# Probability Limit

$Y_{1},Y_{2},\ldots$ are i.i.d. random variables (on some ($\Omega$,F,$P$)) whose common distribution function is $$F(y)=\begin{cases} 0,& y<0\\y^{\alpha}, & 0\leq y\leq 1 \\ 1,& y>1 \end{cases}$$

where $\alpha \in (0,\infty )$. For any $\beta\in (0,\infty )$, determine $P( \left\{w:n^{\beta}Y_{n}(w))\to \infty \text{ as } n \to \infty \right\})$.

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It suffices to find the probability of the complementary event, $$\Pr\left(\varliminf_{n \to \infty} \ n^\beta Y_n < \infty\right) = \Pr(n^\beta Y_n < M \mbox{ infintely often for some} \ M).$$ For fixed $M$, the events $\{[n^\beta Y_n < M]: n \ge 1\}$ are independent and so by Borel Cantelli the probability that they occur infinitely often is $0$ or $1$ according as $\sum P(n^\beta Y_n < M)$ converges or diverges. Knowing what happens for any particular $M$, you can parlay this into an argument that either $\Pr(\varliminf_{n \to \infty} n^\beta Y_n < \infty) = 0$ or $1$ depending on whether you have convergence for all $M$ or not (which in turn will depend on $\alpha$ and $\beta$).