# Show that every finite family of random variables is tight.

A family of random variables $(X_i)_{i \in \mathcal{I}}$ is said to be $tight$ if for every $\epsilon>0$ there exist a compact set $K_\epsilon$ such that

$$\displaystyle\sup_{i \in \mathcal{I}}\mathbb{P}(X_i\notin K_{\epsilon})<\epsilon$$

Show that every finite family of random variables is tight.

So I tried to star with only one random variable, say $X$. So, I have that

$\mathbb{P}(X\notin K_\epsilon)=\mathbb{P}(\{\omega : X(\omega) \notin K_\epsilon \})$

And I get stuck right there. I don't know how to construct my compact set $K_\epsilon$ because I don't know how to work with the event $\{\omega : X(\omega) \notin K_\epsilon \}$. Any help?

Hint: Countable additivity shows that $$\lim_{N\to\infty}P(|X|\le N)=1.$$That takes care of the case of one random variable; the case of finitely many random variables follows.

EDIIT: The discrepancy between this and the other answer is, it seems to me, explained by the fact that there's some confusion over what the word "tight" means. With the standard definition it's a family of probability measures that's tight, not a family of random variables. What I wrote above is correct assuming the definition in the OP; of course it's not correct with the standard definition. The question makes no sense with the standard definition.

The two are related. If $L(X)$ is the distribution of the random variable $X$ then a family of (real-valued) random variables is tight as in the OP if and only if the corresponding family of $L(X)$'s is tight by the standard definition. Of course $L(X)$ is a measure on $\Bbb R$, which does satisfy the conditions mentioned in the other answer,

This isn't true in general; some (fairly light) assumptions are necessary on the topology of your probability space $\Omega$, and your family of events $\mathcal{F}$.

For instance, suppose I let $\Omega = [0,1]$ with the discrete topology, $\mathcal{F}$ to be Lebesgue measurable sets, and $\mathbb{P}$ to be the Lebesgue measure. Under the discrete topology, only finite sets are compact, so for any compact set $K$, $\mathbb{P}(X \notin K) = 1$, implying that $\mathbb{P}$ isn't tight.

Typically, we assume that $\Omega$ is $\sigma$-compact or (sometimes) Polish and take $\mathcal{F}$ to be the Borel sets. Most commonly, $\Omega$ is $\sigma$-compact and $\mathcal{F}$ is the Borel sets, in which case we can use the solution David posted.

• This made no sense to me at first. So I looked it up. Ah, you're assuming the standard definition of "tight", that being a property of a family of probability measures - then of course what you say is correct. But the OP defined the word differently, saying that a family of random variables is "tight" if etc. Let's say that notion is tight_OP. Assuming just that "random variable" means real-valued random variable, there are no extra assumptions needed to show that a finite family of random variables is tight_OP. (A family of RVs is tight_OP if their distributions are a tight family on \$\Bbb R) – David C. Ullrich Nov 5 '15 at 13:35