Bias and MSE of $\hat{\theta} = \min(X_1, \ldots, X_n)$ Let $X_1, ... X_n$ iid with pdf given by $$p_{\theta, r} = r \theta^r x^{- (r+1)} \mathbb{1}\{x \geq \theta\}$$ for $\theta > 0$, and some $r > 2$ that is known. Then $\hat{\theta} = \min(X_1, \ldots, X_n) = X_{(1)}$.
I want to determine the bias and MSE of $\hat{\theta}$, so I need the pdf of $\hat{\theta}$.
If my calculations are correct, the pdf of $\hat{\theta}$ is given by:
$$f_{X_{(1)}} = n(r+1)r^n \theta^{rn} x^{-n(r+1) - 1} \mathbb{1}\{x \geq \theta\}.$$
Wondering if this pdf is correct, and how one would calculate the bias and MSE using this variance? I know that the bias is given by$E[\hat{\theta}] - \theta$, but I end up with a complicated expression, so I believe I am doing something wrong.
 A: $$
p_{\theta, r}(x) \, dx = r \left( \frac x \theta\right)^{- (r+1)}\left( \frac{dx} \theta \right) \mathbb{1}\{x \geq \theta\}.
$$
$$
\text{So for } x\ge\theta, \qquad \Pr(X>x) = \left( \frac x \theta \right)^{-r}
$$
$\min\{X_1,\ldots,X_n\} > x$ if and only if all of $X_1,\ldots, X_n$ are${}> x.$
That has probability $\big( \Pr(X_1>x)\big)^n = \left(\dfrac x \theta \right)^{-nr}.$
Therefore for $x>\theta,$
$$
p_{\min}(x) = \frac d {dx} \Pr(\min \le x) = \frac d {dx} \left( 1 - \left( \frac x \theta \right)^{-nr} \right) = nr\left( \frac x \theta \right)^{-(nr+1)}\cdot\frac 1 \theta.
$$
So
\begin{align}
& \operatorname E(\min) = \int_\theta^\infty x\cdot nr \left( \frac x \theta \right)^{-(nr+1)} \left( \frac{dx}\theta \right) \\[8pt]
= {} & nr\theta \int_\theta^\infty \frac x \theta \cdot \left( \frac x \theta \right)^{-(nr+1)} \, \left( \frac{dx} \theta \right) \\[8pt]
= {} & nr\theta \int_1^\infty u^{-nr} \, du = \theta \cdot \frac {nr} {nr-1} \\[8pt]
= {} & \theta \times \big(\text{something a bit more than 1} \big).
\end{align}
This is to be expected, since the minimum is always greater than $\theta;$ thus so is its expected value, and the fact that $\theta$ appears as a factor of this whole expression is necessary since $\theta$ is a scale parameter.
\begin{align}
& \operatorname E({\min}^2) = \int_\theta^\infty x^2\cdot nr \left( \frac x \theta \right)^{-(nr+1)} \left( \frac{dx}\theta \right) \\[8pt]
= {} & nr\theta^2 \int_\theta^\infty \left(\frac x \theta\right)^2 \cdot \left( \frac x \theta \right)^{-(nr+1)} \, \left( \frac{dx} \theta \right) \\[8pt]
= {} & nr\theta^2 \int_1^\infty u^{-nr+1} \, du = \theta^2 \cdot \frac {nr} {nr-2} \\[8pt]
= {} & \theta^2 \times \big(\text{something a bit more than 1} \big).
\end{align}
So
\begin{align}
& \operatorname{var}(\min) = \theta^2\left( \frac{nr}{nr-2} - \left( \frac{nr}{nr-1} \right)^2 \right) \\[8pt]
\text{and } & \frac{nr}{nr-2} - \left( \frac{nr}{nr-1} \right)^2 \\[8pt]
= {} & \frac{nr(nr-1)^2 - (nr)^2(nr-2)}{(nr-1)^2(nr-2)} \\[8pt]
= {} & \frac{nr}{(nr-1)^2(nr-2)}.
\end{align}
And then:
\begin{align}
& \text{m.s.e.} = \text{variance} + \big( \text{bias} \big)^2 \\[8pt]
= {} & \theta^2 \cdot \frac{nr}{(nr-1)^2(nr-2)} + \left( \theta\cdot \frac 1 {nr-1} \right)^2 \\[8pt]
= {} & \theta^2 \cdot \frac 2 {(nr-1)(nr-2)}.
\end{align}
A: 
I don't understand where $\Pr(X>x)=\Big(\frac{x}{\theta}\Big)^{-r}$ comes from.

Let's focus on your density (that is a known density, the Pareto), $x\geq\theta$:
$$f_X(x)=r\theta^r x^{-(r+1)}$$
the CDF is
$$F_X(x)=\int_\theta^x r\theta^r t^{-(r+1)} \, dt=\cdots=1-\Big(\frac{x}{\theta}\Big)^{-r}$$
Thus
$$P(X>x)=1-F$$
