and analysis tool efficiently, and more general forecast test power.Simulation shows, this method can be used for common small sample.If when the looks of likelihood ratio test resulting P value conclusion and one-dimensional order contingency table are inconsistent, should make P value conclusion consistent according to 0.074 calibration parameter correction P value error based on probability integral transformation with the looks of one-dimensional order contingency table.We are applied to the analysis of stomach trouble with the detection method of likelihood ratio test error, obtain the stomachache severity parameter result consistent with the prediction of contingency table looks of card type.

Description of drawings

Fig. 1 is the card type and the stomachache severity looks of the embodiment of the invention.In Fig. 1, horizontal ordinate is the card type, and ordinate is a probability; Curve 1 is a liver-stomach disharmony; Curve 2 is weakness of the spleen and the stomach; Curve 3 is damp heat in the spleen and the stomach; Curve 3 is a marginal probability.

Fig. 2 is the card type and the stomachache severity looks of the embodiment of the invention.In Fig. 2, horizontal ordinate is the card type, and ordinate is a probability; Curve 1 is a liver-stomach disharmony; Curve 2 is weakness of the spleen and the stomach; Curve 3 is damp heat in the spleen and the stomach; Curve 3 is a marginal probability.

Embodiment

1, independent same distribution

In Bernoulli Jacob (Bernoulli) test occasion, when test number (TN) n is big, calculate also inconvenient.Poisson's theorem is told us, when p≤0.1, can use the Poisson distribution approximate treatment, but doing approximate treatment with normal distribution then the restriction of p≤0.1 can be subjected to, thereby central limit theorem is made accurate Calculation in Bernoulli Jacob's occasion ingenious part can be realized.In a word, central limit theorem will be applied in Bernoulli Jacob's occasion more and more.

This joint comprises law of great numbers (Law of Large Numbers), central limit theorem (Central Limit Theorem) and 3 parts of Rosenblatt conversion.

1.1 law of great numbers

The independent identically distributed suffering law of great numbers of admiring: establish x ₁, x ₂..., x _nBe independence and sequence of random variables, and have mathematical expectation and variance: E (x with same distribution _i)=μ, D (x _i)=σ ²(i=1,2 ..., n), then, have any given ε＞0

$\underset{n\&RightArrow;\&infin;}{\mathrm{lin}}P\left|\right|\frac{1}{n}\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}{x}_i-\mathrm{\&mu;}|<\mathrm{\&epsiv;}|=1---\left(1\right)$

The arithmetic mean convergence in (with)probability that the hot law of great numbers of admiring illustrates independent identically distributed stochastic variable is in its mathematical expectation, and it is for estimating that with arithmetic mean mathematical expectation provides theoretical foundation in the practical application.

1.2 central limit theorem

The central limit theorem of independent same distribution (iid): establish x ₁, x ₂..., x _nBe independence and sequence of random variables, and have mathematical expectation and variance: E (x with same distribution _i)=μ, D (x _i)=σ ²≠ 0 (i=1,2 ..., n), then, have any real number

$\underset{n\&RightArrow;\&infin;}{\mathrm{lin}}P|\frac{\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}{x}_i-\mathrm{\&mu;}}{\sqrt{\mathrm{\&mu;}}\mathrm{\&sigma;}}\&le;x|=\frac{1}{\sqrt{2n}}{\&Integral;}_{-\&infin;}^x{e}^{-\frac{{\mathrm{<figure-callout\; id="2"\; label="t"\; filenames="HSA00000183714800012.png"\; state="\{\{state\}\}">t</figure-callout>}}^2}{2}}\mathrm{dt}=\mathrm{\&Phi;}\left(x\right)---\left(2\right)$

Independent identically distributed central limit theorem has been expressed the special status of normal distribution in theory of probability, although x _iDistribution be arbitrarily, but as long as n fully big, stochastic variable

$\frac{\underset{i=1}{\overset{n}{\mathrm{\&Sigma;}}}{x}_i-\mathrm{\&mu;}}{\sqrt{n}\mathrm{\&sigma;}}\&le;x---\left(3\right)$

