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# 410Hw08ans - STAT 410 Fall 2011 Homework#8(due Friday...

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STAT 410 Homework #8 Fall 2011 (due Friday, October 28, by 3:00 p.m.) 1. 4.3.16 Hint: g184 g185 g183 g168 g169 g167 + + + = 1 2 1 2 n n t n t e o n t for large n . a) M X 1 ( t ) = ( 1 – t ) – 1 , t < 1. M Y n ( t ) = ( ) g184 g184 g185 g183 g168 g168 g169 g167 - 1 X E n n t e = ( ) g184 g184 g185 g183 g168 g168 g169 g167 + + + - n t n t n e e 2 1 X ... X X E = n n t n t e 1 X M g184 g184 g185 g183 g168 g168 g169 g167 g184 g184 g185 g183 g168 g168 g169 g167 - = n n t n t e - - g184 g184 g185 g183 g168 g168 g169 g167 - 1 = n n t n t e n t e - g184 g184 g185 g183 g168 g168 g169 g167 - , 1 < n t . b) g184 g185 g183 g168 g169 g167 + + + = n n t n t e o n t 1 2 1 2 . g159 M Y n ( t ) = n o n n t n t n t n t - g184 g184 g185 g183 g168 g168 g169 g167 g184 g185 g183 g168 g169 g167 + - - + + 1 2 1 2 2 = n o n n t - g184 g184 g185 g183 g168 g168 g169 g167 g184 g185 g183 g168 g169 g167 + - 1 2 1 2 .

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As n , M Y n ( t ) g176 g191 g176 g190 g189 g176 g175 g176 g174 g173 2 2 exp t = M Z ( t ) , where Z has Standard Normal N ( 0, 1 ) distribution. g159 Z Y D n , Z ~ N ( 0, 1 ) . 2. 4.3.17 Hint: Use Theorem 4.3.9. From 4.3.16 , Y n = ( ) 1 X - n n D N ( 0, 1 ) . Let g ( x ) = x . Then g ( x ) is differentiable, g ' ( x ) = x 2 1 , and g ' ( 1 ) = 2 1 . Then, by Theorem 4.3.9, ( ) 1 X - n n ( ) ( ) g184 g185 g183 g168 g169 g167 1 1 , 0 2 ' g N D = g184 g185 g183 g168 g169 g167 4 1 , 0 N .
3. Let Y n be χ 2 ( n ) . What is the limiting distribution of Z n = n n Y - ? Hint: We already know that ( ) n n n 2 Y - = g184 g184 g185 g183 g168 g168 g169 g167 - 1 Y 2 n n n ( ) 1 , 0 N D . Z n = n n Y - = g184 g184 g185 g183 g168 g168 g169 g167 - 1 Y n n n . ( ) n n n 2 Y - = g184 g184 g185 g183 g168 g168 g169 g167 - 1 Y 2 n n n ( ) 1 , 0 N D . g159 g184 g184 g185 g183 g168 g168 g169 g167 - 1 Y n n n ( ) 2 , 0 N D . Let g ( x ) = x . Then g ( x ) is differentiable, g ' ( x ) = x 2 1 , and g ' ( 1 ) = 2 1 . Then, by Theorem 4.3.9, g184 g184 g185 g183 g168 g168 g169 g167 - 1 Y n n n ( ) ( ) g184 g185 g183 g168 g169 g167 2 1 , 0 2 ' g N D = g184 g184 g185 g183 g168 g168 g169 g167 g184 g185 g183 g168 g169 g167 2 2 1 , 0 2 N . g159 Z n = n n Y - = g184 g184 g185 g183 g168 g168 g169 g167 - 1 Y n n n g184 g185 g183 g168 g169 g167 2 1 , 0 N D .

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4. Let X 1 , X 2 , … , X n be a random sample from the distribution with probability density function ( ) ( ) ( ) g537 1 1 g537 g537 ; X x x f - + = , 0 < x < 1, θ > – 1. Recall: g537 ~ = 2 X 1 - is the method of moments estimator of θ . Show that g537 ~ is asymptotically normally distributed ( as n ) . Find the parameters. Beta distribution, α = 1, β = θ + 1. g159 E ( X ) = g533 g302 g302 + = 2 g537 1 + .
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410Hw08ans - STAT 410 Fall 2011 Homework#8(due Friday...

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