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.).



Weighted sum of two random variables ranked by first order stochastic dominance
$\newcommand\de\delta$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.)