The approximate standardized normal distribution N (0,1) that obeys, in other words, when n is very big, independent identically distributed stochastic variable x _iAnd

Approximate Normal Distribution N (n μ, n σ ²).Those many small, stochastic variables of the overall result of enchancement factor effect independently that Here it is, the general rationale of Normal Distribution approx, thereby normal distribution all has great importance in theory with on using.(n p), then when n is very big, has as if x～B

$P(a\&le;x\&le;b)=P(\frac{b-\mathrm{np}}{\sqrt{\mathrm{npq}}}\&le;\frac{x-\mathrm{np}}{\sqrt{\mathrm{npq}}}\&le;\frac{a-\mathrm{np}}{\sqrt{\mathrm{npq}}})$

$\&ap;\mathrm{\&Phi;}\left|\frac{b-\mathrm{np}}{\sqrt{\mathrm{np}(1-p)}}\right|-\mathrm{\&Phi;}\left|\frac{a-\mathrm{np}}{\sqrt{\mathrm{np}(1-p)}}\right|---\left(4\right)$

1.3Rosenblatt conversion

A stochastic process y _t, estimate at time t-1, provide y _tProbability density be f (y _t) with relevant distribution function

Interval prediction is based on contrary distribution letter+number,

${\stackrel{\&OverBar;}{y}}_t=F^{-1}\left(\mathrm{\&alpha;}\right)---\left(5\right)$

For example, 99% the satisfaction in two weeks by a definite date is a quantity

Make

Christoffersen (1998) points out to verify a way of forecast interval, and promptly interval should surpassing or time of α % in violation of rules and regulations, this unlawful practice also should be uncorrelated with the time, and in conjunction with these attributes, variable-definition is

I _t=1, if in violation of rules and regulations

=0, if do not take place in violation of rules and regulations

It should be an independent same distribution Bernoulli sequence that parameter alpha is arranged, because rare violation (by design), check looks at that forming a Bernoulli Jacob whether in violation of rules and regulations needs hundreds of to observe at least, key issue is, Bernoulli Jacob's variable just has only two values (0 and 1), and be worth 1 seldom, the density evaluation method is utilized whole distributions of result, thereby draws a bigger quantity of information from available data.

Not to only limit to note rare violation, just can all realizations of conversion become a series of independent identically distributed stochastic variables.

Specifically, Rosenblatt ( ^[5 ^]Rosenblatt, M.Remarks on a Multivariate Transformation[J] .The Annals of Mathematical Statistics, 1952,23:470-472) Ding Yi conversion

$x_t={\&Integral;}_{-\&infin;}^{y_t}\hat{f}\left(u\right)\mathrm{du}=\hat{F}\left(y_t\right)---\left(6\right)$

Y wherein _tBe afterwards knowledge and

Be the loss density of ex ante forecasting, Rosenblatt shows x _tBe independent same distribution and be uniformly distributed in (0,1).Therefore, if the necessary regular reporting prediction distribution of enterprise, Regulator can use this probability integral transformation, tests and whether violates independence and/or consistance.In addition, its knowledge y no matter _tThe behind distribute, even forecast model

Change in time, this result still sets up.

2 likelihood ratio test frameworks

Below introduced the extension of Rosenblatt conversion, independent same distribution N (0,1) is provided under the null hypothesis variable, this allows convenient and estimates Gauss's likelihood flexibly and make up based on the inspection statistics of likelihood and have good finite sample property.

Be difficult to small data sample check consistency, this check is that the nonparametric that statistical circles is determined is the fact of a straight line with utilizing uniform density.It also is difficult to design the parameter when checking null hypothesis to be U (0,1) stochastic variable.At a nested x of more general pattern _tIndependent same distribution U (0,1) model needs provided support to depend upon unknown parameter, likelihood ratio (Likelihood Ratio, LR) and other statistics because the uncontinuity of objective function does not have usual gradation.

