t308soln - STAT 330 TEST 3 SOLUTIONS 1(a) f (x; ) = (b) L (...

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STAT 330 TEST 3 SOLUTIONS 1( a ) f ( x ; θ )= d dx F ( x ; θ d dx μ x 1 / 3 θ 1 / 3 = 1 3 x 2 / 3 θ 1 / 3 , 0 <x θ . ( b ) L ( θ n Y i =1 f ( x i ; θ n Y i =1 1 3 x 2 / 3 i θ 1 / 3 , 0 i θ = Ã n Y i =1 1 3 x 2 / 3 i ! 1 θ n/ 3 , θ x ( n ) > 0 . 0 x(n) y=L(theta) theta y Figure 1: From the graph it is obvious that L ( θ ) is maximized for θ = x ( n ) . Therefore the M.L. estimator of θ is ˆ θ n = X ( n ) . ( c )( i ) G n ( t P ³ ˆ θ n t ´ = P (max ( X 1 , ..., X n ) t ) = P ( X 1 t,. .., X n t n Y i =1 P ( X i t ) =[ P ( X i t )] n = μ t θ n/ 3 , 0 <t< θ 1
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and lim n →∞ G n ( t )= ( 0 if t< θ 1 if t> θ Therefore ˆ θ n p θ . ( c )( ii ) P ( Y n y P à n à 1 ˆ θ n θ ! y ! = P à ˆ θ n θ 1 y n ! =1 P à ˆ θ n θ θ ³ 1 y n ´ ! " θ ¡ 1 y n ¢ θ # n/ 3 ³ 1 y n ´ n/ 3 , 0 <y<n. Since lim n →∞ ³ 1 y n ´ n/ 3 = e y/ 3 , for all y < therefore lim n →∞ P ( Y n y 0 if y 0 1 e y/ 3 if y> 0 which is the c.d.f. of an EXP (3) random variable. Therefore Y n D Y v EXP (3) .
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t308soln - STAT 330 TEST 3 SOLUTIONS 1(a) f (x; ) = (b) L (...

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