Maximum likelihood and the information bottleneck

服从指数分布的极大似然估计的实际例子1.统计学家使用极大似然估计来估计指数分布的参数。

Statisticians use maximum likelihood estimation toestimate the parameters of the exponential distribution.2.通过估计参数λ，我们可以更好地了解指数分布中事件发生的方式。

By estimating the parameter λ, we can gain a better understanding of how events occur in an exponential distribution.3.极大似然估计是一种常用的统计方法，可用于估计各种类型的分布。

Maximum likelihood estimation is a commonly usedstatistical method that can be used to estimate various types of distributions.4.我们可以使用极大似然估计来估计指数分布的中位数或平均值。

We can use maximum likelihood estimation to estimate the median or mean of the exponential distribution.5.估计出的参数可以用于预测未来事件的发生情况。

The estimated parameters can be used to predict the occurrence of future events.6.极大似然估计需要收集一定数量的数据来进行计算。

Maximum likelihood estimation requires collecting a sufficient amount of data for computation.7.通过比较不同参数值下的似然函数，我们可以找到最可能的参数值。

极大似然估计（Maximum Likelihood Estimation，MLE）是统计学中一种常用的参数估计方法。

它通过寻找参数值，使得观测数据在给定参数下的概率最大化，来估计模型的参数。

极大似然估计常用于拟合概率分布或模型到观测数据，以便更好地理解数据的生成过程或进行预测。

在极大似然估计中，通常使用似然函数（Likelihood Function）来表示观测数据在不同参数下的概率。

似然函数通常记为L(θ|X)，其中θ表示待估计的参数，X表示观测数据。

极大似然估计的目标是找到最大化似然函数的参数θ：$$\hat{\theta}_{\text{MLE}} = \arg \max_{\theta} L(\theta|X)$$拟合指标通常用于评估极大似然估计的质量和模型拟合的好坏。

