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LIKELIHOOD RATIO TESTS

Module by: Ewa Paszek

Summary: This course is a short series of lectures on Introductory Statistics. Topics covered are listed in the Table of Contents. The notes were prepared by Ewa Paszek and Marek Kimmel. The development of this course has been supported by NSF 0203396 grant.

LIKELIHOOD RATIO TESTS

In this paragraph let consider a general test-construction method that is applicable when both the null and alternative hypotheses, say H 0 H 0 and H 1 H 1 are composite.

Let continue to assume that the function form of the p.d.f. is known but that it depends on an unknown parameter or unknown parameters. That is, we assume that the p.d.f. of X is f( x;θ ) f( x;θ ) , where θ θ represents one or more unknown parameters. We let Ω Ω denote the total parameter space, that is, the set of all possible values of the parameter θ θ given by either H 0 H 0 or H 1 H 1 . These hypotheses will be stated as follows: H 0 :θ∈ω, H 1 :θ∈ω', H 0 :θ∈ω, H 1 :θ∈ω',

where ω ω is a subset of Ω Ω and ω' ω' is the complement of ω ω with respect to Ω Ω . The test will be constructed by using a ratio of likelihood functions that have been maximized in ω ω and Ω Ω , respectively. Ina sense, this is a natural generalization of the ratio appearing in the Neyman-Pearson lemma when the two hypotheses were simple.

Definition 1:
The likelihood ratio is the quotient λ= L( ω ^ ) L( Ω ^ ) , λ= L( ω ^ ) L( Ω ^ ) , where L( ω ^ ) L( ω ^ ) is the maximum of the likelihood function with respect to θ θ when θ∈ω θ∈ω and L( Ω ^ ) L( Ω ^ ) is the maximum of the likelihood function with respect to θ θ when θ∈Ω θ∈Ω .

Because λ λ is the quotient of nonnegative functions, λ≥0 λ≥0 . In addition, since ω⊂Ω ω⊂Ω we have that L( ω ^ )≤L( Ω ^ ) L( ω ^ )≤L( Ω ^ ) and hence λ≤1 λ≤1 . Thus 0≤λ≤1 0≤λ≤1 . If the maximum of L in ω ω is much smaller than in Ω Ω , it would seem that the data x 1 , x 2 ,..., x n x 1 , x 2 ,..., x n do not support the hypothesis H 0 :θ∈ω H 0 :θ∈ω . That is a small value of the ratio λ= L( ω ^ ) L( Ω ^ ) , λ= L( ω ^ ) L( Ω ^ ) , would lead to the rejection of H 0 H 0 . On the other hand, a value of the ratio λ λ that is close to one would support the null hypothesis H 0 H 0 . This leads to the following definition 2.

Definition 2:
To test H 0 :θ∈ω H 0 :θ∈ω against H 1 :θ∈ω' H 1 :θ∈ω' , the critical region for the likelihood ratio test is the set of points in the sample space for which λ= L( ω ^ ) L( Ω ^ ) ≤k, λ= L( ω ^ ) L( Ω ^ ) ≤k, where 0<k<1 0<k<1 and k is selected so that the test has a desired significance level α α .

The following example illustrates the definition 2.

Example 1

Assume that the weight X in ounces of a ’10-pound’ bag of sugar is N( μ,5 ) N( μ,5 ) . We shall test the hypothesis H 0 :μ=162 H 0 :μ=162 against the alternative hypothesis H 1 :μ≠162 H 1 :μ≠162 . Thus Ω={ μ:−∞<μ<∞ } Ω={ μ:−∞<μ<∞ } and ω={ 162 } ω={ 162 } . To find the likelihood ratio, we need L( ω ^ ) L( ω ^ ) and L( Ω ^ ) L( Ω ^ ) . When H 0 H 0 is true, μ μ can take on only one value, namely μ μ =162. Thus L( ω ^ )=L( 162 ) L( ω ^ )=L( 162 ) . To find L( Ω ^ ) L( Ω ^ ) , we must find the value of μ μ that minimizes L( μ ) L( μ ) . We recall that μ ^ = x ¯ μ ^ = x ¯ is the maximum likelihood estimate of μ μ . Thus L( Ω ^ )=L( x ¯ ) L( Ω ^ )=L( x ¯ ) and the likelihood ratio λ= L( ω ^ ) L( Ω ^ ) λ= L( ω ^ ) L( Ω ^ ) is given by:

λ= ( 10π ) −n/2 exp⁡[ −( 1 10 ) ∑ 1 n ( x i −162 ) 2 ] ( 10π ) −n/2 exp⁡[ −( 1 10 ) ∑ 1 n ( x i − x ¯ ) 2 ] = exp⁡[ −( 1 10 ) ∑ 1 n ( x i − x ¯ ) 2 −( n 10 ) ( x ¯ −162 ) 2 ] exp⁡[ −( 1 10 ) ∑ 1 n ( x i − x ¯ ) 2 ] =exp⁡[ −( n 10 ) ( x ¯ −162 ) 2 ]. λ= ( 10π ) −n/2 exp⁡[ −( 1 10 ) ∑ 1 n ( x i −162 ) 2 ] ( 10π ) −n/2 exp⁡[ −( 1 10 ) ∑ 1 n ( x i − x ¯ ) 2 ] = exp⁡[ −( 1 10 ) ∑ 1 n ( x i − x ¯ ) 2 −( n 10 ) ( x ¯ −162 ) 2 ] exp⁡[ −( 1 10 ) ∑ 1 n ( x i − x ¯ ) 2 ] =exp⁡[ −( n 10 ) ( x ¯ −162 ) 2 ].

A value of x ¯ x ¯ close to 162 would tend to support H 0 H 0 and in that case λ λ is close to 1. On the other hand, an x ¯ x ¯ that differs from 162 by too much would tend to support H 1 H 1 . A likelihood ratio critical region is given by λ≤k λ≤k , where k is selected so that the significance level of the test is α α . Using this criterion and simplifying the inequality as we do using the Neyman-Person lemma, we have that λ≤k λ≤k is equivalent to each of the following inequalities:

−( n 10 ) ( x ¯ −162 ) 2 ≤k, ( x ¯ −162 ) 2 ≥−( 10 n )ln⁡k, | x ¯ −162 | σ/ n ≥ −( 10/n )ln⁡k σ/ n =c. −( n 10 ) ( x ¯ −162 ) 2 ≤k, ( x ¯ −162 ) 2 ≥−( 10 n )ln⁡k, | x ¯ −162 | σ/ n ≥ −( 10/n )ln⁡k σ/ n =c.

Since Z= X ¯ −162 σ/ n Z= X ¯ −162 σ/ n is N( 0,1 ) N( 0,1 ) when H 0 :μ=162 H 0 :μ=162 is true, let c= z α/2 c= z α/2 . Thus the critical region is C={ x ¯ : | x ¯ −162 | σ/ n ≥ z α/2 } C={ x ¯ : | x ¯ −162 | σ/ n ≥ z α/2 } and, for illustration, if α α =0.05, we have that z 0.025 =1.96. z 0.025 =1.96.

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