21 Testing Problems Testing is certainly another fundamental problem of inference. It is interestingly different from estimation. The accuracy measures are fundamentally different and the appropriate asymptotic theory is also different. Much of the theory of testing has revolved somehow or other on the Neyman-Pearson lemma, which has led to a lot of ancillary developments in other problems in mathematical statistics. Testing has ledto the useful idea of local alternatives, and like maximum likelihood in estimation, likelihood ratio, Wald, and Rao score tests have earned the status of default methods, with a neat and quite unified asymptotic theory. Because of all these reasons, a treatment of testing is essential. We discuss the asymptotic theory of likelihood ratio, Wald, and Rao score tests in this chapter. Principal references for this chapter are Bickel and Doksum (2001), Ferguson (1996), and Sen and Singer (1993). Many other specific references are given in the sections. 21.1 Likelihood Ratio Tests The likelihood ratio test is a general omnibus test applicable, in principle, in most (n) finite dimensional parametric problems. Thus, let X =(X1,...,Xn)bethe n (n) observed data with joint distribution Pθ ¿ µn, θ ∈ Θ, and density fθ(x )= n dPθ /dµn. Here, µn is some appropriate σ-finite measure on Xn, which we assume to be a subset of an Euclidean space. For testing H0 : θ ∈ Θ0 vs. H1 : θ ∈ Θ − Θ0, the likelihood ratio test (LRT) rejects H0 for small values of (n) supθ∈Θ0 fθ(x ) n Λ = (n) supθ∈Θ fθ(x ) The motivation for Λn comes from two sources: (a) The case where H0, H1 are each simple, a most powerful (MP) test is found from Λn by the Neyman-Pearson lemma. (b) The intuitive explanation that for small values of Λn we can better match the observed data with some value of θ outside of Θ0. LRTs are useful because they are omnibus tests and because otherwise optimal tests for a given sample size n are generally hard to find outside of the exponential family. However the LRT is not a universal test. There are important examples where the LRT simply cannot be used, because the null distribution of the LRT test statistic depends on nuisance parameters. Also, the exact distribution of the LRT 319 statistic is very difficult or impossible to find in many problems. Thus, asymptotics become really important. But the asymptotics of the LRT may be nonstandard under nonstandard conditions. Although we only discuss the case of a parametric model with a fixed finite di- mensional parameter space, LRTs and their useful modifications have been studied when the number of parameters grows with the sample size n, and in nonparamet- ric problems. See, e.g., Portnoy(1988),Fan,Hung and Wong(2000), and Fan and Zhang(2001). We start with a series of examples, which illustrate various important aspects of the likelihood ratio method. 21.2 Examples iid 2 Example 21.1. Let X1,X2,... ,Xn ∼ N(µ, σ ) and consider testing H0 : µ =0vs.H1 : µ =06 . Let θ =(µ, σ2). Then, ¡ P ¢ µP ¶n/ n 1 2 2 − i − − 2 supθ∈Θ0 (1/σ )exp 2σ2 i(X µ) i(Xi Xn) n ¡ P ¢ P Λ = n 1 = 2 − i − 2 supθ∈Θ(1/σ )exp 2σ2 i(X µ) i Xi by an elementary calculation of mles of θ under H0 and in the general parameter 2 space. By another elementary calculation, Λn <cis seen to be equivalent to tn >k where √ nXn tn = q P 1 2 i − n n−1 i(X X ) is the t-statistic. In other words, the t-test is the LRT. Also, observe that, 2 2 nXn tn = P 1 2 i − n n−1 i(X X ) P P 2 2 X − (Xi − Xn) = i i P i 1 − 2 n− i(Xi Xn) 1 P − 2 (n 1) i Xi − − = P 2 (n 1) i(Xi − Xn) −2/n =(n − 1)Λn − (n − 1) 320 This implies, µ ¶n/ n − 1 2 n Λ = 2 tn + n − 1 n n − 1 ⇒ log Λn = log 2 t2 + n − 1 µ n ¶ 2 tn ⇒−2logΛn = n log 1+ n − 1 µ µ ¶¶ 2 2 tn tn = n + op n − 1 n − 1 −→L 2 χ1 L under H0 since tn −→ N(0, 1) under H0. Example 21.2. Consider a multinomial distribution MN(n, p1, ··· ,pk). Consider testing H0 : p1 = p2 = ···= pk vs. H1 : H0 is not true. Let n1, ··· ,nk denote the observed cell frequencies. Then by an elementary calcu- lation, µ ¶ ni k n Λn =Πi=1 kni k k X ⇒−log Λn = n(log )+ ni log ni n i=1 The exact distribution of this is a messy discrete object and so asymptotics will be useful. We illustrate the asymptotic distribution for k =2.Inthiscase, − log Λn = n log 2 − n log n + n1 log n1 +(n − n1)log(n − n1) p Let Zn =(n1 − n/2)/ n/4. Then, − log Λn = n log 2 − n log n √ √ √ √ n + nZn n + nZn n − nZn n − nZn + log + log 2 √ 2 2 √ 2 n + nZn √ n − nZn √ = −n log n + log(n + nZn)+ log(n − nZn) √ 2 √ 2 n + nZn Zn n − nZn Zn = log(1 + √ )+ log(1 − √ ) 2 n 2 n √ µ µ ¶¶ 2 2 n + nZn Zn Zn Zn = √ − + op 2 n 2n n 321 √ µ µ ¶¶ 2 2 n − nZn Zn Zn Zn + −√ − + op 2 n 2n n 2 Zn = + op(1). 2 L L − −→ 2 −→ Hence 2logΛn χ1 under H0 as Zn N(0, 1) under H0. Remark: The popular test in this problem is the Pearson chi-square test which rejects H0 for large values of P k − 2 i=1(ni n/k) n/k 2 Interestingly this test statistic too has an asymptotic χ1 distribution as does the LRT statistic. Example 21.3. This example shows that the chi-square asymptotics of the LRT statistic fail when parameters have constraints or somehow there is a boundary phe- nomenon. iid Let X1,X2,... ,Xn ∼ N2(θ, I) where the parameter space is restricted to Θ = {θ = (θ1,θ2):θ1 ≥ 0,θ2 ≥ 0}. Consider testing H0 : θ =0vs. H1 : H0 is not true. T We would write the n realizations as xi =(x1i,x2i) and the sample average would be denoted by x =(x1, x2). The mle of θ is given by, à ! x ∨ 0 θˆ = 1 x2 ∨ 0 Therefore by an elementary calculation, P P − 2 − 2 exp( i x1i/2 i x2i/2) n P P Λ = 2 2 exp(− i(x1i − x1 ∨ 0) /2 − i(x2i − x2 ∨ 0) /2) Case 1: x1 ≤ 0, x2 ≤ 0. In this case Λn =1and−2logΛn =0. ≤ − 2 Case 2: x1 > 0, x2 0. In this case 2logΛn = nx1. ≤ − 2 Case 3: x1 0, x2 > 0. In this case 2logΛn = nx2. 322 − 2 2 Case 4: x1 > 0, x2 > 0. In this case 2logΛn = nx1 + nx2. Now, under H0 each of the above four cases has a probability 1/4 of occurrence. Therefore, under H0, L 1 1 2 1 2 −2logΛn −→ δ{ } + χ + χ , 4 0 2 1 4 2 in the sense of the mixture distribution being its weak limit. Example 21.4. In bio-equivalence trials, a brand name drug and a generic are compared with regard to some important clinical variable, such as average drug concentration in blood over a 24 hour time period. By testing for a difference, the problem is reduced to a single variable, often assumed to be normal. Formally, one iid 2 has X1,X2,... ,Xn ∼ N(µ, σ ) and the bio-equivalence hypothesis is: H0 : |µ|≤² for some specified ²>0. Inference is always significantly harder if known constraints on the parameters are enforced. Casella and Strawderman(1980) is a standard introduction to the normal mean problem with restrictions on the mean; Robertson,Wright and Dykstra (1988) is an almost encyclopedic exposition. Coming back to our example, to derive the LRT, we need the restricted mle and the non-restricted mle. We will assume here that σ2 is known. Then the MLE under H0 is ¯ ¯ µˆH0 = X if |X|≤² = ²sign(X¯)if|X¯| >². Consequently, the LRT statistic Λn satisfies, −2logΛn =0 if |X¯|≤² n = (X¯ − ²)2 if X>²¯ σ2 n = (X¯ + ²)2 if X<¯ −². σ2 Take a fixed µ. Then, µ√ ¶ µ√ ¶ n(² − µ) n(² + µ) Pµ(|X¯|≤²)=Φ +Φ − 1 σ σ µ √ ¶ n(² + µ) Pµ(X<¯ −²)=Φ − σ 323 and µ√ ¶ n(² − µ) Pµ(X>²¯ )=1− Φ σ From these expressions we get: ¯ L Case 1: If −²<µ<²then Pµ(|X|≤²) → 1andso−2logΛn −→ δ{0}. Case 2.a: If µ = −² then each of Pµ(|X¯|≤²)andPµ(X<¯ −²) converges to 1/2. In L 1 1 2 − n −→ this case, 2logΛ 2 δ{0} + 2 χ1, i.e., the mixture of a point mass and a chisquare distribution. L 1 1 2 − n −→ Case 2.b: Similarly if µ = ² then again, 2logΛ 2 δ{0} + 2 χ1. Case 3: If µ>²then Pµ(|X¯|≤²) → 0andPµ(X>²¯ ) → 1. In this case, L 2 2 2 −2logΛn −→ NCχ (1, (µ − ²) /σ )), a noncentral chisquare distribution. L 2 2 2 Case 4: Likewise, if µ<−² then −2logΛn −→ NCχ (1, (µ + ²) /σ )). So, unfortunately, even the null asymptotic distribution of − log Λn depends on the exact value of µ. Example 21.5. Consider the Behrens-Fisher problem with iid ∼ 2 X1,X2,... ,Xm N(µ1,σ1) iid ∼ 2 Y1,Y2,... ,Yn N(µ2,σ2) and all m + n observations are independent. We want to test H0 : µ1 = µ2 vs. H1 : µ1 =6 µ2.