Math 152. Rumbos Fall Solutions to Assignment #12

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1 Math 52. umbos Fall 2009 Solutions to Assignment #2. Suppose that you observe n iid Bernoulli(p) random variables, denoted by X, X 2,..., X n. Find the LT rejection region for the test of H o : p p o versus H : p > p o in terms of the test statistic Y = X i. Solution: The likelihood ratio statistic is Λ(x, x 2,..., x n ) = sup L(p x, x 2,..., x n ) p p o, L(ˆp x, x 2,..., x n ) where L(p x, x 2,..., x n ) = p y ( p) n y, for y = x i, is the likelihood function, and is the MLE for p. Observe that if p o ˆp, then ˆp = y = x n sup L(p x, x 2,..., x n ) = L(ˆp x, x 2,..., x n ) p p o and so Λ(x, x 2,..., x n ) = ; thus, in this case we would not get a rejection region for the LT, : Λ(x, x 2,..., x n ) c, for some 0 < c <. We therefore have that p o < ˆp so that Λ(x, x 2,..., x n ) = L(p o x, x 2,..., x n ) L(ˆp x, x 2,..., x n ) = py o( p o ) n y ˆp y ( ˆp ) n y,

2 Math 52. umbos Fall where ˆp > p o, which can in turn be written as Λ(x, x 2,..., x n ) = ( ˆp p o ) y p o p o ˆp p o n y. Setting t = ˆp p o, follows Λ(t) = we see that Λ can be written as a function of t as for 0 t, t npot ( ) n npot po, for < t p p o t o, since ˆp > p o, where we have used the fact that ˆp = n y so that y = np o t. The graph of Λ(t) can be shown to be like the one sketched as the one shown in Figure on page 3, where we have sketched the case p o = /4 and n = 20 for 0 t 4. The sketch in Figure shows that Λ(t) decreases for t > ; thus, given any positive value of c such that c < and c > p n o, there exists a positive value t 2 such that t 2 >, and Λ(t 2 ) = c, and Λ(t) c for t t 2. Thus, the LT rejection region for the test of H o : H : p > p o is equivalent to ˆp p o t 2, which we could rephrase in terms of Y = : Y t 2 np o, X i as p p o versus for some t 2 with t 2 >. This rejection region can also be phrased as : Y > np o + b,

3 Math 52. umbos Fall Lambda(t) t Figure : Sketch of graph of Λ(t) for p o = /4, n = 20, and 0 t 4

4 Math 52. umbos Fall for some b > 0. The value of b will then be determined by the significance level that we want to impose on the test, and Y is the statistic Y = X i, which counts the number of successes in the sample. 2. Consider the likelihood ratio test for H o : p = p o versus H : p = p, where p o = p, based on a random sample X, X 2,..., X n from a Bernoulli(p) distribution for 0 < p <. Show that, if p > p o, then the likelihood ratio statistic for the test is is a monotonically decreasing function of Y = X i.. Conclude, therefore, that if the test rejects H o at the significance level α for an observed value ˆy of Y, it will also rejects H o at that level for Y > ˆy. Solution: The likelihood ratio statistic in this case is Λ(x, x 2,..., x n ) = py o( p o ) n y p y ( p ), for y = x n y i, which can be written as where Λ(x, x 2,..., x n ) = a n r y, a = p o > 0 and r = p o( p ) p p ( p o ) <, since p > p o. It then follows that the likelihood ratio statistic for the test is is a monotonically decreasing function of Y = X i, since r <. Suppose now that the test rejects H o at the significance level α for an observed value ˆy of Y ; that is Λ(ˆy) c, for some c in (0, ) determined by α. Then, for any value of y which is bigger than ˆy, Λ(y) < Λ(ˆy) c, () since Λ is a decreasing function of y. It then follows from () that the LT will also rejects H o at that level for Y > ˆy.

