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Assignment 3- solution

# Assignment 3- solution - THE UNIVERSITY OF HONG KONG...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT 1302 PROBABILITY AND STATISTICS II (2009-10) Assignment 3 (Sketch Solution) 1. (a) bias( T 1 ) = E [ T 1 ] - μ = ( μ + μ ) / 2 - μ = 0 T 1 is unbiased. MSE( T 1 ) = Var( T 1 ) + bias( T 1 ) 2 = (2 σ 2 ) / 4 + 0 = μ 2 c 2 v / 2. (b) bias( T 2 ) = E [ T 2 ] - μ = (2 μ ) / (2+ c 2 v ) - μ = - c 2 v μ/ (2+ c 2 v ). Thus T 2 is biased unless c v = 0. MSE( T 2 ) = Var( T 2 ) + bias( T 2 ) 2 = (2 μ 2 c 2 v ) / (2 + c 2 v ) 2 + c 4 v μ 2 / (2 + c 2 v ) 2 = μ 2 c 2 v / ( c 2 v + 2). (c) T 2 , for MSE( T 2 ) MSE( T 1 ). 2. (a) For y > 0, cdf of Y is F Y ( y | μ, σ ) = P ( | Z | ≤ y | μ, σ ) = Z y - y ψ ( μ, σ | z ) dz = Z y -∞ ψ ( μ, σ | z ) dz - Z - y -∞ ψ ( μ, σ | z ) dz. The pdf of Y follows by differentiating the above w.r.t. y and noting that ψ ( μ, σ | - y ) = ψ ( - μ, σ | y ). (b) (i) z ( σ ) ψ (0 , σ | z ) = ψ (0 , σ | y ) * y ( σ ) = ψ (0 , σ | y ) + ψ (0 , σ | - y ) ψ (0 , σ | y ). (ii) i Z ( σ ) = - E ( 2 /∂σ 2 ) ln Z ( σ ) = E [ - σ - 2 + 3 Z 2 σ - 4 ] = 2 σ - 2 . (iii) i Y ( σ ) = - E ( 2 /∂σ 2 ) ln * Y ( σ ) = 2 σ - 2 . No information loss. (iv) Yes, according to FC (applied to z ( σ )). (c) (i) ** y ( μ ) = ψ ( μ, 1 | y ) + ψ ( μ, 1 | - y ) = ψ ( μ, 1 | y ) + ψ ( - μ, 1 | y ). (ii) i Z ( μ ) = - E ( 2 /∂μ 2 ) ln ψ ( μ, 1 | Z ) = 1. (iii) i Y ( μ ) = - E ( 2 /∂μ 2 ) ln ** Y ( μ ) = E h 1 - 4 Y 2 ( e μY + e - μY ) 2 i < 1. There is information loss. 3. (a) i ( λ ) = - E £ ( 2 /∂λ 2 ) ln ( λ ) / = E j X j 2 = n ( n + 1)(2 λ ) - 1 . (b) ( ∂/∂λ ) ln ( λ ) = 0 mle ˆ λ n = 2 j X j n ( n + 1) . (Second derivative < 0 confirms maximality.) (i) bias = 2 n ( n +1) j ( ) - λ = 0, (ii) variance = 4 n 2 ( n +1) 2 j ( ) = 2 λ n ( n +1) , (iii) mean squared error = variance. 1

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4. Note that the Fisher information i n contained in an i.i.d. random sample of n observations is simply n times the Fisher information i 1 contained in a single observation, i.e. i n = ni 1 . (a) Putting θ = ln λ , the likelihood (based on X ) is ( θ ) = exp ( θX - e θ ) /X !. Then i 1 = - E ( 2 /∂θ 2 ) ln ( θ ) = e θ and i n = n e θ . (b) Put θ = σ 2 . The likelihood (based on X ) is ( θ ) = (2 πθ ) - 1 / 2 exp {- X 2 / (2 θ ) } . Then i 1 = 1 2 θ 2 and i n = n 2 θ 2 .
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