Berkowitz advocates one and simply is transformed to normal state.At first, conversion is a simple computation, can directly calculate Gauss's likelihood after the conversion and make up LR, to some classifications of pattern failure, the LR check be evenly the strongest (Uniformly Most Powerful, UMP).That is to say that the LR program has higher ability than the check of the fixedly degree of confidence of each value of any other unknown parameter.At last, even it can not be proved to be evenly the most powerful, the LR check often has desirable inspection statistics characteristic and good limited sample behavior (referring to [6] Hogg, R.V., and Craig, A.T.Mathematical Statistics[M] .New York:Macmillan.1965).

The attracting characteristics of another of likelihood test framework are that the researchist has very big decision check for which and how many restrictions.Small sample, but we want tight parametric test probably.

Though people can check average, variance, the degree of bias of independent same distribution U (0,1) data or the like, the performance of these programs usually with provide sample size relevant.

Provide Φ ^-1(.) is contrary Standard Normal Distribution, and then the prediction for any sequence has following result.

Proposition 1: if series

Be as an independent same distribution U (0,1), then

$z_t={\mathrm{\&Phi;}}^{-1}\left[{\&Integral;}_{-\&infin;}^{y_t}f\left(u\right)\mathrm{du}\right]---\left(7\right)$

Be an independent same distribution U (0,1).

The conversion of proposition 1 is to be used for the emulation stochastic variable, and it shows simple an extension of Rosenblatt conversion, and we change and observe the combination income to creating a series

This should be the iid standard normal, is what makes it so useful, is because according to null hypothesis, the data Normal Distribution, and this provides us the convenient tool related with Gauss's likelihood.

In addition, can be in some aspects, the conversion of deal with data has identical explanation as non-switched raw data.Following this notion of proposition official confirmation.

Proposition 2: provide z _tDensity h (z _t) and standardized normal distribution Φ (z _t), then

$\log [f\left(y_t\right)/\hat{f}\left(y_t\right)]=\log [h\left(z_t\right)/\mathrm{\&Phi;}\left(z_t\right)]---\left(8\right)$

Proof: at Φ ^-1Data converted can be written as function of functions

Wherein

Be model prediction, Φ ^-1Be contrary normal distribution, use the Jacobian conversion, z _tDistribution provides

After logarithm and arrangement, the result who obtains requiring.

Proposition 2 regulations, inaccurate density prediction will be retained in the data after the conversion.For example, if Within the specific limits, also will be to make h (z like this _t)＞Φ (z _t) at the respective regions of standard normal.

Not the Rosenblatt conversion, neither further apply the normal state conversion of any distributional assumption basic data; On the contrary, correct density prediction means that converted variable is a normal state.

Suppose given model formation sequence

Because z _tShould be independently to observe and standard normal, diversified check can make up.Particularly null hypothesis can be tested, for example, average that the single order autoregression is selected fully and variance may with (0,1) difference, can write

z _t-μ＝ρ(z _t-1-μ)+ε _t (9)

What 1 null hypothesis of assigning a topic was described is μ=0, ρ=0, and var (ε _t)=1, the definite log-likelihood function relevant with equation (9) is well-known, reprints for convenient here:

$-\frac{1}{2}\log \left(2\mathrm{\&pi;}\right)-\frac{1}{2}\log [{\mathrm{\&sigma;}}^2/(1-{\mathrm{\&rho;}}^2)]-\frac{{(z_1-\mathrm{\&mu;}/(1-\mathrm{\&rho;}))}^2}{2{\mathrm{\&sigma;}}^2/(1-{\mathrm{\&rho;}}^2)}$

$-\frac{T-1}{2}\log \left(2\mathrm{\&pi;}\right)-\frac{T-1}{2}\log \left({\mathrm{\&sigma;}}^2\right)-\underset{t=2}{\overset{T}{\mathrm{\&Sigma;}}}\left(\frac{{(z_t-\mathrm{\&mu;}-\mathrm{\&rho;}{z}_{t-1})}^2}{2{\mathrm{\&sigma;}}^2}\right)---\left(10\right)$

σ wherein ²Be ε _tVariance, for for purpose of brevity, the likelihood that Berkowitz writes has only the unknown parameter function of model, L (μ, σ ², ρ).