以下是一些常见的拟合指标：1. **对数似然（Log-Likelihood）**：对数似然是似然函数取对数的结果，通常用于数值计算和比较不同模型的拟合质量。

对数似然越大，表示模型对观测数据的拟合越好。

2. **AIC（Akaike Information Criterion）**：AIC是一种用于模型选择的指标，它考虑了对数似然和模型参数数量之间的权衡。

AIC越小，表示模型更好，因为它同时考虑了拟合程度和模型的复杂度。

3. **BIC（Bayesian Information Criterion）**：BIC也用于模型选择，类似于AIC，但对模型复杂度的惩罚更严格。

BIC同样越小越好。

4. **假设检验**：假设检验可以用于评估模型参数的显著性。

例如，对于回归模型，可以使用t检验或F检验来检验回归系数是否显著不同于零。

5. **残差分析**：在拟合模型时，对残差进行分析可以帮助评估模型的拟合质量。

残差是观测值与模型预测值之间的差异，正常情况下应该是随机的，没有明显的模式。

这些拟合指标可以帮助你评估极大似然估计的结果以及模型拟合的好坏。

Course 4Fall 2003 Society of Actuaries**BEGINNING OF EXAMINATION** 1. You are given the following information about a stationary AR(2) model:=.(i) ρ105(ii) ρ201=.Determine φ2.(A) –0.2(B) 0.1(C) 0.4(D) 0.7(E) 1.0(i) Losses follow a loglogistic distribution with cumulative distribution function:F x x x b g b g b g =+//θθγγ1(ii)The sample of losses is:10 35 80 86 90 120 158 180 200 210 1500Calculate the estimate of θ by percentile matching, using the 40th and 80th empirically smoothed percentile estimates.(A) Less than 77(B) At least 77, but less than 87(C) At least 87, but less than 97(D) At least 97, but less than 107(E) At least 107(i) The number of claims has a Poisson distribution.(ii) Claim sizes have a Pareto distribution with parameters θ=0.5 and α=6.(iii) The number of claims and claim sizes are independent.(iv) The observed pure premium should be within 2% of the expected pure premium 90% of the time.Determine the expected number of claims needed for full credibility.(A) Less than 7,000(B) At least 7,000, but less than 10,000(C) At least 10,000, but less than 13,000(D) At least 13,000, but less than 16,000(E) At least 16,0004. You study five lives to estimate the time from the onset of a disease to death. The times todeath are:2 3 3 3 7Using a triangular kernel with bandwidth 2, estimate the density function at 2.5.(A) 8/40(B) 12/40(C) 14/40(D) 16/40(E) 17/405. For the model i i i Y X αβε=++, where 1,2,...,10i =, you are given:(i) X i i =R S T1, if the th individual belongs to a specified group 0, otherwise(ii) 40 percent of the individuals belong to the specified group.(iii) The least squares estimate of β is β=4.(iv) ()2ˆˆ92i i Y X αβ−−=∑Calculate the t statistic for testing H 00:β=.(A) 0.9(B) 1.2(C) 1.5(D) 1.8(E) 2.1(i) Losses follow a Single-parameter Pareto distribution with density function:()()1,1f x x xαα+=>, 0 < α < ∞ (ii) A random sample of size five produced three losses with values 3, 6 and 14, and twolosses exceeding 25.Determine the maximum likelihood estimate of α.(A) 0.25(B) 0.30(C) 0.34(D) 0.38(E) 0.42(i) The annual number of claims for a policyholder has a binomial distribution withprobability function:()()221x x p x q q q x −⎛⎞=−⎜⎟⎝⎠, x = 0, 1, 2(ii) The prior distribution is:()34,01q q q π=<<This policyholder had one claim in each of Years 1 and 2.Determine the Bayesian estimate of the number of claims in Year 3.(A) Less than 1.1(B) At least 1.1, but less than 1.3(C) At least 1.3, but less than 1.5(D) At least 1.5, but less than 1.7(E) At least 1.78. For a sample of dental claims 1210,,...,x x x , you are given:(i) 23860 and 4,574,802i i x x ==∑∑(ii) Claims are assumed to follow a lognormal distribution with parameters µ and σ.(iii)µ and σ are estimated using the method of moments.Calculate ∧ for the fitted distribution.(A) Less than 125(B) At least 125, but less than 175(C) At least 175, but less than 225(D) At least 225, but less than 275(E) At least 2759. You are given:(i)Y tij is the loss for the j th insured in the i th group in Year t . (ii)ti Y is the mean loss in the i th group in Year t . (iii)X j i j i ij =R S T0, if the th insured is in the first group (=1)1, if the th insured is in the second group (=2) (iv)21ij ij ij ij Y Y X δφθε=+++, where 1,2i = and 1,2,...,j n = (v)Y Y Y Y 2122111230374041====,,, (vi) ˆ0.75φ=Determine the least-squares estimate of θ.(A) 5.25(B) 5.50(C) 5.75(D) 6.00(E) 6.2510. Two independent samples are combined yielding the following ranks:Sample I: 1, 2, 3, 4, 7, 9, 13, 19, 20Sample II: 5, 6, 8, 10, 11, 12, 14, 15, 16, 17, 18You test the null hypothesis that the two samples are from the same continuous distribution.The variance of the rank sum statistic is:()112n m n m ++Using the classical approximation for the two-tailed rank sum test, determine the p -value.(A) 0.015(B) 0.021(C) 0.105(D) 0.210(E) 0.420(i) Claim counts follow a Poisson distribution with mean θ. (ii) Claim sizes follow an exponential distribution with mean 10θ. (iii) Claim counts and claim sizes are independent, given θ. (iv) The prior distribution has probability density function:b g=5, θ>1πθθCalculate Bühlmann’s k for aggregate losses.(A) Less than 1(B) At least 1, but less than 2(C) At least 2, but less than 3(D) At least 3, but less than 4(E) At least 4(i) A survival study uses a Cox proportional hazards model with covariates Z 1 and Z 2,each taking the value 0 or 1.(ii) The maximum partial likelihood estimate of the coefficient vector is:, .,.ββ12071020e j b g=(iii) The baseline survival function at time t 0 is estimated as .S t 0065b g =.Estimate S t 0b gfor a subject with covariate values 121Z Z ==.(A) 0.34(B) 0.49(C) 0.65(D) 0.74(E) 0.84(i) Z 1 and Z 2 are independent N(0,1) random variables.(ii) a , b , c , d , e , f are constants.(iii) Y a bZ cZ X d eZ f Z =++=++1212 andDetermine ()E Y X .(A) a(B) ()()a b c X d ++−(C) a be cf X d ++−b gb g(D) a be cf e f +++g d /22(E) a be cf e f X d +++−g d g/22(i) Losses on a company’s insurance policies follow a Pareto distribution with probabilitydensity function:()(),0f x x x θθθ=<<∞+(ii) For half of the company’s policies θ=1, while for the other half θ=3.For a randomly selected policy, losses in Year 1 were 5.Determine the posterior probability that losses for this policy in Year 2 will exceed 8.