Boundedness in probability

Suppose we have a set $M$ of random variables on a probability space. Then we defined boundedness of $M$ as,

$M$ is bounded if $\sup_{X\in M}P(|X|>N) \to 0$ as $N \to \infty$.

This definition means that the measure of the set where the elements of $M$ are big is very small, in fact tends to zero. I have three questions to this definition and some further conclusions.

Now suppose we have a unbounded set. My questions are:

1. Unbounded would mean, that for every $N>0$ there exists an $\epsilon >0$ and a $X \in M$ such that $P(|X| > N) \ge \epsilon$. Is this conclusion right?
2. If theres a sequence $(X_n)$ of random variables, unbounded and positive, then there's a subsequence $(X_{n_k})$ and a $\lambda>0$ such that $P(|(X_{n_k})| > k)\ge \lambda$ for every $k \in \mathbb{N}$.

My observations so far to 2. After the comment of Srivatsan (see below), we therefore have:$\exists \epsilon > 0$ sucht that for all $N>0$ exists a $X_n$ such that $P(X_n(\omega) > N) \ge \epsilon$ Put $N=1$, hence there is a $X_{n_1}$ sucht that $P(X_{n_1} > 1 ) \ge \epsilon$. Now put $N=2$, hence there is a $X_i$ such that $P(X_i > 2) \ge \epsilon$. Now the problem is, why do I know that $i> n_1$ ? Otherwise it isn't a subsequence.

hulik

-
"Unbounded" means that there exists an $\varepsilon > 0$ such that for all $N$, there exists $X \in M$ with $P(|X| \geqslant N) \geqslant \varepsilon$. – Srivatsan Dec 19 '11 at 11:40
In the question 2, you probably meant "$P(|X_{n_k}|>k)\geq\lambda$", since with $X_k=k/2$ we get a counter-example. But what does question 3 become? – Davide Giraudo Dec 19 '11 at 12:43
You're right, I edited that. I'm very sorry but I don't know what you mean with: " but what does question 3 become?" – user20869 Dec 19 '11 at 12:46
If I understand well, question 3 is: do we have for almost every $\omega$, and infinitely many $k$, $|X_{n_k}(\omega)|\geq k$. It's true by Borel-Cantelli lemma if $X_{n_k}$ are independent, but not in general, for example taking $\Omega=[0,1]$ with Lebesgue measure and $X_n=(n+1)\mathbf 1_{[0,\lambda)}$. – Davide Giraudo Dec 19 '11 at 12:54
I do not know that the $(X_n)$ are independent. The reason for question 3 was $(1)$. As I worte, I try to prove $(1)$ and I do not see why this is true unless we know that for almost all $\omega$ and infinitly man $k$, it's true that $|X_{n_k}(\omega)| \ge k$. Since I can not assume independence, I have to find a different arguement why $(1)$ is true. – user20869 Dec 19 '11 at 13:00

(1) is not correct. Let $Y$ be a single random variable such that, for every $N$, we have $P(|Y| > N) > 0$; for example, a normal random variable. (I would call such a random variable "unbounded", but that conflicts with your terminology.) Then let $M = \{Y\}$. $M$ is bounded in your sense (exercise: verify this, using continuity of probability) but it is still true that for every $N$ there exists $X \in M$ (namely $X=Y$) and $\epsilon > 0$ (namely $\epsilon = \frac{1}{2}P(|Y|>N)$) such that $P(|X| > N) > \epsilon$.

Srivatsan's comment above gives a corrected statement; I just wanted to show explicitly that the statement in the original question (for every $N>0$ there exists an $\epsilon >0$ and $X \in M$ such that $P(|X|>N) \ge \epsilon$) is not correct.

Regarding (2): You know that there is an $\epsilon$ such that for any $N$ there is an $X_n$ with $P(|X_n|>N) > \epsilon$. In fact, for any $N$ there are infinitely many such $X_n$; thus you can always choose one that occurs later in the sequence than all the ones chosen so far. To see there must be infinitely many such $X_n$, suppose there were only finitely many and show that in fact we would have to have $\sup_n P(|X|>N) \to 0$ as $N \to \infty$. (Warmup: what if there were only one such $X_n$? Now what if there were only two?)

Hint: Fix a random variable $X$. This is a measurable function $X : \Omega \to \mathbb{R}$; for every $\omega \in \Omega$, $|X(\omega)|$ is some real number, and so there is an integer (depending on $\omega$) which is greater than it. It follows that $\bigcap_{N \in \mathbb{N}} \{|X| \ge N\} = \emptyset$. Now "continuity from above" (which follows from countable additivity) implies that $\lim_{N \to \infty} P(|X| \ge N) = 0$. This is the key step you need.

-
Thanks Nate Eldrege for you answer! with " 1 is not correct" you mean : for every $N >0$ there exists an $\epsilon >0$ and $X \in M$ such that $P(|X|> N)\ge \epsilon$, right? As you can see in the comments above, Srivatsan corrected this already. At this point I just have trouble with 2), see also my observations so far for 2) above. – user20869 Dec 20 '11 at 18:44
@hulik: Right. I've edited to address this, and also your question (2). – Nate Eldredge Dec 20 '11 at 20:27
Eldrege: Thanks for answering 2)! I have a question about that: We know it exists $\epsilon$ such that for every $N>0$ there's a $X_n$ with $P(|X_n|>N) > \epsilon$. You claim that for every $N$ there are infinitely many $X_n$ with this property. So let $N>0$ and suppose there's just one of this $X_n$. First it's clear: $\mathbf1\{|X_n|>N\} \ge \mathbf1\{|X_n|>N-1\} \dots \mathbf1\{|X_n|>1\}$ So if I can prove, that there's no $X_n$ such that for this fixed $X_n$ : $P(|X_n|> N ) > \epsilon$ for all $N$ then I'm done. But I do not see why this could not be the case.... – user20869 Dec 21 '11 at 8:44
Perhaps there's a element $\tilde{X}$ of the sequence which is on a set (with large enough measure) as big as I want. If I can prove that it doesn't hold for 1) then using induction I can prove your claim. However, I do not see how to prove this for one, aso mentioned above. Again thanks for your help! – user20869 Dec 21 '11 at 8:47
@hulik: I added another hint. – Nate Eldredge Dec 21 '11 at 19:02