The LR independence test of observed value can reduce

${\mathrm{LR}}_{\mathrm{ind}}=-2\left(L(\widehat{\mathrm{\&mu;}},{\widehat{\mathrm{\&sigma;}}}^2,0\right)-L\left(\widehat{\mathrm{\&mu;}},{\widehat{\mathrm{\&sigma;}}}^2,\widehat{\mathrm{\&rho;}})\right),---\left(11\right)$

Wherein cap is expressed as estimated value, and this test statistics is the degree that metric data is supported the non-zero parameter, and in null hypothesis, test statistics is distributed as χ ²(1), the side's of card degree of freedom is 1, and mode that can be common is carried out reasoning like this.

The shortcoming of LR check is West ([7] West, K.D.Asymptotic Inference About Predictive Ability[J] .Econometrica, 1996,64:1067-1084) emphasized: the prediction of small sample estimation model generation sometimes may be subjected to the uncertainty influence of parameter.

Below provide the specific embodiment of the detection method of likelihood ratio test error.

Certain city two families (influenza vaccines) inoculation point A, B do business rivalry, and supposing to add up to every day has n position client, and this n position client independently and at random selects the influenza vaccines of each inoculation point mutually, the vaccine problem of storing up of each inoculation point of consideration.

This problem is

N Bernoulli trials problem, so its mathematical model is: establish the stochastic variable X value 0,1 of parent, X=1 represents to inoculate the influenza vaccines that A is ordered, and X=0 represents to inoculate the influenza vaccines that B is ordered, and then the probability function of X is

$f(X=x;p)={\left(\frac{1}{2}\right)}^x{\left(\frac{1}{2}\right)}^{1-x},x=\mathrm{0,1}---\left(12\right)$

Take out n sample X thus at random ₁, X ₂..., X _n(X _i=0,1 i=1,2 ..., n).

If stochastic variable Y=X ₁+ X ₂+ ... + X _nThe number that inoculation A order among the expression n position client props up that (S＜n), the probability of S people or above inoculation A point influenza vaccines are P (Y 〉=S) if the influenza vaccines of inoculation point A prepare S

Because of the distribution function of Y is

$h(Y=y;n,p)=\left(\begin{array}{c}n\\ y\end{array}\right){\left(\frac{1}{2}\right)}^y{\left(\frac{1}{2}\right)}^{n-y},y=\mathrm{1,2},...,n---\left(13\right)$

So can be by following its right value of asking

$P(Y\&GreaterEqual;S)=\underset{y=S}{\overset{n}{\mathrm{\&Sigma;}}}\left(\begin{array}{c}n\\ y\end{array}\right){\left(\frac{1}{2}\right)}^y{\left(\frac{1}{2}\right)}^{n-y}---\left(14\right)$

, but, can utilize following its approximate value of asking of central limit theorem for related with Gauss's likelihood.

If

Know that by central limit theorem its distribution is similar to

Cause

So

$t=\frac{\stackrel{\&OverBar;}{X}-\frac{1}{2}}{\sqrt{\frac{1}{4n}}}---\left(15\right)$

Distribution be similar to N (0,1), because of

$Y\&GreaterEqual;S\&cong;\stackrel{\&OverBar;}{X}\&GreaterEqual;\frac{S}{n}\&cong;t\&GreaterEqual;\frac{2S-n}{\sqrt{n}}---\left(16\right)$

So

$P\{Y\&GreaterEqual;S\}=P(t\&GreaterEqual;\frac{2S-n}{\sqrt{n}})~\frac{1}{\sqrt{2\mathrm{\&pi;}}}{\&Integral;}_{\frac{2s-n}{\sqrt{n}}}^{\&infin;}{e}^{-\frac{{\mathrm{<figure-callout\; id="2"\; label="t"\; filenames="HSA00000183714800012.png"\; state="\{\{state\}\}">t</figure-callout>}}^2}{2}\mathrm{dt}}$

$=0.5-\frac{1}{\sqrt{2\mathrm{\&pi;}}}{\&Integral;}_0^{\frac{2s-n}{\sqrt{n}}}{e}^{-\frac{{\mathrm{<figure-callout\; id="2"\; label="t"\; filenames="HSA00000183714800012.png"\; state="\{\{state\}\}">t</figure-callout>}}_2}{2}}\mathrm{dt}=0.5-\mathrm{\&Phi;}\left(\frac{2S-n}{\sqrt{n}}\right)---\left(17\right)$