(A) 0.11(B) 0.15(C) 0.19(D) 0.21(E) 0.2715. You are given total claims for two policyholders:Year1 2 3 4PolicyholderX 730 800 650 700Y 655 650 625 750Using the nonparametric empirical Bayes method, determine the Bühlmann credibilitypremium for Policyholder Y.(A) 655(B) 670(C) 687(D) 703(E) 71916. A particular line of business has three types of claims. The historical probability and thenumber of claims for each type in the current year are:Type HistoricalProbabilityNumber of Claimsin Current YearA 0.2744 112B 0.3512 180C 0.3744 138You test the null hypothesis that the probability of each type of claim in the current year is the same as the historical probability.Calculate the chi-square goodness-of-fit test statistic.(A) Less than 9(B) At least 9, but less than 10(C) At least 10, but less than 11(D) At least 11, but less than 12(E) At least 1217. Which of the following is false?(A) If the characteristics of a stochastic process change over time, then the process isnonstationary.(B) Representing a nonstationary time series by a simple algebraic model is often difficult.(C) Differences of a homogeneous nonstationary time series will always be nonstationary.(D) If a time series is stationary, then its mean, variance and, for any lag k, covariancemust also be stationary.(E) If the autocorrelation function for a time series is zero (or close to zero) for all lagsk>0, then no model can provide useful minimum mean-square-error forecasts offuture values other than the mean.18. The information associated with the maximum likelihood estimator of a parameter θ is 4n,where n is the number of observations.Calculate the asymptotic variance of the maximum likelihood estimator of 2θ.(A) 12n(B) 1n(C) 4n(D) 8n(E) 16n19. You are given:(i) The probability that an insured will have at least one loss during any year is p.(ii) The prior distribution for p is uniform on []0,0.5.(iii) An insured is observed for 8 years and has at least one loss every year.Determine the posterior probability that the insured will have at least one loss during Year 9.(A) 0.450(B) 0.475(C) 0.500(D) 0.550(E) 0.62520. At the beginning of each of the past 5 years, an actuary has forecast the annual claims for agroup of insureds. The table below shows the forecasts (X) and the actual claims (Y). Atwo-variable linear regression model is used to analyze the data.t X t Y t1 475 2542 254 4633 463 5154 515 5675 567 605You are given:(i) The null hypothesis is0:0,1Hαβ==.(ii) The unrestricted model fit yields ESS = 69,843.Which of the following is true regarding the F test of the null hypothesis?(A) The null hypothesis is not rejected at the 0.05 significance level.(B) The null hypothesis is rejected at the 0.05 significance level, but not at the 0.01 level.(C) The numerator has 3 degrees of freedom.(D) The denominator has 2 degrees of freedom.(E) TheF statistic cannot be determined from the information given.21-22. Use the following information for questions 21 and 22.For a survival study with censored and truncated data, you are given:Time (t) Number at Riskat Time t Failures at Time t1 30 52 27 93 32 64 25 55 20 4 21. The probability of failing at or before Time 4, given survival past Time 1, is31q.Calculate Greenwood’s approximation of the variance of 31 q.(A) 0.0067(B) 0.0073(C) 0.0080(D) 0.0091(E) 0.010521-22. (Repeated for convenience) Use the following information for questions 21 and 22.For a survival study with censored and truncated data, you are given:Time (t) Number at Riskat Time t Failures at Time t1 30 52 27 93 32 64 25 55 20 4 22. Calculate the 95% log-transformed confidence interval for H3b g, based on the Nelson-Aalenestimate.(A) (0.30,0.89)(B) (0.31,1.54)(C) (0.39,0.99)(D) (0.44,1.07)(E) (0.56,0.79)(i) Two risks have the following severity distributions:Amount of Claim Probability of ClaimAmount for Risk 1Probability of ClaimAmount for Risk 2250 0.5 0.72,500 0.3 0.260,000 0.2 0.1(ii) Risk 1 is twice as likely to be observed as Risk 2.A claim of 250 is observed.Determine the Bühlmann credibility estimate of the second claim amount from the same risk.(A) Less than 10,200(B) At least 10,200, but less than 10,400(C) At least 10,400, but less than 10,600(D) At least 10,600, but less than 10,800(E) At least 10,800(i) A sample x x x 1210,,,… is drawn from a distribution with probability density function:1211exp()exp(), 0[]x x x θθσσ−+−<<∞(ii)θσ>(iii) x x i i ==∑∑15050002 andEstimate θ by matching the first two sample moments to the corresponding population quantities.(A) 9(B) 10(C) 15(D) 20(E) 2125. You are given the following time-series model:115.028.0−−−++=t t t t y y εεWhich of the following statements about this model is false?(A) 10.4ρ=(B) 1,2,3,4,....k k ρρ<=(C) The model is ARMA(1,1).(D) The model is stationary.(E) The mean, µ, is 2.26. You are given a sample of two values, 5 and 9.You estimate Var(X ) using the estimator g (X 1, X 2) = 21().2i X X −∑Determine the bootstrap approximation to the mean square error of g .(A) 1(B) 2(C) 4(D) 8(E) 1627. You are given:(i) The number of claims incurred in a month by any insured has a Poisson distributionwith mean λ.(ii) The claim frequencies of different insureds are independent.(iii) The prior distribution is gamma with probability density function:()()6100100120efλλλλ−=(iv) Month Number of Insureds NumberofClaims1 100 62 150 83 200 114 300 ?Determine the Bühlmann-Straub credibility estimate of the number of claims in Month 4.