Random variables, inequalities involving them and medians thereof

Question

I have three facts which I know to be true and I am wondering if together they logically imply a fourth one (or not!).

1. Consider $F_x$ and $G_y$ two continuous, uni-modal distributions on $\mathbb{R}$ and two random variables $X$ and $Y$ with $X\sim F_x$ and $Y\sim G_y$. I have two continuous functions $a:\mathbb{R}\to\mathbb{R}_+$ and $b:\mathbb{R}\to\mathbb{R}_+$ that only cross each other once.
2. Consider $S_x$ and $S_y$ two subsets of $\mathbb{R}$: $\int_{S_x}f_x(x)dx>1/2$ and $\int_{S_y}g_y(y)dy>1/2$.
3. I also know that $\forall x\in S_x$ and $\forall y\in S_y$ it holds that: $$a(x)-b(x)<a(y)-b(y)$$

Now, do 1-2-3 above imply that:

$$\mbox{med}_{X\sim F_x}(a(X)-b(X))\leq\mbox{med}_{Y\sim G_y}(a(Y)-b(Y))$$

Edit:

$X$ and $Y$ are independent random variables. $S_x$ and $S_y$ are compact sets.

Answer 1

There are some distractions in the problem statement, so I have taken the opportunity to clean it up and simplify it. In so doing some conditions were relaxed, thereby generalizing the situation. (It appeared necessary to assume the function $a-b$ is measurable, though.)

The result comes down to this: if you know that more than half the probability of one random variable ($u(X)$) is concentrated among small numbers and more than half the probability of another random variable ($u(Y)$) is concentrated among larger numbers, then you can conclude that the median of the first does not exceed the median of the second. If that's not completely obvious, then read on.

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Let $F$ and $G$ be any distributions (instead of $F_x$ and $G_y$, but without their restrictions). Let $u:\mathbb{R}\to \mathbb{R}$ be any measurable function (playing the role of $a-b$, but without any restrictions). Let $X\sim F$ and $Y\sim G$ be independent random variables. Suppose there exist any sets $S_F$ and $S_G$ (compact or not) such that

1. $\Pr(X\in S_F) \gt 1/2 \lt \Pr(Y\in S_G)$ and

2. $\sup(u(S_F)) \le \inf(u(S_G))$.

The question asks whether we can conclude $\newcommand{\m}{\text{med}} \m(u(X)) \le \m(u(Y))$.

To address this, pick $\lambda$ with $\sup(u(S_F)) \le \lambda \le \inf(u(S_G))$ and note that (2) implies

$$ \{x\,|\, u(x) \le \lambda\} \supset S_F.$$

Therefore, by taking probabilities and applying (1),

$$\Pr(u(X) \le \lambda) \ge \Pr(X \in S_F) \ge 1/2.$$

It is immediate that $\m(u(X)) \le \lambda$.

The counterpart of this argument for $Y$ shows $\m(u(Y)) \ge \lambda$.

Together these results entail $\m(u(X))\le \lambda \le \m(u(Y))$, QED.