Bias, SE and MSE of Uniform Distribution Let $X_1,\ldots,X_n$ be an i.i.d.  sequence of  Uniform $(\mu,2\mu)$ and let an estimator be $\hat{\mu} = \frac{1}{2} \max\{X_1,\ldots,X_n\}$. Find the
bias, SE, and MSE of this estimator.
Hint: Let $U_1,\ldots,U_n$ be i.i.d.  Uniform $(0,1)$ and $Y = \max\{U_1, \ldots,U_n\}$.
It is known that $Y$ and $(\max\{X_1,\ldots,X_n\} - \mu)/\mu$ have the same distribution. 
So, this implies that $E[Y] = E[(\max\{X_1,\ldots,X_n\} - \mu)/\mu]$ and $E[Y^2] = E[(\max\{X_1,\ldots,X_n\} - \mu)^2/\mu^2]$.
The first two moments of Y are given by
$E[Y] = n/(n+1)$ and $E[Y^2] = n/(n+2)$ .
 A: You've got
$$
E[Y] = \frac { (E[\max\{X_1,\ldots,X_n\}]) - \mu} \mu
$$
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.
If you know that $\operatorname{var}(Y) = E[Y^2] - (E[Y])^2$ then you can find $\operatorname{var}(Y)$. And
$$
\operatorname{var}(Y) = \operatorname{var} \frac{\max\{X_1,\ldots,X_n\} - \mu}\mu  = \frac 1 {\mu^2} \operatorname{var}(\max \{X_1,\ldots,X_n\}). \tag 2
$$
Hence you can find the variance of $\max\{X_1,\ldots,X_n\}$.  Recall that the variance plus the square of the bias is the mean squared error.
PS: You said in comments "I can't seem to see how this is linked to the original question."  How it is linked to the original question is by algebra.
You have
\begin{align}
\operatorname{var}(Y) & = E[Y^2] = (E[Y])^2 = \frac n {n+2} - \left( \frac n {n+1} \right)^2 \\[10pt]
& = \frac{n(n+1)^2 - n^2(n+2)}{(n+1)^2(n+2)} = \frac n {(n+1)^2(n+2)} \tag 3
\end{align}
From $(2)$ and $(3)$ we get
$$
\operatorname{var} \left(\frac 1 2 \max\{X_1,\ldots,X_n\} \right) = \frac {n\mu^2} {4(n+1)^2(n+2)}. \tag 4
$$
(One expects the variance to be proportional to $\mu^2$ because $\mu$ is a scale parameter.)
If we're going to use $\dfrac 1 2 \max$ as an estimator of $\mu$, then our estimator of $(4)$ will be
$$
\frac {n\max^2} {4(n+1)^2(n+2)},
$$
so we can take that to be the mean squared error.
Solve $(1)$ for $\mu$ and conclude that
$$
E\left[\frac 1 2 \max\right] = \frac{2n+1}{2n+2}\mu.
$$
That is the expected value of the estimator, so the bias is that minus $\mu$:
$$
\frac{2n+1}{2n+2}\mu - \mu = \frac{-1}{2n+2}\mu.
$$
Next:
\begin{align}
\text{mean squared error} & = \text{variance} + \left(\text{bias}\right)^2 \\[10pt]
& = \frac {n\mu^2} {4(n+1)^2(n+2)} + \left( \frac{-1}{2n+2}\mu \right)^2 \\[10pt]
& = \frac {\mu^2} {2(n+1)(n+2)}.
\end{align}