(A) 16.7(B) 16.9(C) 17.3(D) 17.6(E) 18.028. You fit a Pareto distribution to a sample of 200 claim amounts and use the likelihood ratio testto test the hypothesis that 1.5α= and 7.8θ=.You are given:(i) The maximum likelihood estimates are α= 1.4 and θ = 7.6.(ii) The natural logarithm of the likelihood function evaluated at the maximum likelihoodestimates is −817.92.(iii) ()ln 7.8607.64i x +=∑Determine the result of the test.(A) Reject at the 0.005 significance level.(B) Reject at the 0.010 significance level, but not at the 0.005 level.(C) Reject at the 0.025 significance level, but not at the 0.010 level.(D) Reject at the 0.050 significance level, but not at the 0.025 level.(E) Do not reject at the 0.050 significance level.29. You are given:(i) The model is Y X i i i =+βε, i = 1, 2, 3.(ii)i X i Var εi b g11 12 2 93 316 (iii)The ordinary least squares residuals are εβi i i Y X =−, i = 1, 2, 3.Determine E X X X ,,ε12123d i.(A) 1.0(B) 1.8(C) 2.7(D) 3.7(E) 7.630. For a sample of 15 losses, you are given:(i)Interval Observed Number ofLosses(0, 2] 5(2, 5] 5(5, ∞) 5 (ii) Losses follow the uniform distribution on 0,θb g.Estimate θ by minimizing the function()231j jjjE OO=−∑, where j E is the expected number oflosses in the j th interval andjO is the observed number of losses in the j th interval.(A) 6.0(B) 6.4(C) 6.8(D) 7.2(E) 7.631. You are given:(i) The probability that an insured will have exactly one claim is θ.(ii) The prior distribution of θ has probability density function:πθθθb g=<<3201,A randomly chosen insured is observed to have exactly one claim.Determine the posterior probability that θ is greater than 0.60.(A) 0.54(B) 0.58(C) 0.63(D) 0.67(E) 0.7232. The distribution of accidents for 84 randomly selected policies is as follows:Number of Accidents Number of Policies0 321 262 123 74 45 26 1Total 84 Which of the following models best represents these data?binomial(A) Negativeuniform(B) Discrete(C) Poisson(D) Binomial(E) Either Poisson or Binomial33. A time series yt follows an ARIMA(1,1,1) model with φ107=., θ103=−. and σε210=..Determine the variance of the forecast error two steps ahead.(A) 1(B)5(C) 8(D)10(E) 12(i) Low-hazard risks have an exponential claim size distribution with mean θ. (ii) Medium-hazard risks have an exponential claim size distribution with mean 2θ. (iii) High-hazard risks have an exponential claim size distribution with mean 3θ. (iv) No claims from low-hazard risks are observed.(v) Three claims from medium-hazard risks are observed, of sizes 1, 2 and 3. (vi) One claim from a high-hazard risk is observed, of size 15.Determine the maximum likelihood estimate of θ.(A) 1(B) 2(C) 3(D) 4(E) 5(i)partial X =pure premium calculated from partially credible data(ii)partial E X µ⎡⎤=⎣⎦ (iii) Fluctuations are limited to ±k µ of the mean with probability P(iv) Z = credibility factorWhich of the following is equal to P ?(A) partial Pr k X k µµµµ⎡⎤−≤≤+⎣⎦(B) partial Pr +Z k Z X Z k µµ⎡⎤−≤≤⎣⎦(C) partial Pr +Z Z X Z µµµµ⎡⎤−≤≤⎣⎦(D) ()partial Pr 111k Z X Z k µ⎡⎤−≤+−≤+⎣⎦(E) ()partial Pr 1k Z X Z k µµµµµ⎡⎤−≤+−≤+⎣⎦36. For the model 1223344i i i i i Y X X X ββββε=++++, you are given:(i) N = 15(ii)(iii) ESS =28282.Calculate the standard error of 32ˆˆββ−.(A) 6.4(B) 6.8(C) 7.1(D) 7.5(E) 7.837. You are given:Assume a uniform distribution of claim sizes within each interval.Estimate E X X 2150c h g −∧.(A)Less than 200(B)At least 200, but less than 300(C)At least 300, but less than 400(D)At least 400, but less than 500(E)At least 50038. Which of the following statements about moving average models is false?(A) Both unweighted and exponentially weighted moving average (EWMA) models canbe used to forecast future values of a time series.(B) Forecasts using unweighted moving average models are determined by applying equalweights to a specified number of past observations of the time series.(C) Forecasts using EWMA models may not be true averages because the weights appliedto the past observations do not necessarily sum to one.(D) Forecasts using both unweighted and EWMA models are adaptive because theyautomatically adjust themselves to the most recently available data.(E) Using an EWMA model, the two-period forecast is the same as the one-periodforecast.39. You are given:(i) Each risk has at most one claim each year.(ii)Type of Risk Prior Probability Annual Claim ProbabilityI 0.7 0.1II 0.2 0.2III 0.1 0.4 One randomly chosen risk has three claims during Years 1-6.Determine the posterior probability of a claim for this risk in Year 7.(A) 0.22(B) 0.28(C) 0.33(D) 0.40(E) 0.4640. You are given the following about 100 insurance policies in a study of time to policysurrender:(i) The study was designed in such a way that for every policy that was surrendered, ar, is always equal to 100.new policy was added, meaning that the risk set,j(ii) Policies are surrendered only at the end of a policy year.(iii) The number of policies surrendered at the end of each policy year was observed to be:1 at the end of the 1st policy year2 at the end of the 2nd policy year3 at the end of the 3rd policy yearn at the end of the n th policy year(iv) The Nelson-Aalen empirical estimate of the cumulative distribution function at time n, F, is 0.542.)(ˆnWhat is the value of n?(A) 8(B) 9(C) 10(D) 11(E) 12**END OF EXAMINATION**Course 4, Fall 2003PRELIMINARY ANSWER KEYQuestion # Answer Question # Answer 1 A21 A 2 E22 D 3 E23 D 4 B24 D 5 D25 E 6 A26 D 7 C27 B 8 D28 C 9 E29 B 10 D30 E 11 C31 E 12 A32 A 13 E33 B 14 D34 B 15 C35 E 16 B36 C 17 C37 C 18 B38 C 19 A39 B 20 A40 E。

