Weighted sum of two random variables ranked by first order stochastic dominance

Question

Suppose $X$ and $Y$ are two non-negative, independent random variables such that $X \succcurlyeq_{st} Y$. That is, $X$ first-order stochastically dominates $Y$. Suppose that $X$ and $Y$ have smooth CDFs that admit a density.

Let $a, b > 0$ be two constants such that $a > b$. Is the following true?

$$a X + b Y \succcurlyeq_{st} a Y + b X$$

I suspect that the answer is no because various results in Shaked and Shanthikumar to this effect talk about random variables ranked under the reversed-hazard-rate order, and not FOSD (Theorem 4.A.36,4.A.37 e.g.).

Answer 1

$\newcommand\de\delta$Indeed, a counterexample is as follows: $a=2$, $b=1$, $$X\sim\frac12(\de_0+\de_2),\quad Y\sim\frac12(\de_0+\de_1),$$ where $\de_x$ is the Dirac measure supported on the singleton set $\{x\}$.

Then $X\succcurlyeq_{st}Y$, but $$a X + b Y \not\succcurlyeq_{st} a Y + b X,$$ because $P(aX+bY\ge 2)=\frac12\not\ge\frac34=P(aY+bX\ge 2)$.

(Your smoothness condition is inessential, since any distribution can be appropriately approximated by a smooth distribution.)