Estimators, sufficiency, consistency, and bias 
A random variable is said to have the Pareto distribution with parameters $\alpha$ and $\beta$, $P(\alpha, \beta)$, if its cumulative distribution function is given by
$$F(x)= 1 - (\frac{\beta}{x})^{\alpha},$$ if $x \geq \beta$, and $0$ otherwise.
a) If $X_1, ... ,X_n$ are iid $P(\alpha, \beta)$ find the probability density function of min$(X_1, ... , X_n)$.

Let's rewrite the random variables as $Y_1, ... ,Y_n$, where $Y_1 < Y_2 < ... < Y_n$.
Then $F_{Y_1}(y)  = P(Y_1 \leq y, ... , Y_n \leq y)$
$$=  1 - P(Y_1 \geq y, ... , Y_n \geq y)$$ $$= 1 - P(Y_1 \geq y) ... P( Y_n \geq y)$$ $$= 1 - [P(Y_1 \geq y)]^n$$
So it suffices to find $P(Y_1 \geq y)$. But $P(Y_1 \geq y) = 1 - P(Y_1 \leq y) = 1 - (1 - (\frac{\beta}{y})^{\alpha}) = (\frac{\beta}{y})^{\alpha}$
So, $f_{Y_1}(y) = (\alpha n)\beta^{\alpha n}y^{-\alpha n -1}$

b) If $X_1, ... ,X_n$ are iid $P(\alpha = 3, \beta)$, show that $\beta = min (X_1, ... ,X_n)$ is:
i. a biased estimator for $\beta$. Compute the bias of $\hat{\beta}$

$E(Y_1) = \int^{\infty}_{\beta} 3n \beta^{3n} y^{-3n} dy$ $$=lim_{u \rightarrow \infty} [\int^u_{\beta} 3n \beta^{3n} y^{-3n} dy]$$ $$=lim_{u \rightarrow \infty} [\frac{3n}{-3n+1} \beta^{3n} y^{-3n+1} ]^u_{\beta}$$ $$= \frac{-3n}{-3n+1 \beta}$$
Now we compute the bias
$E(Y_1 - \beta)$ $$= E(Y_1) - E(\beta)$$ $$= [\frac{-3n}{-3n+1 \beta}] - [\int^{\infty}_{\beta} 3n\beta^{3n+1}y^{-3n-1} dy] $$ $$= [\frac{-3n}{-3n+1 \beta}] - [\frac{3n \beta^{3n+1} y^{-3n}}{-3n}]^{\infty}_{\beta}= [\frac{-3n}{-3n+1 \beta}] +\beta$$

ii. a  consistent estimator for $\beta$

This one seems confusing to me. According to the textbook, an estimator is consistent if it approaches the estimated value in probability. However, we have $\lim_{n \rightarrow \infty} P[|Y_1 - \beta|  < \epsilon]$...which doesn't seem right to me, because there is no $n$ in the expression $|Y_1 - \beta|$. So I'm guessing if I'm on the wrong track...

iii. a sufficient statistic for $\beta$.

It is sufficient by Neyman's theorem, because we can write the pdf as $f_{Y_1}(y) = [3n] [\beta^{3n}y^{-3n-1}]$.
Are my answers correct? If not, can you give me a hint? Also, can you help me with part ii?
Thanks in advance
 A: There is a mistake in computing the expected value of the minimum order statistic. Specifically,
$$E(Y_1) = \int^{\infty}_{\beta} 3n \beta^{3n} y^{-3n} dy = [\frac{3n}{-3n+1} \beta^{3n} y^{-3n+1} ]^{\infty}_{\beta}$$
$$= 0 - \frac{3n}{-3n+1} \beta^{3n} \beta^{-3n+1} = \frac{-3n}{-3n+1} \beta $$
which can also be written 
$$E(Y_1) = \frac 1{1-\frac 13 n^{-1}}\beta $$
I will provide an alternative way to prove consistency, and leave for you to deal with the usual way.  A set of sufficient conditions for consistency is
$$ \lim_{n\rightarrow \infty} E(Y_1) = \beta,\qquad  \lim_{n\rightarrow \infty} \operatorname{Var}(Y_1) = 0$$ 
(they are only sufficient because the variance of an estimator may not exist). We have
$$E(Y^2_1) = \int^{\infty}_{\beta} 3n \beta^{3n} y^{-3n+1} dy = [\frac{3n}{-3n+2} \beta^{3n} y^{-3n+2} ]^{\infty}_{\beta} =- \frac{3n}{-3n+2} \beta^{3n}\beta^{-3n+2} = \frac{-3n}{-3n+2}\beta^2$$
and so 
$$\operatorname{Var}(Y_1) = E(Y^2_1) - [E(Y_1)]^2 = \frac{-3n}{-3n+2}\beta^2 - \left(\frac{-3n}{-3n+1}\right)^2 \beta^2$$
$$= \left [\frac 1{1-\frac 23 n^{-1}}- \left(\frac 1{1-\frac 13 n^{-1}}\right)^2\right]\beta^2  = \left [\frac {(1/9)n^{-2}} {\left(1-\frac 23 n^{-1}\right)\cdot  \left(1-\frac 13 n^{-1}\right)^2}\right]\beta^2$$
Then it is evident that the sufficient conditions for consistency are satisfied. Intuitively, as the sample size goes to infinity, it "becomes certain" that the smallest realized value will be equal to the theoretical minimum, and so the minimum order statistic stops being a random variable and it becomes a constant (hence the zero variance).
