X1,...Xn~N(mu,sigma^2) (sigma^2 known) is the estimator mu=(average(x))^2-((sigma^2)/n) an unbiased estimator to mu^2 ? cheers
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yes, since sigma is known AND normality has nothing to do with this
why is it unbiased, how did you work it out ?
$\displaystyle E(\bar X^2) =V(\bar X) + (E\bar X)^2={\sigma^2\over n}+\mu^2$
just one question if I might: an estimator T is unbiased if E(T)=parameter where in the solution we calculated E(T) ?
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