{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

mont4e_sm_ch07_supplemental

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ∏...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online