(the accurate single argument normal distribution of bidding $\mathrm{\&Phi;}\left(x\right)=\frac{1}{\sqrt{2\mathrm{\&pi;}}}{\&Integral;}_0^x{e}^{-\frac{{\mathrm{<figure-callout\; id="2"\; label="t"\; filenames="HSA00000183714800012.png"\; state="\{\{state\}\}">t</figure-callout>}}^2}{2}}\mathrm{dt})$

Suitably getting S makes

The time then represent at most only 1 people that 100 philtrums are not inoculated into, or the rarest 99 people of 100 philtrums are inoculated into influenza vaccines.

For example inoculate number n=1000 every day when (two inoculations point adds up to the inoculation number)

$0.5-\mathrm{\&Phi;}\left(\frac{2S-1000}{31.6}\right)<0.01---\left(18\right)$

Then

$\mathrm{\&Phi;}\left(\frac{2S-1000}{31.6}\right)>0.49---\left(19\right)$

By

Can look into gaussian distribution table gets

$\frac{2S-1000}{31.6}=2.33---\left(20\right)$

The S that satisfies formula (19) is $\frac{2S-1000}{31.6}>2.33,S>536,$

Then must be as if inoculating number n=1000 (two inoculations point adds up to) every day, A inoculation point desires to make minimum 99 people of 100 philtrums to be inoculated into influenza vaccines, must set and store up the same if B inoculation point of minimum 537 of vaccine (so get S=537) and desire to make minimum 99 people of 100 philtrums to be inoculated into influenza vaccines also must to set and store up minimum 537 of vaccine (set vaccine 537 then can) so two an inoculations total need be established vaccine 537+537=1074 props up, cause two inoculation point clients only 1000 people are called 0.074 calibration parameter so lose 74 (0.074) vaccines by business rivalry.

Below provide experimental result and analysis.

When if the looks of likelihood ratio test resulting P value conclusion and one-dimensional order contingency table are inconsistent, should be according to 0.074 calibration parameter correction P value error of the 3rd joint statement, make P value conclusion consistent with the looks of one-dimensional order contingency table, whether our the desire severity of relatively having a stomachache is relevant with the card type, measured 227 sampled points, must table 1 data, be the one-dimensional order contingency table, table 1 comprises the stomachache severity (0-3: normal, slight of frequency J=4 kind varying level, moderate, severe) find in I=3 kind card type; When the grid of table all is bigger than 6, large sample theory is suitable for contingency table for the scoring check, if we the frequency of every row (card type) divided by total corresponding line, we obtain to have relative probability to show the table 2 of 3 * 4 contingency tables.

Utilize these data p _Ij/ p _I.And p _.jDraw, can obtain the looks of serious symptom degree, the looks (OK) of different card types are presented in the line chart of Fig. 1, the marginal probability and the conditional probability of indication stomachache severity, away from go and show that capable classification is relevant with the row classification.

Table 1 stomachache severity frequency

Table 2 stomachache severity relative probability

In Fig. 1 level " moderate ", between dotted line and solid line (DAMPH) maximum disparity is arranged, almost reach 0.2, stomachache severity 〉=2 of deduction DEFSS are compared with DAMPH less probability.Therefore the severity of having a stomachache is relevant with the card type.Utilize card side's independence test, whether be associated χ with the stomachache order of severity with true authentication-type ²Value is 25.90, P=0.0002, and expression stomachache severity is relevant with the card type.Proportion of utilization calculates than digital-to-analogue type, L ₁Be equivalent to-297.9847, if b ₁=b ₂=0, L ₀=-300.9095 ,-2 (L ₀-L ₁) obtain D2 and equal 5.8496, relevant χ ²Critical value will be checked H ₀: β ₁=β ₂=0, find So P value 0.0536, null hypothesis is not rejected, and draws such conclusion, β ₁And β ₂Be zero simultaneously, the constituent ratio looks of these conclusions and Fig. 1 are inconsistent, and according to behind the 0.074 above-mentioned calibration parameter round-off error, the P value is 0.0497, and then the constituent ratio looks with Fig. 1 are consistent.In addition, analyze β ₁=0 or β ₂=0 or both neither be zero.Table 3 is the results by maximal possibility estimation (MLE).

