mont4e_sm_ch07_supplemental

7 56 n l n xi 1 i 1 n ln l n ln 1 ln

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Unformatted text preview: ∂a 0 = 2n2 a − 2n1 + 2n1a 2a(n2 + n1 ) = 2n1 a(n2 + n1 ) = n1 a= n1 n2 + n1 7-55 n −x ⎛ 1 ⎞ ∑θ n 2 L(θ ) = ⎜ ∏ xi ⎜ 3⎟ e ⎟ ⎝ 2θ ⎠ i =1 n i i =1 n nx ⎛1⎞ i ln L(θ ) = n ln⎜ ⎜ 3 ⎟ + 2 ∑ ln xi − ∑ θ ⎟ ⎝ 2θ ⎠ i =1 i =1 ∂ ln L(θ ) − 3n n xi = +∑ 2 θ ∂θ i =1θ Making the last equation equal to zero and solving for theta, we obtain: n ˆ Θ= ∑ xi i =1 3n as the maximum likelihood estimate. 7-56 n L(θ ) = θ n ∏ xiθ −1 i =1 n ln L(θ ) = n lnθ + (θ − 1) ∑ ln( xi ) i =1 ∂ ln L(θ ) n n = + ∑ ln( xi ) ∂θ θ i =1 making the last equation equal to zero and solving for theta, we obtain the maximum likelihood estimate. ˆ Θ= −n n ∑ ln( xi ) i =1 7-57 7-17 L(θ ) = 1 θ n n ∏...
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