5 Math 52. umbos Fall We wish to use an LT to test the hypothesis H o : μ = μ o against the alternative H : μ = μ o based on a random sample, X, X 2,..., X n, from a normal(μ, ) distribution. (a) Give the maximum likelihood estimator, ˆμ, for μ based on the sample. Solution: The likelihood function in this problem is n L(μ x, x 2,..., x n ) = e (x i μ) 2 /2 for μ. (2π) n/2 To find the MLE for μ, it suffices to maximize the natural logarithm of the likelihood function l(μ) = 2 (x i μ) 2 n ln(2π), for μ. 2 Taking derivatives we get l (μ) = (x i μ) = x i nμ, and l (μ) = (x i μ) = n < 0. It then follows that ˆμ = x i = x is the only critical point of n l and l (ˆμ) < 0. Thus, the likelihood function is maximized at ˆμ = x, the sample mean. (b) Give the likelihood ratio statistic for the test. Solution: The likelihood ratio statistic is Λ(x, x 2,..., x n ) = L(μ o x, x 2,..., x n ) L(ˆμ x, x 2,..., x n ), where ˆμ = x is the MLE for μ. We then have that Λ(x, x 2,..., x n ) = e n (x i μ o) 2 /2 e n (x i x) 2 /2, (2)

6 Math 52. umbos Fall where (x i μ o ) 2 = (x i x + x μ o ) 2 = (x i x) 2 + (x μ o ) 2 = (x i x) 2 + n(x μ o ) 2, since 2(x i x)(x μ o ) = 2(x μ o ) (x i x) = 0. Hence, from (2) we have that Λ(x, x 2,..., x n ) = e n(x μo)2 /2, (3) where x denotes the sample mean. (c) Express the LT rejection region in terms of the sample mean X n. Solution: The LT rejection region is given by : Λ(x, x 2,..., x n ) c, for some c with 0 < c <. It then follows from equation (3) in part (b) of this problem that H o is rejected if e n(x μo)2 /2 c, or, taking natural logarithm on both sides of the last inequality, n(x μ o ) 2 /2 ln c, or or n(x μ o ) 2 2 ln c, n x μo 2 ln c b > 0. Thus, the LT will reject H o if n Xn μ o b, for some b > 0 determined by the significance level α.

7 Math 52. umbos Fall Let X, X 2,..., X n denote a random sample from a uniform(0, θ) distribution for some parameter θ > 0. (a) Give the likelihood function L(θ x, x 2,..., x n ). Solution: The pdf for each of the X i s is if 0 < x < θ; θ f(x θ) = 0 otherwise. Thus, the likelihood function is if 0 < x θ L(θ x, x 2,..., x n ) = n, x 2,..., x n < θ; 0 otherwise. (4) (b) Give the maximum likelihood estimator for θ. Solution: Observe that the likelihood function in (4) is a decreasing function of θ. Thus, L(θ x, x 2,..., x n ) will the largest when θ is the smallest value it can take, ˆθ, based on the sample. This value is the maximum of the values x, x 2,..., x n, because, if x i > θ for some i, then L(θ x, x 2,..., x n ) = 0 according to the definition of the likelihood function given (4). It then follows that the MLE for θ is ˆθ = max{x, X 2,..., X n }. 5. Let denote the rejection region for an LT of H o : θ = θ o versus H : θ = θ based on a random sample, X, X 2,..., X n, from continuous distribution with pdf f(x θ). Let L(θ x, x 2,..., x n ) denote the likelihood function. Suppose the LT has significance level α. (a) Explain why α = L(θ o x, x 2,..., x n ) dx dx 2 dx n.

8 Math 52. umbos Fall Answer: The significance level α is the probability the the LT will reject H o when H o is true; in other words, when θ = θ o. Thus, α = P((x, x 2,..., x n ) = f(x, x 2,..., x n θ o ) dx dx 2 dx n, where is the rejection region of the LT and f(x, x 2,..., x n θ o ) is the joint distribution of the sample for the case in which H o is true. It then follows that α = L(θ o x, x 2,..., x n ) dx dx 2 dx n, by the definition of the likelihood function. (b) Explain why the power of the test is γ(θ ) = L(θ x, x 2,..., x n ) dx dx 2 dx n. Answer: The power of the test, γ(θ ), is the probability the the LT will reject H o when H o is false; in other words, when θ = θ. Thus, γ(θ ) = f(x, x 2,..., x n θ ) dx dx 2 dx n, which yields γ(θ ) = L(θ x, x 2,..., x n ) dx dx 2 dx n. by the definition of the likelihood function. (c) Explain why α cγ(θ ), where c is the critical value used in the definition of the rejection region,, for the LT. Solution: Since, Λ(x, x 2,..., x n ) c on,, for some c (0, ) determined by α, it follows that L(θ 0 x, x 2,..., x n ) cl(θ x, x 2,..., x n )

9 Math 52. umbos Fall for all (x, x 2,..., x n ). Consequently, L(θ 0 x, x 2,..., x n ) cl(θ x, x 2,..., x n ) from which the result follows in view of parts (a) and (b) of this problem.

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