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Suppose I have two positive random variables $X$ and $Y$ such that $X\leq Y$. Is it true that $Pr(X>y)\leq Pr(Y>y)$ for all $y\in [0,\infty]$?

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If $P\{X \leq Y\} = 1$, then all the probability mass lies in the region of the plane between the $y$ axis and the line $y=x$. Now, the event $\{Y > y\}$ is the disjoint union of three events: $$\{0 < X \leq y, Y > y\}, ~ \{ Y \geq X > y\}, ~ \{X > Y > y\}$$ of which the third has probability $0$ for all $y > 0$, and so $$\begin{align*}P\{Y > y\} &= P\{0 < X \leq y, Y > y\} + P\{ Y \geq X > y\}\\ &\geq P\{ Y \geq X > y\}. \end{align*}$$ On the other hand, the event $\{X > y\}$ is the disjoint union of three events: $$\{ Y \geq X > y\}, ~ \{X > Y > y\}, ~ \{X > y, Y \leq y\}$$ of which the latter two have probability $0$ for all $y > 0$, and so $$ P\{X > y\} = P\{ Y \geq X > y\} \leq P\{Y > y\}$$

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I take it that $X\leq Y$ means that $P(X\leq Y)=1$. Under this assumption we have $$ P(X>y)=P(X>y,X\leq Y)\leq P(Y>y,X\leq Y)=P(Y>y),\quad y\geq 0, $$ since $$ \{X>y,X\leq Y\}\subseteq \{Y>y,X\leq Y\}. $$

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