st422%20hw03%20solution

# st422%20hw03%20solution - Solutions Homework 3 3 1 2 1 2 1...

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Unformatted text preview: Solutions: Homework 3: 3 1 2 1 2 1 2 3 3 1 2 1 8.2 : ) ˆ ˆ ˆ ( ) ( (1 ) ) ˆ ˆ = aE( ) (1 ) ( ) = (a+1-a) = b) ˆ ˆ ˆ Given and are independent, to choose 'a' that would minimize Var( ) ˆ ˆ ˆ ( ) ( (1 ) ) ˆ ˆ , a E E a a a E V V a a θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ = + - + - = + - 2 1 2 2 2 3 1 2 2 2 2 2 1 2 3 2 2 1 2 2 2 2 2 1 2 ˆ ˆ are independent .Thus, cov(a ,(1 ) ) ˆ ˆ ˆ ( ) ( ) (1 ) ( ) = a (1 ) ˆ ( ) , 2 2(1 ) Also, to check if this is the minimum value of 'a', d a V a V a V a dV Thus da a a a θ θ θ θ θ σ σ θ σ σ σ σ σ- = = + - + - = ⇒-- = ⇒ = + 2 2 2 3 1 2 2 2 2 3 2 2 1 2 ˆ ( ) 2( ) 0, da ˆ a= minimizes V( ) V θ σ σ σ θ σ σ = + > ⇒ + 1 1 1 1 1 2 1 2 2 1 2 3 1 2 3 1 2 4 1 2 3 (1) (1 8.4 ) Given Y exp( ), to find estimators which are unbiased: ˆ ˆ ( ) ( ) ˆ ˆ ( ) / 2 ( ) [ ( ) ( )]/ 2 2 / ˆ ˆ ( 2 ) /3 ( ) [ ( ) 2 ( )]/3 ( 2 ) / ˆ min( , , ) ( a Y E E Y Y Y E E Y E Y Y Y E E Y E Y Y Y Y Y f y θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ = ⇒ = = = + ⇒ = + = = = + ⇒ = + = + = = = ∼ (1) 1 ) 4 4 5 4 1 ) ( )(1 ( ) Here n=3 f(y) = (1/ )exp(-y/ ) y>0 = 0 elsewhere F(y)=P(Y y)=1-exp(-y/ ) y>0 (y) = (3/ )exp(-3y/ ) ˆ ˆ exp( /3) E( )= /3 ˆ ˆ ( ) (1/ ) ( ) / ˆ , n y n i i nf y F y f Y E n E Y n n Thus θ θ θ θ θ θ θ θ θ θ θ...
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st422%20hw03%20solution - Solutions Homework 3 3 1 2 1 2 1...

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