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Iosif Pinelis
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$\newcommand\de\delta$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

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.)

$\newcommand\de\delta$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 $P(aX+bY\ge 2)=\frac12\not\ge\frac34=P(aY+bX\ge 2)$.

$\newcommand\de\delta$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.)

Source Link
Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229

$\newcommand\de\delta$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 $P(aX+bY\ge 2)=\frac12\not\ge\frac34=P(aY+bX\ge 2)$.