Package‘sEparaTe’August18,2023Title Maximum Likelihood Estimation and Likelihood Ratio TestFunctions for Separable Variance-Covariance StructuresVersion0.3.2Maintainer Timothy Schwinghamer<***************************.CA>Description Maximum likelihood estimation of the parametersof matrix and3rd-order tensor normal distributions with unstructuredfactor variance covariance matrices,two procedures,and for unbiasedmodified likelihood ratio testing of simple and double separabilityfor variance-covariance structures,two procedures.References:Dutilleul P.(1999)<doi:10.1080/00949659908811970>,Manceur AM,Dutilleul P.(2013)<doi:10.1016/j.cam.2012.09.017>,and Manceur AM,DutilleulP.(2013)<doi:10.1016/j.spl.2012.10.020>.Depends R(>=4.3.0)License MIT+file LICENSEEncoding UTF-8LazyData trueRoxygenNote7.2.0NeedsCompilation noAuthor Ameur Manceur[aut],Timothy Schwinghamer[aut,cre],Pierre Dutilleul[aut,cph]Repository CRANDate/Publication2023-08-1807:50:02UTCR topics documented:data2d (2)data3d (2)lrt2d_svc (3)lrt3d_svc (5)mle2d_svc (7)12data3d mle3d_svc (8)sEparaTe (10)Index11 data2d Two dimensional data setDescriptionAn i.i.d.random sample of size7from a2x3matrix normal distribution,for a small numerical example of the use of the functions mle2d_svc and lrt2d_svc from the sEparaTe packageUsagedata2dFormatA frame(excluding the headings)with42lines of data and4variables:K an integer ranging from1to7,the size of an i.i.d.random sample from a2x3matrix normal distributionId1an integer ranging from1to2,the number of rows of the matrix normal distributionId2an integer ranging from1to3,the number of columns of the matrix normal distribution value2d the sample data for the observed variabledata3d Three dimensional data setDescriptionAn i.i.d.random sample of size13from a2x3x2tensor normal distribution,for a small numerical example of the use of the functions mle3d_svc and lrt3d_svc from the sEparaTe packageUsagedata3dFormatA frame(excluding the headings)with156lines of data and5variables:K an integer ranging from1to13,the size of an i.i.d.random sample from a2x3x2tensor matrix normal distributionId3an integer ranging from1to2,the number of rows of the3rd-order tensor normal distribution Id4an integer ranging from1to3,the number of columns of the3rd-order tensor normal distribu-tionId5an integer ranging from1to2,the number of edges of the3rd-order tensor normal distribution value3d the sample data for the observed variablelrt2d_svc Unbiased modified likelihood ratio test for simple separability of avariance-covariance matrix.DescriptionA likelihood ratio test(LRT)for simple separability of a variance-covariance matrix,modified tobe unbiased infinite samples.The modification is a penalty-based homothetic transformation of the LRT statistic.The penalty value is optimized for a given mean model,which is left unstruc-tured here.In the required function,the Id1and Id2variables correspond to the row and column subscripts,respectively;“value2d”refers to the observed variable.Usagelrt2d_svc(value2d,Id1,Id2,subject,data_2d,eps,maxiter,startmat,sign.level,n.simul)Argumentsvalue2d from the formula value2d~Id1+Id2Id1from the formula value2d~Id1+Id2Id2from the formula value2d~Id1+Id2subject the replicate,also called the subject or individual,thefirst column in the matrix (2d)datafiledata_2d the name of the matrix dataeps the threshold in the stopping criterion for the iterative mle algorithm(estimation) maxiter the maximum number of iterations for the mle algorithm(estimation)startmat the value of the second factor variance-covariance matrix used for initialization,i.e.,to start the mle algorithm(estimation)and obtain the initial estimate of thefirst factor variance-covariance matrixsign.level the significance level,or rejection rate in the testing of the null hypothesis of simple separability for a variance-covariance structure,when the unbiased mod-ified LRT is used,i.e.,the critical value in the chi-square test is derived bysimulations from the sampling distribution of the LRT statistic n.simul the number of simulations used to build the sampling distribution of the LRT statistic under the null hypothesis,using the same characteristics as the i.i.d.random sample from a matrix normal distributionOutput“Convergence”,TRUE or FALSE“chi.df”,the theoretical number of degrees of freedom of the asymptotic chi-square distribution that would apply to the unmodified LRT statistic for simple separability of a variance-covariance structure“Lambda”,the observed value of the unmodified LRT statistic“critical.value”,the critical value at the specified significance level for the chi-square distribution with“chi.df”degrees of freedom“mbda”will indicate whether or not the null hypothesis of separability was rejected, based on the theoretical LRT statistic“Simulation.critical.value”,the critical value at the specified significance level that is derived from the sampling distribution of the unbiased modified LRT statistic“mbda.simulation”,the decision(acceptance/rejection)regarding the null hypothesis of simple separability,made using the theoretical(biased unmodified)LRT“Penalty”,the optimized penalty value used in the homothetic transformation between the biased unmodified and unbiased modified LRT statistics“U1hat”,the estimated variance-covariance matrix for the rows“Standardized_U1hat”,the standardized estimated variance-covariance matrix for the rows;the standardization is performed by dividing each entry of U1hat by entry(1,1)of U1hat“U2hat”,the estimated variance-covariance matrix for the columns“Standardized_U2hat”,the standardized estimated variance-covariance matrix for the columns;the standardization is performed by multiplying each entry