Tightness of random variables

Question

in my probability theory course, we defined a sequence of random variables $(X_i)_{i=1}^{\infty}$ to be tight if for all $\epsilon >0$, there is a constant $M$ s.th. $P(|X_n|>M) < \epsilon$ for all $n \in \mathbb{N}$.

I have seen the following criteria for tightness/non-tightness and I was wondering whether they are true or not:

• $(X_i)_{i=1}^{\infty}$ tight if there exists $M$ s.th. lim$_{n\to \infty} P(|X_n|>M) = 0$
• $(X_i)_{i=1}^{\infty}$ not tight if for all $M$ lim$_{n\to \infty} P(|X_n|>M)>0$

I am quite sure that the first one is right (we get the condition that the probability is smaller than $\epsilon$ for all but finitely many n and can take the maximum of all the remaining $M$ necessary for the finitely many n). For the second one I am not so sure, I was thinking that maybe we need some uniform bound, i.e. lim$_{n\to \infty}P(|X_n|>M) \geq \epsilon>0$. I know that the criterion is right if the limit is equal to 1.

Answer 1

Your first proposal is not equivalent to tightness. A family consisting of just copies of a single $N(0,1)$ random variable is tight but does not satisfy that definition. However your first proposal is sufficient, as your argument shows.

As for your second proposal, if you replace $\lim$ with $\limsup$ (which is necessary because the limit you're asking for doesn't generally even exist) then you recover the negation of the first proposal. This is a common pitfall; $\neg \left ( \lim_{n \to \infty} a_n=a \right )$ is really equivalent to $\limsup_{n \to \infty} |a_n-a|>0$, not $\lim_{n \to \infty} a_n \neq a$.