Taking log on both sides we get, Σ In (L(a = 2, B)) =-2n·In(B)+ In Ex, Now partially differentiating on both sides with respect to ß and equating to zero we get, 8 In (L(a = 2, ß)) SB Ex, 2n = 0 2n 2...

where does the x go?

Taking log on both sides we get,<br>Σ<br>In (L(a = 2, B)) =-2n·In(B)+ In Ex,<br>Now partially differentiating on both sides with respect to ß and equating to zero we get,<br>8 In (L(a = 2, ß))<br>SB<br>Ex,<br>2n<br>= 0<br>2n<br>2<br>Comment<br>Step 5 of 5 A<br>Hence the maximum likelihood estimator of ß is<br>

Extracted text: Taking log on both sides we get, Σ In (L(a = 2, B)) =-2n·In(B)+ In Ex, Now partially differentiating on both sides with respect to ß and equating to zero we get, 8 In (L(a = 2, ß)) SB Ex, 2n = 0 2n 2 Comment Step 5 of 5 A Hence the maximum likelihood estimator of ß is

Jun 02, 2022

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Taking log on both sides we get, Σ In (L(a = 2, B)) =-2n·In(B)+ In Ex, Now partially differentiating on both sides with respect to ß and equating to zero we get, 8 In (L(a = 2, ß)) SB Ex, 2n = 0 2n 2...

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