| Mixed linear models arc popular in biology,medicine,economy,finance,environmentalism, design of industry. So in resent 30 years, many statisticians are interest in the statistical inference for linear mixed model. This paper is an review on the inference for mixed linear model, it is organized as follows: In chapter 1,we introduce the background of the mixed linear model; In chapter 2, the estimation of fixed domino offect is introduced; In chapter 3, we introduced some way to estimate the variance components,which are analysis of variance estimate,maximum likelihood estimate,restricted maximum likelihood estimate,minimum norm quadratic unbised estimate;In chapter 4, we introduce the test of variance components; In chapter 5, the method of spetral decomposition estimate is introduced, we analysis the property of spetral decomposition estimate and the condition for which minimum norm quadratic unbised estimate,variance estimate and spetral decomposition estimate are equivalency, meanwhile, we obtain the reveal equation for maximum likelihood estimate and restricted maximum likelihood estimate. The general form of linear mixed iswhere y is n×1 vector,X is the known design matrix,βis the fixed domino offect, Ui is known design matrix,Eξi=0, i=1,2,…,k.Theorem 1. For mixed linear model(l), suppose the combine distribution of e,ξis symmetry with origin,(?)2=(?)2(y) is a estimate ofσ2 which is a idol function of y, then for any estimatable function c'βand E(c'(?)((?)2)) <∞,then E(c'(?)((?)2))=c'β.The method of analysis of variance is as follows:Step 1:For a variance components model, we first take random effect as fixed effect,then we compute the square of the effect for model. Step 2: Then we compute the mean of the square, (in this time ,the random effect is viewed as fixed), they are linear function of the variance components.Step 3: Let the square equal to their mean, we obtain a linear equation of the variance component, which we can get a estimation from it.The maximum likelihood function is:Fan binghi obtain some simple equation for the model by QR decompose,now we will introduce the basic results:Step1:We can ontain a linear equation for the parameters by ANOVA estimation methods,SSe is the square of irregular.Step2:we choose a proper correctitude matrix Q,such thatWe let(?)i=Q'iy,i=1,…,k.z2=Q'k+1y,then where Rijis defined in(3.20),e(?)i=Q'iε,i=1,…,k,e2=Qk+1ε,then z'(?)iz(?)i= SSξi,z'2z2=SSe.Step3:we let z'(?)iz(?)i=E(z'(?)iz(?)i),z'2z2=E(z'2z2),thusTheorem 2. For function c'β,ξi-N(0,σi2Iqi),then(1)c'β*(i)-N(c'β,λic'(X'MiX)-),c'β*(i) andc'β*(j),(i≠j)are indenpdent.(2) Letri= rank(Mi)=rank(MiX),thenriλi*-Xri2,i=1,…,k.(3)λi*and c'β*(i) are indepdent.The following theorem give some easy conditions under which SDE ,ANOVAE and MINQUE are the same.Theorem 3. Under the condition of 4.6, then MINQUE,SDE and ANOVAE ofσ12,…,σk2 are the same.Theorem 4. Under the condition of 4.6,then the likelihood equation ofβandσ2 arc idol.The mixed linear model with two variance is popular in biology,medicine,economy etc.so this model is important in linear mixed model, we consider the following model where y is n×1 data, X and Z is n x p and n×q designed matrix,βis the fixed responds,u is q×1 random vector,it distribution is N(0,σu2),εis n×1 vector,its distribution is N(0,σ2In),εand u is independent.Theorem 5. For mixed model(5),supposePXZZ' is symmetry ,ang exitsα> 0,such that ZZ'=αPZ,thenc'(?),(?)2,(?)2 are the UMVUE ofc'β,σu2,σε2.Under the condition of Theorem 7, we can give the exactitude probability that (?)u2 is negative, which is as follows:Theorem 6. Under the condition of Theorem 7,whereτ=σu2/σε2,r1= rank(X : Z)-rank(X),r2=n-rank(X:Z),Fm,nis F distribution.Theorem 7. For model(5),suppose M(X) (?) M(Z)and existα>0,such that ZZ'=αPz,then the 1-αconfidence interval for c'βiswhere (?)=(X'X)-X'y,k=rank(Z)-rank(X).Now we consider the hypothesis test problem:we consider the following statistics:Theorem 8. Under the condition of Theorem 7, is the UMPUT of (6),the power function isTheorem 9 Letτ1 andτ2 is defined in (5.60)(5.61,let r*= ((2r0+2r1)/(r0r1 + 2r0))1/2,then(1) forr1=1or2,thenτ,(?)ε2(0, r)=Ois a nonnegative estimation.(2) for r1<2,r0<2,then for anyc∈(0,1],τ∈[τ1,0],(?)ε2(c,τ)is a nonnegative estimation..(3) for r1 > 2,r0≤2,then for any c∈[(?),1],[τ1,0],(?)ε2(c,r)is a nonnegative estimation..forr1>2,r0>2,andr0(r1-2)2≤2(r0+r1)(r1+2),then for anyc∈[(?),r*],τ∈[τ1,0],(?)ε2(c,τ)is a nonnegative estimation..
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