Find RV distribution and prove its convergence to zero

Question

Let $X_n$ be a sequence of independent and uniformly distributed over $[0; 1]$ random variables. The task is to find distribution of the following random variable: $m_n=\min\{X_1, X_2, \ldots, X_n\}$ and also proof that it converges almost surely to zero, in other words: $$\Pr(\lim_{n \rightarrow \infty}m_n=0)=1$$

Answer 1

For any $x\in[0,1]$ we have $$ \left\{\min_{1\leqslant i\leqslant n} X_i \geqslant x \right\} = \bigcap_{i=1}^n \{X_i\geqslant x\}. $$ It follows that \begin{align} \mathbb P(m_n\geqslant x) &= \mathbb P\left(\bigcap_{i=1}^n \{X_i\geqslant x\} \right)\\ &= \prod_{i=1}^n \mathbb P(X_i\geqslant x)\\ &= \mathbb P(X_1\geqslant x)^n\\ &= (1-x)^n, \end{align} and so the distribution function of $m_n$ is $$ F_n = (1 - (1-x)^n)\cdot\mathsf 1_{[0,1]}(x) + \mathsf 1_{(1,\infty)}(x) $$ and the density $$ f_n = n(1-x)^{n-1}\cdot\mathsf 1_{[0,1]}(x). $$ Recall that $Z_n\to Z$ almost surely if and only if $$ \mathbb P\left(\limsup_{n\to\infty} \{|Z_n-Z|>\varepsilon\}\right) = 0 $$ for all $\varepsilon>0$. Let $\varepsilon\in(0,1)$, then $$ \sum_{n=1}^\infty \mathbb P(m_n>\varepsilon) = \sum_{n=1}^\infty (1-\varepsilon)^n = \frac{1-\varepsilon}\varepsilon<\infty, $$ so by the Borel-Cantelli lemma $$ \mathbb P\left(\limsup_{n\to\infty} \{X_n>\varepsilon\}\right)=0, $$ and hence $m_n\to 0$ almost surely.

Answer 2

$$ \min\{X_1,\ldots,X_n\} \ge x \text{ if and only if } \big[X_1\ge x\ \& \cdots \ \&\ X_n\ge x\big]. $$ If you can find the probability of that, then that gives you the distribution of the minimum.