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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?

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    $\begingroup$ 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
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    $\begingroup$ 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
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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

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

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