Show that $Y_1$, the first order statistic is a consistent estimator of the parameter $\alpha $ of a uniform distribution with $\beta = \alpha+1$
Here $f_{Y(1)}(y_1)= \begin{cases} n*[1-(y_{(1)}-\alpha)]^{n-1}, & {\alpha <Y_{(1)}< \beta} \\ 0, & \text{otherwise} \\ \end{cases}$
By definition If $P[|Y_{(1)}-\alpha|<\epsilon]$ approaches 1 as n goes to infinity for all $\epsilon >0$ then $Y_1 $is a consistent estimator for $\alpha. $ Thus,
$P[|Y_{(1)}-\alpha|<\epsilon]$=$P[\alpha - \epsilon< Y_{(1)}<\alpha +\epsilon]$.
My question is with the limits of integration.
What is the lower and upper limits of $\int n*[1-(y_{(1)}-\alpha)]^{n-1} dy_{(1)}$ be? should it be from $\alpha$ to $\alpha + \epsilon$ Or from $\alpha $ to $\beta$.
I think it should be from because this should work for any value of $\epsilon $ and if I put limit as $\alpha$ to $\alpha + \epsilon$ then this would violate $\beta = \alpha+1$ for a particular $\epsilon$.
Please tell me what the limits should be to show that this is a consistent estimator

  • $\begingroup$ How exactly do you define probabilities "$P$" in terms of integrals in the first place? Giving some thought to that might reveal the answer to you. $\endgroup$
    – whuber
    Commented Apr 7, 2014 at 15:53
  • $\begingroup$ Sorry I don't understand what you mean $\endgroup$
    – clarkson
    Commented Apr 8, 2014 at 1:34

1 Answer 1


You must take into consideration the fact that you are examining a distribution with a closed interval as support. This means that $$\alpha \le Y_{(1)} \le \beta = \alpha +1 \Rightarrow 0\le Y_{(1)} -\alpha \le 1$$ so you already now that $\sup|Y_{(1)}-\alpha| = 1$ and so for $\epsilon >1 $ the condition required for consistency holds.

So you need only to prove

$$\lim_{n\rightarrow \infty}P[|Y_{(1)}-\alpha|<\epsilon] =1\,\qquad 0<\epsilon \le 1 $$

Furthermore, since $0\le Y_{(1)} -\alpha \le 1$ we have that $$P[|Y_{(1)}-\alpha|<\epsilon]= P[Y_{(1)}-\alpha<\epsilon] = P[Y_{(1)}<\alpha +\epsilon]$$

Then use the cumulative distribution function to express this probability in the usual way (no $\epsilon$ under consideration here will send you out of the support) and consider the limit as $n\rightarrow \infty$, for $0<\epsilon \le 1$.

  • $\begingroup$ Thanks for the great explanation .In a similar way I was able to show that $Y_{(n)}$ is also a consistent estimator for $\beta$.There $\alpha - \beta <= Y_{(n)}-\beta<=0$ and thus $0<=|Y_{(n)}-\beta|<=1$.Thus $sup |Y_{(n)}-\beta|=1$.And for $0<\epsilon<1 $ case $P[|Y_{(n)}-\beta|<\epsilon]$ since $Y_{(n)}-\beta$<=0 this probability becomes $P[-(Y_{(n)}-\beta)<\epsilon]$ $\endgroup$
    – clarkson
    Commented Apr 8, 2014 at 2:22

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