Generalized Likelihood Ratio Test Example：广义似然比检验的例子.pdfVIP

generalized likelihood ratio test example a.k.a. too much for in class but certainly worth making sure you can do each step! consider testing h0: 0 vs. ha: 0 for an random sample form a population that is normally distributed (where 2 is unknown). one way to do this is to construct the likelihood ratio test where p(|h0 is true)=. using the definition of the likelihood ratio test on page 308, and plugging in the normal p.d.f., we get: () () = = = = = = n i i hh n i i h n i i hh n i i h n hh n h x x xf xf xxf xxf a a a 1 2 2 1 2 2 1 1 1 1 2 exp 2 1 max 2 exp 2 1 max ),|(max ),|(max ),|,.(max ),|,.(max 0 0 0 0 0 0 the next step is to find what the maximum is in both the numerator and denominator. the denominator is the easy place to start because it is over the set of all possible and values. in other words we just need to find the mle. for the numerator we have to deal with the restriction on . denominator: to find the mle we need to take the derivative with respect to both of the parameters, so () ()() () = = = = = = = n i i n i i n i i n i i xn n xx lik x lik 1 2 2 1 2 2 1 2 2 1 2 2 2 )2log( 2 log 22 1 log 2 exp 2 1 loglog 2 exp 2 1 ()() = = = = n i mlei n i i n i i xnx xx 11 2 1 2 0) 1( 2 2 ()() ()() n xx n x xnxn n i i n i mlei n i mle i n i i = = += = = = 1 2 1 2 1 3 2 1 3 2 )2( 2 numerator: notice in the numerator we need to find the maximum subject to the null hypothesis being true. that is, since must be 0, mleh0 must be 0. so we do the math just like we did for the denominator and get that xis the h0mle as long as it is 0. if it is then the maximum must happen at the boundary, and be either 0 or - (which would mean the maximum would never be reached). notice what happens if we plug in - for ; the likelihood is zero. therefore the h0mle is 0 if x0. to get the h0mle for we simply follow the math above and plug in the value of h0mle for when needed. this gives: if 0 x , x mleh = 0 and () n xx n i i mleh = =1 2 0 if 0 x , 0 0 = mleh and () n x n i i mleh = =1 2 0 0 plugging the values in: first notice that =1 (and we accept) if 0 x because the maximums plugged into the numerator and denominator are the same. the tricky case then is if 0 x. so lets carry through the product and just plug the values in: () () () () = = = = = = 2 1 2 2 1 2 1 2 2 1 2 2 2 exp 2 1 max 2 exp 2 1 max 2 exp 2 1 max 2 exp 2 1 max 0 0 0 0 n i i n hh n i i n h n i i hh n i i h x x x x a a () () () () () () () () () () 2 1 2 0 1 2 2 1 2 0 1 2 1 2 1 2 1 2 1 2 0 1 2 0 1 2 0 2 exp 2 exp 2 exp 2 1 2 exp 2 1 n n i i n i i n n i i n i i n i i n i i n n i i n i i n i i n n i i x xx n n x xx n xx xx n xx n x x n x = = = = = = = = = = = = = now we need to find the so that () () = = = = 2 1 2 0 1 2 n n i i n i i x xx p this doesnt look friendly at all but if we remember that the xs are normal we might be able to change this into a distribution we can work with (t? 2? f?). notice that the numerator looks very close to the standard deviation what you would find in the denominator of a t-test. also notice that the denominator is at least suggestive of the 0 x you would find in the numerator of the t-test statistic. since this is are only lead, lets see if we can work it out () () () () () () 1 2 1 2 0 2 1 2 1 2 0 2 1 2 0 1 2 + = = = = = = n i i n i i n n i i n i i n n i i n i i xx xxx xx x x xx () () ()()() () )(2 1 2 11 2 0 1 2 0 2 1 2 1 2 0 + + = = = = n i i n i n i n i ii n i i n i i xx xxxxxx xx xxx note that xand 0 are constant with respect to n, so that the second term is zero, and the sum turns into a multiplier of n in the third term: ()() () () () 1 1 2 2 0 1 2 1 2 0 2 + + = = n i i n i i n i i xx xn xx xnxx () () () () () () v n i i v n i i n i i n xx x xx xn xx xn 1 1 2 01 1 2 2 0 1 2 2 0 = = () () () vv n i in s x n n xx x = 0 1 2 0 1 this is just the classic t-statistic! (and recall that if 0 x then we would accept.) finally! so the test for h0: 0 vs. ha: 0 is reject if question: how would the above math change if we had used h0: =0 vs. ha: 0 ? what about theorem a? notice that theorem a on page 310 says that 2log should be chi- squared with degrees of freedom equal to dim-dimo. does that work here? notice that it probably shouldnt for two different reasons! reason 1: = possible values of and is two-dimensional but so is o because it is half of a plane (a line is one-dimensional). so the degrees of freedom should be 2-2=0 which makes no sense! reason 2: if 0 x we get that =1 so that , so that 2log = 0. there is thus some probability that 2log = 0. but a chi-squared distribution is continuous and so has to have 0 probability of taking any given value! we get something called a chi-bar distribution that is a mixture of a discrete distribution at zero and part of a chi-squared distribution. yuck! however, if we had used the hypotheses h0: =0 vs. ha: 0 it would have worked out just fine. in this case is two-dimensional, but o has a fixed value of so that it is only one- dimensional. if everything works out we should get that 2log will be chi-squared with one degree of freedom. () () 2 1 2 0 1 2 n n i i n i i x xx = = = so, using some of the math we saw earlier () () () () n n i i n i i n n i i n i i xx x x xx = = = = = = 1 2 1 2 0 2 1 2 0 1 2 loglog2log2 = () n n i i xx xn + =1 2 2 0) ( 1log () 2 1 0 t n s x () n n n n n i i n t n n s x n n s x n n n xx x += += += += = 1 1log 1 1log 1 )( 1log 1 1 )(