of U2hat by entry(1,1)of U1hat“Shat”,the sample variance-covariance matrix computed from the vectorized data matrices ReferencesManceur AM,Dutilleul P.2013.Unbiased modified likelihood ratio tests for simple and double separability of a variance-covariance structure.Statistics and Probability Letters83:631-636.Examplesoutput<-lrt2d_svc(data2d$value2d,data2d$Id1,data2d$Id2,data2d$K,data_2d=data2d,n.simul=100)outputlrt3d_svc An unbiased modified likelihood ratio test for double separability of avariance-covariance structure.DescriptionA likelihood ratio test(LRT)for double separability of a variance-covariance structure,modified tobe unbiased infinite samples.The modification is a penalty-based homothetic transformation of the LRT statistic.The penalty value is optimized for a given mean model,which is left unstructured here.In the required function,the Id3,Id4and Id5variables correspond to the row,column and edge subscripts,respectively;“value3d”refers to the observed variable.Usagelrt3d_svc(value3d,Id3,Id4,Id5,subject,data_3d,eps,maxiter,startmatU2,startmatU3,sign.level,n.simul)Argumentsvalue3d from the formula value3d~Id3+Id4+Id5Id3from the formula value3d~Id3+Id4+Id5Id4from the formula value3d~Id3+Id4+Id5Id5from the formula value3d~Id3+Id4+Id5subject the replicate,also called individualdata_3d the name of the tensor dataeps the threshold in the stopping criterion for the iterative mle algorithm(estimation) maxiter the maximum number of iterations for the mle algorithm(estimation)startmatU2the value of the second factor variance-covariance matrix used for initialization startmatU3the value of the third factor variance-covariance matrix used for initialization,i.e.,startmatU3together with startmatU2are used to start the mle algorithm(estimation)and obtain the initial estimate of thefirst factor variance-covariancematrix U1sign.level the significance level,or rejection rate in the testing of the null hypothesis of simple separability for a variance-covariance structure,when the unbiased mod-ified LRT is used,i.e.,the critical value in the chi-square test is derived bysimulations from the sampling distribution of the LRT statistic n.simul the number of simulations used to build the sampling distribution of the LRT statistic under the null hypothesis,using the same characteristics as the i.i.d.random sample from a tensor normal distributionOutput“Convergence”,TRUE or FALSE“chi.df”,the theoretical number of degrees of freedom of the asymptotic chi-square distribution that would apply to the unmodified LRT statistic for double separability of a variance-covariance structure“Lambda”,the observed value of the unmodified LRT statistic“critical.value”,the critical value at the specified significance level for the chi-square distribution with“chi.df”degrees of freedom“mbda”,the decision(acceptance/rejection)regarding the null hypothesis of double sep-arability,made using the theoretical(biased unmodified)LRT“Simulation.critical.value”,the critical value at the specified significance level that is derived from the sampling distribution of the unbiased modified LRT statistic“mbda.simulation”,the decision(acceptance/rejection)regarding the null hypothesis of double separability,made using the unbiased modified LRT“Penalty”,the optimized penalty value used in the homothetic transformation between the biased unmodified and unbiased modified LRT statistics“U1hat”,the estimated variance-covariance matrix for the rows“Standardized_U1hat”,the standardized estimated variance-covariance matrix for the rows;the standardization is performed by dividing each entry of U1hat by entry(1,1)of U1hat“U2hat”,the estimated variance-covariance matrix for the columns“Standardized_U2hat”,the standardized estimated variance-covariance matrix for the columns;the standardization is performed by multiplying each entry of U2hat by entry(1,1)of U1hat“U3hat”,the estimated variance-covariance matrix for the edges“Shat”,the sample variance-covariance matrix computed from the vectorized data tensorsReferencesManceur AM,Dutilleul P.2013.Unbiased modified likelihood ratio tests for simple and double separability of a variance-covariance structure.Statistics and Probability Letters83:631-636.Examplesoutput<-lrt3d_svc(data3d$value3d,data3d$Id3,data3d$Id4,data3d$Id5,data3d$K,data_3d=data3d,n.simul=100)outputmle2d_svc Maximum likelihood estimation of the parameters of a matrix normaldistributionDescriptionMaximum likelihood estimation for the parameters of a matrix normal distribution X,which is char-acterized by a simply separable variance-covariance structure.In the general case,which is the case considered here,two unstructured factor variance-covariance matrices determine the covariability of random matrix entries,depending on the row(one factor matrix)and the column(the other factor matrix)where two X-entries are.In the required function,the Id1and Id2variables correspond to the row and column subscripts,respectively;“value2d”indicates the observed variable.Usagemle2d_svc(value2d,Id1,Id2,subject,data_2d,eps,maxiter,startmat)Argumentsvalue2d from the formula value2d~Id1+Id2Id1from the formula value2d~Id1+Id2Id2from the formula value2d~Id1+Id2subject the replicate,also called individualdata_2d the name of the matrix dataeps the threshold in the stopping criterion for the iterative mle algorithmmaxiter the maximum number of iterations for the iterative mle algorithmstartmat the value of the second factor variance-covariance matrix used for initializa-tion,i.e.,to start the algorithm and obtain the initial estimate of thefirst factorvariance-covariance matrixOutput“Convergence”,TRUE or FALSE“Iter”,will indicate the number of iterations needed for