Find RV distribution and prove its convergence to zero 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: $$\Pr(\lim_{n \rightarrow \infty}m_n=0)=1$$
 A: $$
\min\{X_1,\ldots,X_n\} \ge x \text{ if and only if } \big[X_1\ge x\ \& \cdots \ \&\ X_n\ge x\big].
$$
If you can find the probability of that, then that gives you the distribution of the minimum.
A: 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.
