# Tag Info

You've got $E[Y] = E[(\max\{X_1,\ldots,X_n\} - \mu)/\mu]$ and you should know that that implies $$E[Y] = \frac { (E[\max\{X_1,\ldots,X_n\}]) - \mu} \mu$$ and you've got $E[Y] = n/(n+1)$, so you have $$\frac n {n+1} = \frac { (E[\max\{X_1,\ldots,X_n\}]) - \mu} \mu. \tag 1$$ From that you get $E[\max\{X_1,\ldots,X_n\}]$ and then from that you get the bias. ...
Community wiki answer based on the comments to allow the answer to be accepted: In this case the parameter happens to be the mean. The unbiased estimator $\hat\theta$ of the parameter is the usual unbiased estimator of the mean, whose variance is $1/N$ times the population variance, so estimating its variance is equivalent to estimating the population ...