Let $(X_{n})_{n}$ be a sequence of random variables that are identically distributed on $\mathcal{U}(0,1)$. Furthermore, let $\alpha > 1$.

Show that $n^{\alpha}X_{n} \xrightarrow{n \to \infty} +\infty$ almost surely

My idea: Let $\epsilon > 0$

$\sum_{n \in \mathbb N}P(n^{\alpha}X_{n} \leq \epsilon)=\sum_{n \in \mathbb N}P(X_{n} \leq \frac{\epsilon}{n^{\alpha}})=\sum_{n \in \mathbb N}\frac{\epsilon}{n^{\alpha}}<\infty$ since $\alpha > \infty$

Therefore, by Borel-Cantelli, $P(\limsup_{n \to \infty}\{n^{\alpha}X_{n} \leq \epsilon\})=0$

This means, $\exists N \in \mathbb N$ so that $n^{\alpha}X_{n} > \epsilon$ for all $ n \geq N$ almost surely BUT $\epsilon>0$ was chosen arbitrarily and therefore $(n^{\alpha}X_{n})_{n\in \mathbb N}$ is not bounded a.s.

$\Rightarrow$ $n^{\alpha}X_{n} \xrightarrow{n \to \infty} +\infty$ a.s.

Is my reasoning sound?


1 Answer 1


Your logic is essentially fine.

I would probably rename $\epsilon$ to something else, since ultimately you want to think of it as being a large positive number.

Also, when you write $$\sum_{n \in \mathbb N}P(X_{n} \leq \frac{\epsilon}{n^{\alpha}})=\sum_{n \in \mathbb N}\frac{\epsilon}{n^{\alpha}}$$

This should be $\le$ instead of $=$, because it is possible for $\epsilon/n^\alpha$ to be larger than 1, in which case $P(X_n \le \epsilon/n^\alpha)$ is simply 1. But you still get what you want.


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