Unbiased Estimator for b: - Sum of ln(xi)/n

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DavidLiew
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If I want shows that [tex]\hat{b}[/tex] is an unbiased estimator for the b
where [tex]\hat{b}[/tex] = - [tex]\sum[/tex] ln xi /n
f(x)= [tex]\frac{1}{b}[/tex] e(1-b/b)
 
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DavidLiew said:
If I want shows that [tex]\hat{b}[/tex] is an unbiased estimator for the b
where [tex]\hat{b}[/tex] = - [tex]\sum[/tex] ln xi /n
f(x)= [tex]\frac{1}{b}[/tex] e(1-b/b)


Is that meant to be (1-b)/b? If its not then you'll get 1/b which is basically a uniform distribution,given that the domain is accurate.