Dr. Hackney STA Solutions pg 98

Dr. Hackney STA Solutions pg 98 - Chapter 7 Point...

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Chapter 7 Point Estimation 7.1 For each value of x , the MLE ˆ θ is the value of θ that maximizes f ( x | θ ). These values are in the following table. x 0 1 2 3 4 ˆ θ 1 1 2 or 3 3 3 At x = 2, f ( x | 2) = f ( x | 3) = 1 / 4 are both maxima, so both ˆ θ = 2 or ˆ θ = 3 are MLEs. 7.2 a. L ( β | x ) = n Y i =1 1 Γ( α ) β α x α - 1 i e - x i = 1 Γ( α ) n β " n Y i =1 x i # α - 1 e - Σ i x i log L ( β | x ) = - log Γ( α ) n - log β + ( α - 1) log " n Y i =1 x i # - i x i β log L ∂β = - β + i x i β 2 Set the partial derivative equal to 0 and solve for β to obtain ˆ β = i x i / ( ). To check that this is a maximum, calculate 2 log L ∂β 2 ± ± ± ± β = ˆ β = β 2 - 2 i x i β 3 ± ± ± ± β = ˆ β = ( ) 3 ( i x i ) 2 - 2( ) 3 ( i x i ) 2 = - ( ) 3 ( i x i ) 2 < 0 . Because ˆ β is the unique point where the derivative is 0 and it is a local maximum, it is a global maximum. That is, ˆ β is the MLE. b. Now the likelihood function is L ( α,β | x ) = 1 Γ( α ) n β " n Y i =1 x i # α - 1 e - Σ i x i , the same as in part (a) except α and β are both variables. There is no analytic form for the
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