Let $U_1,U_2,...U_n,$ be iid samples from uniform distribution $U(0,1)$. And the order Statistic:
\begin{equation}
U_{n,1}\leq U_{n,2}\leq ...\leq U_{n,n},
\end{equation}
(1) Proof $U_{n,1}\rightarrow0$ a.s.
(2) Given $m$, calculate the limitation of $U_{n,m}.$
My ideas so far:
For question (1), the density function of $U_{n,1}$ is $f_{U_{n,1}}(x)=n(1-x)^{n-1}.$
And if $x\neq0,$ then $\lim_{n \to \infty} f_{U_{n,1}}(x)=0.$
Which means $U_{n,1}\rightarrow0$ a.s.
And question (2) can be solved in the same way.
I don't know whether my proof is right. If not, how to proof r.v.s $X_n\rightarrow0$ a.s. directly?
Thanks in advance for any tips or help in general.
For the a.s. convergence of $U_{n,1}$, you may use the Borel-Cantelli lemma. Specifically, for $0<\epsilon\le 1$, $$ \sum_{n\ge 1}\mathsf{P}(U_{n,1}>\epsilon)=\sum_{n\ge 1}(1-\epsilon)^n<\infty. $$ Thus, $\mathsf{P}(U_{n,1}>\epsilon\text{ i.o.})=0$.