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?