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So I have a random variable $X$ that takes values in the non-negative integers. I want it to have one of the following properties: either $$\lim_{n\to\infty} \Pr(X\ge n\mid X\ge n/2)=0$$ or there exists $N$ such that for all $n>N$ $$\Pr(X\ge n\mid X\ge n/2) \le 1/4.$$ (Obviously the latter follows from the former. I really only need the latter but the former might be easier to work with.)

I'm looking for a condition on $X$ that yields this conclusion. Ideally, it would be a condition on the moments of $X$ and/or a condition on the behavior of the unconditional probability $\Pr(X\ge n)$ as $n$ gets large.

I was hoping $E(X)<\infty$ might work but the following is a counterexample: let $X$ be such that $\Pr(X\ge n\mid X\ge n/2) = 1/3$ for all $n$ and $\Pr(X\ge 1)=p>0$. Then $\Pr(X\ge 2^n) = 3^{-n}p$ for all $n$ and so $$\sum_{n=0}^\infty 2^n \Pr(X\ge 2^n) = 3p.$$ By the Cauchy Condensation test $$E(X) = \sum_{n=1}^\infty \Pr(X\ge n)$$ converges (in fact $3p/2 \le E(X) \le 3p < \infty$) which confirms that $X$ is a counterexample.

Now I'm wondering whether $E(X)<\infty$ together with $V(X)<\infty$ might be sufficient but I'm not sure how to approach that problem. I'd appreciate any help.

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A variant on your example: Take a random variable such that $P(X=k^k)=k^{-k^2}$ for all integers $k \geq 2$, and otherwise $X=0$. Now all the moments of $X$ are bounded, but your desired conclusion still fails. – Kevin Costello Apr 10 '13 at 19:22
Thanks @KevinCostello that is very helpful. I see that all of the moments of $X$ are bounded in your example but can you say a little bit more about why the desired conclusion fails? – Remco Apr 11 '13 at 14:22
Just because the set of possible $X$ is so sparse -- if $X$ is at least $k^k+1$, it is automatically at least $(k+1)^{k+1} > 2(k^k+1)$, so for $n$ of that form the conditional probability is $1$. – Kevin Costello Apr 11 '13 at 18:49

The following might offer intuition, though it is very simple. I write this as I think about the problem. Notice \begin{equation*} \mathbb{P}(X \ge n |X \ge n/2) = \mathbb{P}(X \ge n | X \ge \lceil n/2 \rceil) = \frac{\mathbb{P}(X \ge n, X \ge \lceil n/2\rceil)}{\mathbb{P}(X \ge \lceil n/2 \rceil)} = \frac{\mathbb{P}(X \ge n)}{\mathbb{P}(X \ge \lceil n/2 \rceil)} =: \frac{p_n}{p_{\lceil n/2 \rceil}} \end{equation*} meaning you need random variable $X$ distributed so that $p_n \downarrow 0$ faster than $p_{\lceil n/2 \rceil} \downarrow 0$ as $n \uparrow \infty$ (or $p_n = 0$ for some $n$), where, as usual, $\lceil \cdot \rceil$ is the ceiling function, taking the smallest integer larger than the real number it evaluates.

Indeed, this should immediately suggest that bounded in $\mathbb{L}^1$ is not enough, because the random variable $X$ could just be getting very large on sets of decreasing mass.

Writing $a_n:= p_{\lceil n/2 \rceil} - p_n$ and rewriting the last term above as $$\frac{p_n}{p_{\lceil n/2 \rceil}} = \frac{p_n}{a_n + p_n},$$ we see that we need only $a_n$ increasing in $n$, or $$\mathbb{P}(X \ge \lceil n/2 \rceil) - \mathbb{P}(X \ge n) = \mathbb{P}(\lceil n/2 \rceil \le X < n)$$ increasing in $n$. So though the windows $\left[\lceil n/2 \rceil, n\right)$ are expanding, $n$ is also marching off to infinity. But then $$\sum_{n \ge 1: n \text{ even}} a_n = \sum_{n \ge 1: n \text{ even}} \mathbb{P}(\lceil n/2 \rceil \le X < n) \le \sum_{n \ge 1} \mathbb{P}(X \ge n) \le \mathbb{E}[X] + 1$$ and we said $a_n \uparrow \infty$, so it must be that $\mathbb{E}[X] = \infty$. Perhaps similar reasoning might help you consider cases $\mathbb{E}[X^p]$ for $p > 1$.

Obviously $X$ essentially bounded satisfies your criteria, but that may be too binding for your interest.

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