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My question is:

If $\{X_n\}_{n\in\mathbb{N}}$ represents a class of uniformly integrable random variables, is $\{X_n(w)\}_{n\in\mathbb{N}}$ a bounded set for a fixed sample path $w$?

My thoughts:

Uniform integrability implies that $$ \lim_{K\to\infty} \sup_{n\in\mathbb{N}} E[|X_n| : |X_n| > K] = 0. $$ So we know that the probability of larger values becomes smaller. The limit above goes to 0 but this doesn't mean it will reach 0.

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No.

If, say, $X_n$ is i.i.d. $N(0,1)$, most paths will have this property: for $x\in\mathbb R$, $\mbox{card}(\{i\le n: X_n(\omega)>x\})/n\to 1-\Phi(x)$, by the SLLN. That means, for almost all $\omega$, the terms in $X_n(\omega)$ exceed $x$ infinitely often, that is, for infinitely many values of $n$. Hence, for almost all $\omega$, the sequence $X_n(\omega)$ is unbounded.

Here is a different argument along these lines, based on the same $X_n$: There is an increasing unbounded sequence $a_n$, such that $\sum_{n\ge1} (1-\Phi(a_n)) = \infty$. (Such as $a_n = \sqrt{(2-\epsilon)\log n}$, for $\epsilon\in(0,1)$, as can be seen with a bit of calculus. Another example: the solution, say, of $1-\Phi(a_n) = 1/n$.) By the second Borel-Cantelli lemma, the divergence of $\sum_n P(X_n>a_n) $ implies that with probability one, $X_n>a_n$ for infinitely many $n$.

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    $\begingroup$ By the same reasoning, for almost all $\omega$, the set $\{X_n(\omega)\mid n\in\mathbb N\}$ is dense in $\mathbb R$. $\endgroup$
    – Did
    Nov 25, 2017 at 10:59

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