From table 3, refusal β ₁=0 and β ₂=0, the P value is respectively 0.021 and 0.038, and this is that to accuse of type relevant with stomachache seriousness, and INCRD relatively DEFSS be (1,0) and (0,1) as indieating variable x, therefore

L _j(1，0)＝θ _j+β ₁ (21)

The parameter of table 3 card type and stomachache severity

And

L _j(0，1)＝θ _j+β ₂ (22)

Be respectively INCRD and the DEFSS logarithm than number ratio, the logarithm than number ratio of DAMPH is then

L _j(0，0)＝θ _j (23)

Represent the ratio number of severity≤j of INCRD to compare logarithm greater than the ratio number of DAMPH than logarithm.In other words, patient's INCRD stomachache symptom is normally so not serious;

Represent the ratio number of severity≤j of DEFSS to compare logarithm greater than the ratio number of DAMPH than logarithm.In other words, patient's DEFSS stomachache symptom is normally so more not serious than DAMPH; These conclusions are consistent with the constituent ratio looks of Fig. 1.

But, be revised as 22 and 5 (table 4) respectively as the patient that will go up the severity grade weakness of the spleen and the stomach in the example 2,3, then adopt χ ²Check is analyzed, and whether is associated χ with the stomachache order of severity with true authentication-type ²Value is 27.31, P=0.0001, and expression stomachache severity is relevant with the card type.

Table 4 stomachache severity frequency

The parameter of table 5 card type and stomachache severity

Proportion of utilization calculates than digital-to-analogue type, L ₁Be equivalent to-297.0911, if b ₁=b ₂=0, L ₀=-300.0885 ,-2 (L ₀-L ₁) obtain D2 and equal 5.9948, relevant χ ²Critical value will be checked H ₀: β ₁=β ₂=0, find

So P value 0.0499, null hypothesis is rejected, and draws such conclusion, β ₁And β ₂Not zero simultaneously.In addition, analyze β ₁=0 or β ₂=0 or both neither be zero, table 5 is the results by maximal possibility estimation (MLE).

From table 5, refusal β ₁=0 and β ₂=0, the P value is respectively 0.021 and 0.033, this be accuse of type with the stomachache seriousness relevant, these conclusions are consistent with the constituent ratio looks of Fig. 2.

The prediction of all science is extremely important, comprises the traditional Chinese medical science, and likelihood ratio test is differentiated, and the suitable foot of its prediction efficiency and model is inseparable, and failure means the deficiency of forecast model.The present invention has developed the inference procedure of the predicated error of the prediction of relevant contingency table looks and likelihood ratio test, be used in the prediction of one-dimensional order contingency table and small sample, its objective is the analysis tool that prediction accuracy and efficient are provided, and more general forecast test power.Simulation shows, this program can be used for common small sample.

As if LR check framework is flexibly, intuitively, and the method for inspection provides very good power of test characteristic, the shortcoming of LR check is the prediction that the small sample estimation model produces, and may be subjected to the uncertainty influence of parameter.The present invention proposes a kind of new calibration parameter method and assesses this prediction; If when the looks of likelihood ratio test resulting P value conclusion and one-dimensional order contingency table are inconsistent, should make P value conclusion consistent according to 0.074 above-mentioned calibration parameter correction P value error with the looks of one-dimensional order contingency table.

