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 $\limsup\limits_{n\to \infty} X_n = \infty$ almost surely. Hint: Use sets of the form $\{ X_n \geq b_n \}$ for a suitable $b_n \to \infty$.

I've tried this so far (unsure + might be incorrect): I have to show almost sure convergence, so using this Definition $$P(\limsup_{n\to \infty} X_n = X ) = 1$$ Lim Sup equals to the following (not sure about handling with infinity in the next step) $$P(\inf_{k \in \mathbb n} \sup_{n\geq k}{X_n} = X ) = 1$$ If I now insert the $\{ X_n \geq b_n\}$: for $b_n := \sqrt n$ and use continuity from above from the lim sup, it equals(?) to $$\lim_{n \to \infty} P(\{X_n \geq b_n \} ) = 1 \Rightarrow \lim_{n \to \infty} \frac{n-\sqrt n}{n} = 1 \Rightarrow \text{almost sure convergence}$$ Is this (somehow) correct?

  • 1
    $\begingroup$ I don't understand your logic in the last line. You would like to show that $X_n \geq b_n$ holds for infinitely many $n$ with probability one, and using the independence of events $\{X_n \geq b_n\}$ you can safely use the second Borel-Cantelli lemma. $\endgroup$ – Sangchul Lee Mar 11 '18 at 1:41
  • $\begingroup$ I used the continuous uniform distribution for $P( \{X_n \geq b_n \}) =\mu(\{X_n \geq b_n \})/\mu(n)$, is that also fine? $\endgroup$ – Tearsdontfalls Mar 11 '18 at 2:07
  • 1
    $\begingroup$ What I am concerned is not the computation of $\lim_{n\to\infty} \mathbb{P}[X_n \geq b_n] = 1$ for $b_n = \sqrt{n}$. Rather, I am questioning on how $\lim_{n\to\infty} \mathbb{P}[X_n \geq b_n] = 1$ is related to $P[\limsup_{n\to\infty} X_n = \infty] = 1$. $\endgroup$ – Sangchul Lee Mar 11 '18 at 2:12
  • $\begingroup$ Ah, me too. Thought that $\{X_n \geq b_n \}$ represents the supremum part, so that inf remains and is continuous from above, therefore I can move the limit out of the $P(..)$. $\endgroup$ – Tearsdontfalls Mar 11 '18 at 2:29

If $b_n \to \infty$, then we find that

$$ \{ \limsup_{n\to\infty} X_n = \infty \} \supseteq \{ X_n \geq b_n \text{ infinitely often} \} = \bigcap_{N\geq 1} \bigcup_{n\geq N} \{ X_n \geq b_n \}. $$

So it suffices to show that $\mathbb{P}[X_n \geq b_n \text{ infinitely often}] = 1$ for some suitable choice of $(b_n)$. Now

\begin{align*} 1 - \mathbb{P} \left[ \bigcap_{N\geq 1} \bigcup_{n\geq N} \{ X_n \geq b_n \} \right] &= \mathbb{P} \left[ \bigcup_{N\geq 1} \bigcap_{n\geq N} \{ X_n < b_n \} \right] \\ &\leq \sum_{N=1}^{\infty} \mathbb{P} \left[\bigcap_{n\geq N} \{ X_n < b_n \} \right] \\ &= \sum_{N=1}^{\infty} \prod_{n\geq N} \mathbb{P}[X_n < b_n ] \end{align*}

Now if $(b_n)$ is chosen so that $b_n \to \infty$ and $\mathbb{P}[X_n \geq b_n] \to 1$, then you can easily check that this bound is exactly $0$, hence proving the desired claim.

Addendum. For a better resolution, let $M_n = \max\{X_1,\cdots,X_n\}$ and notice that

$$ \mathbb{P}[M_n \leq x] = \prod_{i=1}^{n} \mathbb{P}[X_n \leq x] = \prod_{i=1}^{n} \min\left\{ 1, \frac{x}{n} \right\} = \dfrac{x^{n-\lfloor x\rfloor} \cdot\lfloor x\rfloor!}{n!} \qquad \text{for } x \in (0, n). $$

Now if we plug $x = n - z\sqrt{n}$, then for each fixed $z \geq 0$ it is not hard to prove that

$$ \mathbb{P}[M_n \leq n - z\sqrt{n}] \xrightarrow[n\to\infty]{} e^{-z^2/2}. $$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.