Derive the bias and MSE of the estimator $\hat{\beta}$ Let $Y_1, Y_2, \ldots, Y_n$ denote a random sample of size $n$ from a population whose density is given by 
$$
f(y)=
\begin{cases} 
      3\beta^3 y^{-4} & \beta \leq y \\
      0 & \text{elsewhere}
   \end{cases}
$$
where $\beta > 0$ is unknown. (This is a Pareto distribution) Consider the estimator $\hat{\beta} = \min(Y_1, Y_2, \ldots, Y_n)$.
a. Derive the bias of the estimator $\hat{\beta}$.
b. Derive MSE($\hat{\beta}$).
For part a
Since this is an estimator that is dealing with random variables, and I only need the minimum, then I can represent this as ordered statistics. Namely, $\hat{\beta} = Y_{(1)}$.
Using the formula for ordered statistics, the pdf is:
$$
g_{(1)}(Y_{(1)}) = n(1-F(y))^{n-1}f(y)
$$
My first question arises here, I am not sure how I am supposed to find $F(y)$ (and I am assuming that $f(y)$ is the given distribution above). 
Checking against the solution, I know that the pdf is supposed to be:
$$
g_{(1)}(Y_{(1)}) = 3n\beta^{3n}y^{(-3n-1)}
$$
After solving getting the pdf of $Y_{(1)}$, I can then use it in the bias formula.
$$
Bias(\hat{\beta}) = E[\hat{\beta}] - \beta \\
Bias(\hat{\beta}) = E[3n\beta^{3n}y^{(-3n-1)}] - \beta
$$
My second question arises here. I'm not really sure how I am supposed to find the expectation. Namely, I don't know how I am supposed to treat the $y$ variable. I think that the $\beta$ variable is supposed to be treated the same way as a random variable, and that $n$ is supposed to be treated as a constant.
After solving for the expectation, I should be able to get the Bias.
For part b
$$
MSE(\hat{\beta}) = Bias(\hat{\beta})^2 + Var(\hat{\beta})
$$
The bias part of this formula should follow from part a. This would leave just the variance to calculate. I would expect to use the variance formula:
$$
Var(\hat{\beta}) = E[\hat{\beta}^2] - E[\hat{\beta}]^2
$$
And I would also expect that both expectations in this formula could be calculated using the same methods in part a. Is this correct?
 A: I got this question recently. This question is from Mathematical Statistics - Wackerly - Exercise 8.1.

Let $Y_{1}, Y_{2}, \ldots, Y_{n}$ denote a random sample of size $n$ from a population whose density is given by
$$
f(y)=\left\{\begin{array}{ll}
3 \beta^{3} y^{-4} & , \quad \beta \leq y \\
0 & , \text { en cualquier otro caso }
\end{array}\right.
$$
where $\beta>0$ is unknown. (This is one of the Pareto distributions introduced in Exercise 6.18.) Consider the estimator $\hat{\beta}=\min \left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)$

This my approach:
a.
Considering distribution function from Pareto distribution we have:
$$F(y)=\begin{cases}0,& y<\beta\\1-\left(\frac{\beta}{y}\right)^\alpha, & y\geq \beta \end{cases}$$
Which density function (derivative) is:
$$f(y)=\begin{cases}0,& y<\beta\\
\alpha \beta ^\alpha y^{-(\alpha +1)}, & y\geq \beta \end{cases}$$
We proceed to calculate $Y_{(1)}$ density function (with $\alpha=3$, our problem):
\begin{align}
\begin{split}
    g_{(1)}(y)&=n\left[1-F(y)\right]^{n-1}f(y) \\ 
              &=n\left[1-\left( 1-\left(\frac{\beta}{y}\right)^3 \right)\right]^{n-1} 3 \beta ^3 y^{-(3+1)}\\
              &= n\left[\beta^{3(n-1)}y^{-3(n-1)}\right]3 \beta ^3 y^{-(4)}\\
              &= 3n\beta^{3n}y^{-(3n+1)}, \qquad y\geq \beta 
\end{split}
\end{align}
Now, we'll check this definition:

Let $\hat{\theta}$ be a point estimator for a parameter $\theta$. Then $\hat{\theta}$ is an unbiased estimator if $E(\hat{\theta})=\theta$. If $E(\hat{\theta})\neq \theta$, $\hat{\theta}$ is biased.

\begin{align*}
    E(Y_{(1)}) &= \int_{\infty}^{\infty} y f_{(1)}(y) \ d y\\
               &= \int_{\beta}^{\infty} y 3n\beta^{3n}y^{-(3n+1)} \ dy\\
               &= 3n\beta^{3n}\lim_{h\to \infty}\int_{\beta}^{h} y^{-3n} \ dy\\
               &= 3n\beta^{3n}\lim_{h\to \infty}\left[\frac{y^{-3n+1}}{-3n+1}\right]_\beta^h\\
               &= 3n\beta^{3n}\left[-\frac{\beta^{-3n+1}}{-3n+1}\right]\\
               &= \left(\frac{3n}{3n-1}\right)\beta
\end{align*}
So we know, by definition, $\hat{\theta}$ is a biased estimator. Now we were asked to find the bias, which is easily to calculate considering this definition:

$$B(\hat{\theta}) = E(\hat{\theta})-\theta$$

Therefore:
\begin{align*}
    B(\hat{\beta})&=E(\hat{\beta})-\beta\\
                   &= \left(\frac{3n}{3n-1}\right)\beta - \beta \\
                   &= \left(\frac{1}{3n-1}\right)
\end{align*}
b.
We proceed to calculate $E(\hat{\beta}^2)$ (which later will be necessary):
\begin{align*}
E(\hat{\beta}^2)= E(Y_{(1)}^2) &= \int_{\infty}^{\infty} y^2 f_{(1)}(y) \ d y\\
               &= \int_{\beta}^{\infty} y^2 3n\beta^{3n}y^{-(3n+1)} \ dy\\
               &= 3n\beta^{3n}\lim_{h\to \infty}\int_{\beta}^{h} y^{-3n+1} \ dy\\
               &= 3n\beta^{3n}\lim_{h\to \infty}\left[\frac{y^{-3n+2}}{-3n+2}\right]_\beta^h\\
               &= 3n\beta^{3n}\left[-\frac{\beta^{-3n+2}}{-3n+2}\right]\\
               &= \left(\frac{3n}{3n-2}\right)\beta^2
\end{align*}
Finally, we have:
\begin{align*}
    MSE(\hat{\beta})&=E[(\hat{\beta}-\beta)^2]\\
                    &= E[\hat{\beta}^2-2\hat{\beta}\beta +\beta^2]\\
                    &= E(\hat{\beta}^2)-2\beta E(\hat{\beta})+ E(\beta^2)\\
                    &= E(\hat{\beta}^2)-2\beta E(\hat{\beta})+ \beta^2\\
                    &= \left(\frac{3n}{3n-2}\right)\beta^2 -2\beta^2\left(\frac{3n}{3n-1}\right) +\beta^2\\
                    &= \left(\frac{3n}{3n-2}\right)\beta^2 -\left(\frac{3n}{3n-1}\right)\beta^2\\
                    &= \left(\frac{3n}{3n-2}-\frac{3n}{3n-1}\right)\beta^2\\
                    &= \left(\frac{2}{(3n-1)(3n-2)}\right)\beta^2
\end{align*}
A nicest version of this problem (in Spanish) is at this repository on Github: Parcial 3 - Estadística Matemática.
