# Convergence in probability of minimum

This is a homework problem. Suppose we have a random sample $X_1,\ldots,X_n \overset{iid}{\sim} F$ with density $f(x) = 2(x-\theta)$ for $x\in (\theta,\theta+1)$. Let $X_{(1)} = \min{\{X_1,\ldots,X_n\}}$. Show $X_{(1)} \overset{P}{\longrightarrow} \theta$.

My attempt:

A few calculations first:

$F(x) = (x-\theta)^2$ for $x\in(\theta,\theta+1)$

$f_{X_{(1)}} = n(1-F(x))^{n-1}f(x) = 2n(x-\theta)(1-(x-\theta)^2)^{n-1}$

\begin{align*} P(|X_{(1)} -\theta| > \epsilon) &= P(|X_1-\theta|> \epsilon, \ldots, |X_n-\theta|> \epsilon)\\ &= P(|X_1-\theta|> \epsilon)^n\\ &= \end{align*}

Not exactly sure where to go from here. Any suggestions would be appreciated.

• Have you noticed $P(|X_{(1)} -\theta| > \epsilon) = P(X_{(1)} -\theta > \epsilon)$ ? – Stéphane Laurent Nov 4 '13 at 20:57
• Do you know of an inequality that allows to bound your last equation from above? – Michael M Nov 4 '13 at 21:32
• I see how $P(|X_{(1)}-\theta|>\epsilon) = P(X_{(1)}-\theta>\epsilon)$ but I don't see how this probability goes to zero. – bdeonovic Nov 4 '13 at 22:16
• @michael-mayer perhaps you are referring to chebychev inequality? I don't see how that can help. – bdeonovic Nov 4 '13 at 22:45
• Your solution is obviously easier than mine! – Michael M Nov 5 '13 at 12:45

## 1 Answer

I think I got it from the hints you guys gave:

\begin{align*} P(|X_{(1)} -\theta| > \epsilon) &= P(X_{(1)} -\theta > \epsilon)\\ &= P(X_{(1)} > \epsilon + \theta) \\ &= P(X_1 > \epsilon+\theta)^n\\ &= (1-\epsilon^2)^n \end{align*}

and $\lim_{n\to\infty}{(1-\epsilon^2)^n}=0$ for $\epsilon \in (0,1)$.