the mle algorithm to converge“Xmeanhat”,the estimated mean matrix(i.e.,the sample mean)“First”,the row subscript,or the second column in the datafile“U1hat”,the estimated variance-covariance matrix for the rows“Standardized.U1hat”,the standardized estimated variance-covariance matrix for the rows;the stan-dardization is performed by dividing each entry of U1hat by entry(1,1)of U1hat“Second”,the column subscript,or the third column in the datafile“U2hat”,the estimated variance-covariance matrix for the columns“Standardized.U2hat”,the standardized estimated variance-covariance matrix for the columns;the standardization is performed by multiplying each entry of U2hat by entry(1,1)of U1hat“Shat”,is the sample variance-covariance matrix computed from of the vectorized data matrices ReferencesDutilleul P.1990.Apport en analyse spectrale d’un periodogramme modifie et modelisation des series chronologiques avec repetitions en vue de leur comparaison en frequence.D.Sc.Dissertation, Universite catholique de Louvain,Departement de mathematique.Dutilleul P.1999.The mle algorithm for the matrix normal distribution.Journal of Statistical Computation and Simulation64:105-123.Examplesoutput<-mle2d_svc(data2d$value2d,data2d$Id1,data2d$Id2,data2d$K,data_2d=data2d) outputmle3d_svc Maximum likelihood estimation of the parameters of a3rd-order ten-sor normal distributionDescriptionMaximum likelihood estimation for the parameters of a3rd-order tensor normal distribution X, which is characterized by a doubly separable variance-covariance structure.In the general case, which is the case considered here,three unstructured factor variance-covariance matrices determine the covariability of random tensor entries,depending on the row(one factor matrix),the column (another factor matrix)and the edge(remaining factor matrix)where two X-entries are.In the required function,the Id3,Id4and Id5variables correspond to the row,column and edge subscripts, respectively;“value3d”indicates the observed variable.Usagemle3d_svc(value3d,Id3,Id4,Id5,subject,data_3d,eps,maxiter,startmatU2,startmatU3)Argumentsvalue3d from the formula value3d~Id3+Id4+Id5Id3from the formula value3d~Id3+Id4+Id5Id4from the formula value3d~Id3+Id4+Id5Id5from the formula value3d~Id3+Id4+Id5subject the replicate,also called individualdata_3d the name of the tensor dataeps the threshold in the stopping criterion for the iterative mle algorithmmaxiter the maximum number of iterations for the iterative mle algorithmstartmatU2the value of the second factor variance covariance matrix used for initialization startmatU3the value of the third factor variance covariance matrix used for initialization,i.e.,startmatU3together with startmatU2are used to start the algorithm andobtain the initial estimate of thefirst factor variance covariance matrix U1Output“Convergence”,TRUE or FALSE“Iter”,the number of iterations needed for the mle algorithm to converge“Xmeanhat”,the estimated mean tensor(i.e.,the sample mean)“First”,the row subscript,or the second column in the datafile“U1hat”,the estimated variance-covariance matrix for the rows“Standardized.U1hat”,the standardized estimated variance-covariance matrix for the rows;the stan-dardization is performed by dividing each entry of U1hat by entry(1,1)of U1hat“Second”,the column subscript,or the third column in the datafile“U2hat”,the estimated variance-covariance matrix for the columns“Standardized.U2hat”,the standardized estimated variance-covariance matrix for the columns;the standardization is performed by multiplying each entry of U2hat by entry(1,1)of U1hat“Third”,the edge subscript,or the fourth column in the datafile“U3hat”,the estimated variance-covariance matrix for the edges“Shat”,the sample variance-covariance matrix computed from the vectorized data tensorsReferenceManceur AM,Dutilleul P.2013.Maximum likelihood estimation for the tensor normal distribution: Algorithm,minimum sample size,and empirical bias and dispersion.Journal of Computational and Applied Mathematics239:37-49.10sEparaTeExamplesoutput<-mle3d_svc(data3d$value3d,data3d$Id3,data3d$Id4,data3d$Id5,data3d$K,data_3d=data3d) outputsEparaTe MLE and LRT functions for separable variance-covariance structuresDescriptionA package for maximum likelihood estimation(MLE)of the parameters of matrix and3rd-ordertensor normal distributions with unstructured factor variance-covariance matrices(two procedures),and for unbiased modified likelihood ratio testing(LRT)of simple and double separability forvariance-covariance structures(two procedures).Functionsmle2d_svc,for maximum likelihood estimation of the parameters of a matrix normal distributionmle3d_svc,for maximum likelihood estimation of the parameters of a3rd-order tensor normaldistributionlrt2d_svc,for the unbiased modified likelihood ratio test of simple separability for a variance-covariance structurelrt3d_svc,for the unbiased modified likelihood ratio test of double separability for a variance-covariance structureDatadata2d,a two-dimensional data setdata3d,a three-dimensional data setReferencesDutilleul P.1999.The mle algorithm for the matrix normal distribution.Journal of StatisticalComputation and Simulation64:105-123.Manceur AM,Dutilleul P.2013.Maximum likelihood estimation for the tensor normal distribution:Algorithm,minimum sample size,and empirical bias and dispersion.Journal of Computational andApplied Mathematics239:37-49.Manceur AM,Dutilleul P.2013.Unbiased modified likelihood ratio tests for simple and doubleseparability of a variance covariance structure.Statistics and Probability Letters83:631-636.Index∗datasetsdata2d,2data3d,2data2d,2data3d,2lrt2d_svc,3lrt3d_svc,5mle2d_svc,7mle3d_svc,8sEparaTe,1011。

