Dr. Hackney STA Solutions pg 95

# Dr. Hackney STA Solutions pg 95 - ∂ 2 ∂θ 2 log f T,U |...

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Second Edition 6-9 where F 2 n, 2 n is an F random variable with 2 n degrees of freedom in the numerator and denominator. This follows since 2 Y i and 2 X i θ are all independent exponential(1), or χ 2 2 . Diﬀerentiating (in t ) and simplifying gives the density of T as f T ( t ) = Γ(2 n ) Γ( n ) 2 2 t ± t 2 t 2 + θ 2 ² n ± θ 2 t 2 + θ 2 ² n , and the second derivative (in θ ) of the log density is 2 n t 4 + 2 t 2 θ 2 - θ 4 θ 2 ( t 2 + θ 2 ) 2 = 2 n θ 2 ± 1 - 2 ( t 2 2 + 1) 2 ² , and the information in T is 2 n θ 2 " 1 - 2E ± 1 T 2 2 + 1 ² 2 # = 2 n θ 2 1 - 2E ³ 1 F 2 2 n, 2 n + 1 ! 2 . The expected value is E ³ 1 F 2 2 n, 2 n + 1 ! 2 = Γ(2 n ) Γ( n ) 2 Z 0 1 (1 + w ) 2 w n - 1 (1 + w ) 2 n = Γ(2 n ) Γ( n ) 2 Γ( n )Γ( n + 2) Γ(2 n + 2) = n + 1 2(2 n + 1) . Substituting this above gives the information in T as 2 n θ 2 ´ 1 - 2 n + 1 2(2 n + 1) µ = I ( θ ) n 2 n + 1 , which is not the answer reported by Joshi and Nabar. (ii) Let W = i X i and V = i Y i . In each pair, X i and Y i are independent, so W and V are independent. X i exponential(1 ); hence, W gamma( n, 1 ). Y i exponential( θ ); hence, V gamma( n,θ ). Use this joint distribution of ( W,V ) to derive the joint pdf of ( T,U ) as f ( t,u | θ ) = 2 [Γ( n )] 2 t u 2 n - 1 exp ± - t - ut θ ² , u > 0 , t > 0 . Now, the information in ( T,U ) is - E ±
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Unformatted text preview: ∂ 2 ∂θ 2 log f ( T,U | θ ) ² =-E ±-2 UT θ 3 ² = E ± 2 V θ 3 ² = 2 nθ θ 3 = 2 n θ 2 . (iii) The pdf of the sample is f ( x , y ) = exp [-θ ( ∑ i x i )-( ∑ i y i ) /θ ] . Hence, ( W,V ) deﬁned as in part (ii) is suﬃcient. ( T,U ) is a one-to-one function of ( W,V ), hence ( T,U ) is also suﬃcient. But, E U 2 = E WV = ( n/θ )( nθ ) = n 2 does not depend on θ . So E( U 2-n 2 ) = 0 for all θ , and ( T,U ) is not complete. 6.39 a. The transformation from Celsius to Fahrenheit is y = 9 x/ 5 + 32. Hence, 5 9 ( T * ( y )-32) = 5 9 (( . 5)( y ) + ( . 5)(212)-32) = 5 9 (( . 5)(9 x/ 5 + 32) + ( . 5)(212)-32) = ( . 5) x + 50 = T ( x ) . b. T ( x ) = ( . 5) x + 50 6 = ( . 5) x + 106 = T * ( x ). Thus, we do not have equivariance....
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