Or use np lemma a reject ho if f x τ 4 f x τ 1 k

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Or use NP lemma. a) Reject Ho if f ( x | τ = 4) f ( x | τ = 1) >k. The LHS = 4 n 2 n producttext n i =1 (1 - e - x 2 i ) 4 - 1 producttext n i =1 (1 - e - x 2 i ) = 2 n n productdisplay i =1 (1 - e - x 2 i ) 2 . So reject Ho if n productdisplay i =1 (1 - e - x 2 i ) 2 >k prime or n productdisplay i =1 (1 - e - x 2 i ) >c or n summationdisplay i =1 log(1 - e - x 2 i ) >d
CHAPTER 11. STUFF FOR STUDENTS 420 where α = P τ =2 ( n productdisplay i =1 (1 - e - x 2 i ) >c ) . b) Replace 4 - 1 by τ 1 - 1 where τ 1 > 2. Then reject H 0 if n productdisplay i =1 (1 - e - x 2 i ) τ 1 - 2 >k prime which gives the same test as in a). 7.22. By exponential family theory, the UMP test rejects H 0 if T ( x ) = - n i =1 1 x i >k where P θ =1 ( T ( X ) >k ) = α . Alternatively, use the Neyman Pearson lemma: a) reject Ho if f ( x | θ = 2) f ( x | θ = 1) >k prime . The LHS = 2 n exp( - 2 1 x i ) exp( - 1 x i ) . So reject Ho if 2 n exp[( - 2 + 1) summationdisplay 1 x i ] >k prime or if - 1 x i >k where P 1 ( - 1 x i >k ) = α. b) In the above argument, reject H 0 if 2 n exp[( - θ 1 + 1) summationdisplay 1 x i ] >k prime or if - 1 x i >k where P 1 ( - 1 x i >k ) = α for any θ 1 > 1. Hence the UMP test is the same as in a). 7.23 a) We reject H 0 iff f 1 ( x ) f 0 ( x ) >k . Thus we reject H 0 iff 2 x 2(1 - x ) >k . That is 1 - x x < k 1 , that is 1 x < k 2 , that is x > k 3 . Now 0 . 1 = P ( X > k 3 ) when f ( x ) = f 0 ( x ), so k 3 = 1 - 0 . 1. 7.24. a) Let k = [2 πσ 1 σ 2 (1 - ρ 2 ) 1 / 2 ]. Then the likelihood L ( θ ) = 1 k n exp parenleftBigg - 1 2(1 - ρ 2 ) n summationdisplay i =1 bracketleftBigg parenleftbigg x i - μ 1 σ 1 parenrightbigg 2 - 2 ρ parenleftbigg x i - μ 1 σ 1 parenrightbiggparenleftbigg y i - μ 2 σ 2 parenrightbigg + parenleftbigg y i - μ 2 σ 2 parenrightbigg 2 bracketrightBiggparenrightBigg .
CHAPTER 11. STUFF FOR STUDENTS 421 Hence L ( ˆ θ ) = 1 [2 π ˆ σ 1 ˆ σ 2 (1 - ˆ ρ 2 ) 1 / 2 ] n exp parenleftbigg - 1 2(1 - ˆ ρ 2 ) [ T 1 - ρT 2 + T 3 ] parenrightbigg = 1 [2 π ˆ σ 1 ˆ σ 2 (1 - ˆ ρ 2 ) 1 / 2 ] n exp( - n ) and L ( ˆ θ 0 ) = 1 [2 π ˆ σ 1 ˆ σ 2 ] n exp parenleftbigg - 1 2 [ T 1 + T 3 ] parenrightbigg = 1 [2 π ˆ σ 1 ˆ σ 2 ] n exp( - n ) . Thus λ ( x , y ) = L ( ˆ θ 0 ) L ( ˆ θ ) = (1 - ˆ ρ 2 ) n/ 2 . So reject H 0 if λ ( x , y ) c where α = sup θ Θ o P ( λ ( X , Y ) c ) . Here Θ o is the set of θ = ( μ 1 2 1 2 ) such that the μ i are real, σ i > 0 and ρ = 0, ie, such that X i and Y i are independent. b) Since the unrestricted MLE has one more free parameter than the restricted MLE, - 2 log( λ ( X , Y )) χ 2 1 , and the approximate LRT rejects H 0 if - 2 log λ ( x , y ) 2 1 , 1 - α where P ( χ 2 1 2 1 , 1 - α ) = α . 8.1 c) The histograms should become more like a normal distribution as n increases from 1 to 200. In particular, when n = 1 the histogram should be right skewed while for n = 200 the histogram should be nearly symmetric. Also the scale on the horizontal axis should decrease as n increases. d) Now Y N (0 , 1 /n ). Hence the histograms should all be roughly symmetric, but the scale on the horizontal axis should be from about - 3 / n to 3 / n . 8.3. a) E ( X ) = 3 θ θ +1 , thus n ( X - E ( X )) D N (0 ,V ( X )), where V ( X ) = 9 θ ( θ +2)( θ +1) 2 . Let g ( y ) = y 3 - y , thus g prime ( y ) = 3 (3 - y ) 2 . Using the delta method, n ( T n - θ ) D N (0 , θ ( θ +1) 2 θ +2 ) .

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