稳健最大似然(mlr)估计方法The robust maximum likelihood estimation method (mlr) is a statistical technique used to estimate the parameters of a model while accounting for outliers or extreme values in the data. This method is particularly useful in situations where traditional maximum likelihood estimation may be prone to bias or inefficiency due to the presence of outliers. By incorporating robust estimators, such as the Huber or Tukey biweight functions, mlr helps to reduce the impact of outliers on parameter estimates, resulting in more reliable and accurate model fitting.稳健最大似然估计方法（mlr）是一种统计技术，用于在考虑数据中的异常值或极端值的情况下估计模型的参数。

这种方法特别适用于传统最大似然估计可能因为异常值存在而产生偏差或低效的情况。

通过结合稳健估计器，如Huber或Tukey双加权函数，mlr有助于减少异常值对参数估计的影响，从而实现更可靠和准确的模型拟合。

One of the key advantages of the robust maximum likelihood estimation method is its ability to provide consistent estimates even in the presence of outliers. Traditional maximum likelihoodestimation methods are sensitive to outliers, which can greatly impact the accuracy of parameter estimates and the overall fit of the model. By using robust estimators, mlr is able to downweight the influence of outliers while still maintaining the consistency of parameter estimates, resulting in more robust and reliable inference.稳健最大似然估计方法的关键优势之一是它能够在存在异常值的情况下提供一致的估计。

最⼤似然估计（Maximumlikelihoodestimation）最⼤似然估计提供了⼀种给定观察数据来评估模型参数的⽅法，即：“模型已定，参数未知”。

简单⽽⾔，假设我们要统计全国⼈⼝的⾝⾼，⾸先假设这个⾝⾼服从服从正态分布，但是该分布的均值与⽅差未知。

我们没有⼈⼒与物⼒去统计全国每个⼈的⾝⾼，但是可以通过采样，获取部分⼈的⾝⾼，然后通过最⼤似然估计来获取上述假设中的正态分布的均值与⽅差。

最⼤似然估计中采样需满⾜⼀个很重要的假设，就是所有的采样都是独⽴同分布的。

下⾯我们具体描述⼀下最⼤似然估计：⾸先，假设为独⽴同分布的采样，θ为模型参数,f为我们所使⽤的模型，遵循我们上述的独⽴同分布假设。

参数为θ的模型f产⽣上述采样可表⽰为回到上⾯的“模型已定，参数未知”的说法，此时，我们已知的为，未知为θ，故似然定义为: 在实际应⽤中常⽤的是两边取对数，得到公式如下： 其中称为对数似然，⽽称为平均对数似然。

⽽我们平时所称的最⼤似然为最⼤的对数平均似然，即：举个别⼈博客中的例⼦，假如有⼀个罐⼦，⾥⾯有⿊⽩两种颜⾊的球，数⽬多少不知，两种颜⾊的⽐例也不知。

我们想知道罐中⽩球和⿊球的⽐例，但我们不能把罐中的球全部拿出来数。

现在我们可以每次任意从已经摇匀的罐中拿⼀个球出来，记录球的颜⾊，然后把拿出来的球再放回罐中。

这个过程可以重复，我们可以⽤记录的球的颜⾊来估计罐中⿊⽩球的⽐例。

假如在前⾯的⼀百次重复记录中，有七⼗次是⽩球，请问罐中⽩球所占的⽐例最有可能是多少？很多⼈马上就有答案了：70%。

⽽其后的理论⽀撑是什么呢？我们假设罐中⽩球的⽐例是p，那么⿊球的⽐例就是1-p。

因为每抽⼀个球出来，在记录颜⾊之后，我们把抽出的球放回了罐中并摇匀，所以每次抽出来的球的颜⾊服从同⼀独⽴分布。

这⾥我们把⼀次抽出来球的颜⾊称为⼀次抽样。

题⽬中在⼀百次抽样中，七⼗次是⽩球的概率是P(Data | M)，这⾥Data是所有的数据，M是所给出的模型，表⽰每次抽出来的球是⽩⾊的概率为p。

简述极大似然估计法的原理The principle of maximum likelihood estimation (MLE) is a widely used method in statistics and machine learning for estimating the parameters of a statistical model. In essence, MLE attempts to find the values of the parameters that maximize the likelihood of observing the data given a specific model. In other words, MLE seeks the most plausible values of the parameters that make the observed data the most likely.极大似然估计法的原理是统计学和机器学习中广泛使用的一种方法，用于估计统计模型的参数。

在本质上，极大似然估计试图找到最大化观察到的数据在特定模型下的可能性的参数值。

换句话说，极大似然估计寻求最合理的参数值，使观察到的数据最有可能。

To illustrate the concept of maximum likelihood estimation, let's consider a simple example of flipping a coin. Suppose we have a fair coin, and we want to estimate the probability of obtaining heads. We can model this situation using a Bernoulli distribution, where the parameter of interest is the probability of success (, getting heads). By observing multiple coin flips, we can calculate the likelihood ofthe observed outcomes under different values of the parameter and choose the value that maximizes this likelihood.为了说明极大似然估计的概念，让我们考虑一个简单的抛硬币的例子。