# Show $\liminf\limits_{n\to \infty} X_n = 0$ almost surely for $X_n$ uniform on $(0,n)$ and $(X_n)$ independent

Let $(X_n)_{n \in \mathbb N}$ a sequence of independent random variables, where $X_n$ is uniformly continuous distributed across $(0,n)$. Show that $\liminf_{n\to \infty} X_n = 0$ almost surely. Hint: Use sets of the form $\{ X_n \leq c_n \}$ for a suitable $c_n \to 0$.

I've tried this so far: (Probably wrong because first line takes a too large subset) $$\liminf \{X_n \leq c_n\} \subseteq \{\liminf X_n \leq c_n\}$$ $$(\liminf \{X_n \leq c_n\})^c=\limsup \{X_n > c_n\}$$ Therefore: $$P(\liminf X_n =0) = 1-P(\limsup \{X_n > c_n\})$$ Using Borel Cantelli Lemma for $P(\limsup \{X_n > c_n\}$ for independent $\{X_n> c_n\}$: $$\sum_{n=0} ^\infty {P(\{X_n > c_n \})} = \sum_{n=0} ^\infty {\frac{n -c_n}{n}} = \sum_{n=0} ^\infty {1 -\frac{c_n}{n}} \to \infty$$ if $c_n \to 0$, therefore that would contradict the exercise by: $$P(\liminf X_n =0) = 1-1=0$$

2nd Try: $$P(\liminf \{X_n \leq c_n\}) = P(\bigcap \bigcup \{X_n \leq c_n\}) \stackrel{decreasing}{=}\lim P(\bigcup_{m \geq n} \{X_m \leq c_m\})$$ $$=\lim 1 - \prod_{m=n}^\infty {(1 - 1/m)} \to 1$$

• You are confusing the events $$\{\liminf X_n \leq c_n\}$$ and $$\liminf \{X_n \leq c_n\}$$ If you think hard about what the symbols mean, you should see these are not equal. – Did Jul 10 '18 at 20:01
• Could you please tell us what you try to prove in your "2nd try" ? Is my answer not clear ? – Surb Jul 11 '18 at 13:56
• Sorry, tried to prove the exercise in a different way. Wrote that in the same time while you answered. – Tearsdontfalls Jul 11 '18 at 18:58

Let $\varepsilon>0$ and set $$A_n:=A_n(\varepsilon)=\{X_n<\varepsilon\}.$$ Since $X_n$ are independent, then so is $A_n$'s. Since $\mathbb P(A_n)=\frac{\varepsilon}{n}$, we have that $$\sum_{n=1}^\infty \mathbb P(A_n)=\infty .$$

Using second Borel-Cantelli lemma, we get $$\mathbb P\{A_n\ \ i.o\}=\mathbb P\{X_n<\varepsilon\ \ i.o.\}=1.$$ Therefore, $$\mathbb P\left\{\liminf_{n\to \infty }X_n\leq \varepsilon\right\}=1.$$ Since $\varepsilon>0$ is unspecified, the claim follow.

Small exercise :-)

If you want, you can prove that if $X_n$ are i.i.d. uniform $[0,n^\alpha ]$ for a $\alpha >1$, then $$\liminf_{n\to \infty }X_n=+\infty \ \ \ a.s.$$

• I think it would be more correct to first write $$\{\liminf_{n\to\infty} X_n \leq \varepsilon\} = \{ X_n \leq \varepsilon \, \text{i.o.}\}$$ as from your answer it sounds like you claim $$\{\liminf_{n\to\infty} X_n < \varepsilon\} = \{ X_n < \varepsilon \, \text{i.o.}\}$$ , which is not true. – Calculon Jul 12 '18 at 8:06
• @Calculon: Not at all (we don't have equality). But Indeed, there is just a small typo that I corrected. – Surb Jul 12 '18 at 8:20
• Not sure what you mean by "we don't have equality". The first equality I wrote is true, right? – Calculon Jul 12 '18 at 8:22
• @Calculon: Not really, It should be more $\{X_n\leq \varepsilon \ \ i.o.\}\subset \{\liminf_{n\to \infty }X_n\leq \varepsilon\}$. But I can also use $\{X_n<\varepsilon\ \ i.o.\}$. But indeed, at the end I should have wrote $\mathbb P\{\liminf_{n\to \infty }X_n\leq \varepsilon\}$ and not $\mathbb P\{\liminf_{n\to \infty }X_n< \varepsilon\}$ – Surb Jul 12 '18 at 8:24
• @Calculon: I can't read your statement. Take $X_n(\omega )=\frac{1}{n}+\frac{1}{2}$ for all $\omega$. We have that $$\liminf_{n\to \infty }X_n(\omega )\leq \frac{1}{2}$$ for all $\omega$, but $$\left\{X_n\leq \frac{1}{2}\ \ i.o.\right\}=\emptyset.$$ – Surb Jul 12 '18 at 9:15