I've been trying to prove the following inequality:
$$ \mathbb{E}[X \mid X > S] \geq \mathbb{E}[X \mid X>S, Y>S] $$
where $X$, $Y$, $S$ are mutually independent real-valued random variables (let's make them absolutely continuous to simplify). The intuition why it might be true is simple enough: if you condition on $Y>S$, the posterior on $S$ is shifted “downwards”, and then the conditioning on $X>S$ is not as informative and raises the expected value of $X$ less.
To make it simple, I have not succeeded proving this and I'm starting to wonder if it's in fact true, so I would like to know if anybody has an idea.
For completeness, my most promising attempt used the correlation inequality. First we write:
$$ \mathbb{E}[X \mid X>S, Y>S] = \frac{1}{P(X>S, Y>S)}\int_{-\infty}^{+\infty}p_S(s)P(Y>s)\int_s^{+\infty} p_X(x) x \,dx \,ds $$
whence:
$$ \mathbb{E}[X \mid X>S, Y>S] = \frac{1}{P(X>S, Y>S)}\int_{-\infty}^{+\infty}p_S(s)P(Y>s, X>s)\mathbb{E}[X \mid X>s] \,ds $$
Take $f(s) = \mathbb{P}(X > s, Y > s)$ and $g(s) = \mathbb{E}[X \mid X > s]$. Because $f$ is decreasing and $g$ is increasing, we get:
$$ \mathbb{E}[X \mid X>S, Y>S] \leq \frac{1}{\dots} \left(\int_{-\infty}^{+\infty} p_S(s) P(Y>s, X>s) \,ds\right) \left(\int_{-\infty}^{+\infty} p_S(s) \mathbb{E}[X \mid X>s] \,ds \right) $$
i.e.:
$$ \mathbb{E}[X \mid X>S, Y>S] \leq \int_{-\infty}^{+\infty} \frac{p_S(s)}{\mathbb{P}(X > s)} \int_s^{+\infty} p_X(x) x \,dx \,ds $$
This is frustratingly close but not quite it; if you apply the correlation inequality again you get something that is too lax (by Jensen's inequality), and I'm worrying what I got here is too lax already, or that I made a mistake.
(Why do I care? It happens to be relevant to a Bayesian model of certain linguistic phenomena. Basically, if this is true, following a certain model, you predict that upon hearing “Alice and Bridget are tall”, people will not imagine Alice to be as tall as if they hear “Alice is tall”.)
Ok, so a small variation on my initial attempt works.
Let's rewrite it as: $$ \mathbb{E}[X|X > S, Y > S] = \frac{1}{\mathbb{P}(Y>S|X>S)} \int p_S(s) \frac{\mathbb{P}(X>s)}{\mathbb{P}(X>S)} \mathbb{P}(Y>s) \mathbb{E}[X|X>s]\,ds $$
We apply the correlation inequality, crucially taking $p_S(s) \frac{\mathbb{P}(X>s)}{\mathbb{P}(X>S)}$ to be the density (this is the density of $S$ conditional on $X > S$). We get:
$$ \mathbb{E}[X|X > S, Y > S] \leq \frac{1}{\dots} \left(\int p_S(s) \frac{\mathbb{P}(X>s)}{\mathbb{P}(X>S)} \mathbb{P}(Y>s) \,ds\right) \left( \int p_S(s) \frac{\mathbb{P}(X>s)}{\mathbb{P}(X>S)} \mathbb{E}[X|X>s]\,ds \right) $$
The first integral is $\mathbb{P}(Y>S|X>S)$. The second one is $\mathbb{E}[X|X>S]$. Yay!