Claims (1)

1. the detection method of likelihood ratio test error is characterized in that may further comprise the steps:

Step 1) is set mathematical model: establish the stochastic variable X value 0,1 of parent, then the probability function of X is:

$f(X=x;p)={\left(\frac{1}{2}\right)}^x{\left(\frac{1}{2}\right)}^{1-x},x=\mathrm{0,1}---\left(12\right)$

Take out n sample X thus at random ₁, X ₂..., X _n(X _i=0,1 i=1,2 ..., n), in the formula (12), it is 0 or 1 that x represents possible outcome, it is 0 probability (each test is all identical) that p represents the x possible outcome;

Step 2) establishes stochastic variable Y=X ₁+ X ₂+ ... + X _n, because of the distribution function of Y is

$h(Y=y;n,p)=\left(\begin{array}{c}n\\ y\end{array}\right){\left(\frac{1}{2}\right)}^y{\left(\frac{1}{2}\right)}^{n-y},y=\mathrm{1,2},...,n---\left(13\right)$

In the formula (13), h represents in the binomial experiment, tests that to have be for y time the probability distribution of 0 stochastic variable X for n. time; It is 0 number of times that y represents the result, and n represents test number (TN), and p represents each 0 probability that occurs;

So can be by following its right value of asking

$P(Y\&GreaterEqual;S)=\underset{y=S}{\overset{n}{\mathrm{\&Sigma;}}}\left(\begin{array}{c}n\\ y\end{array}\right){\left(\frac{1}{2}\right)}^y{\left(\frac{1}{2}\right)}^{n-y}---\left(14\right)$

In the formula (14), P represent n test to have S time at least (S＜n) is 0 probability, and S represents that to have S time at least be 0, and n represents test number (TN), and it is 0 number of times that y represents the result;

Step 3) can utilize following its approximate value of asking of central limit theorem to be for related with Gauss's likelihood:

$\stackrel{\&OverBar;}{X}=\frac{1}{n}\underset{k=1}{\overset{n}{\mathrm{\&Sigma;}}}{X}_k,$

Wherein, n representative sample size;

Step 4) knows that by central limit theorem its distribution is similar to

Wherein, N represents normal distribution, n representative sample size,

Cause

P is 0 probability, and q is 1 probability), so

$t=\frac{\stackrel{\&OverBar;}{X}-\frac{1}{2}}{\sqrt{\frac{1}{4n}}}---\left(15\right)$

Distribution be similar to N (0,1), wherein on behalf of t, t distribute,

The representative sample size is the average of the stochastic variable X of n, and n representative sample size is n,

Cause

$Y\&GreaterEqual;S\&cong;\stackrel{\&OverBar;}{X}\&GreaterEqual;\frac{S}{n}\&cong;t\&GreaterEqual;\frac{2S-n}{\sqrt{n}}---\left(16\right)$

In the formula (16), it is 0 number of times that y represents the result, S represent the result be 0 have at least S time (S＜n), n representative sample size,

So $P\{Y\&GreaterEqual;S\}=P(t\&GreaterEqual;\frac{2S-n}{\sqrt{n}})~\frac{1}{\sqrt{2\mathrm{\&pi;}}}{\&Integral;}_{\frac{2s-n}{\sqrt{n}}}^{\&infin;}{e}^{-\frac{t^2}{2}\mathrm{dt}}$

$=0.5-\frac{1}{\sqrt{2\mathrm{\&pi;}}}{\&Integral;}_0^{\frac{2s-n}{\sqrt{n}}}{e}^{-\frac{t_2}{2}}\mathrm{dt}=0.5-\mathrm{\&Phi;}\left(\frac{2S-n}{\sqrt{n}}\right)---\left(17\right)$

Wherein, the accurate single argument normal distribution of bidding $\mathrm{\&Phi;}\left(x\right)=\frac{1}{\sqrt{2\mathrm{\&pi;}}}{\&Integral;}_0^x{e}^{-\frac{t^2}{2}}\mathrm{dt});$

Step 5) when n=∞ (n=1000), by

Can look into gaussian distribution table (the most important continuous probability distribution table of statistics) gets

$\frac{2S-1000}{31.6}=2.33---\left(20\right)$

Compare with n again after solving S, can draw calibration parameter;

Step 6) is when the looks of likelihood ratio test resulting P value conclusion and one-dimensional order contingency table are inconsistent, and the calibration parameter correction P value error according to based on probability integral transformation makes P value conclusion consistent with the looks of one-dimensional order contingency table.

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