Can I express $\mathbb{P}(Y\leq 2\leq X\leq 3)$ as $F_{Y}(2)[F_{X}(3) - F_{X}(2)]$?

Question

If I know that X follows CDF $F_1$ and Y follows CDF $F_2$.

Can I express Prob($Y \leq 2 \leq X \leq 3$ )= $F_2(2)[F_1(3)-F_1(2)]$?

(EXTRA: the only relation between $X$ and $Y$ is $Y \leq X$. But with event $Y \leq 2 \leq X \leq 3$ I believe that relation is already satisfied.)

Answer 1

If $X$ and $Y$ are independent, then you can express the proposed probability as suggested: \begin{align*} \mathbb{P}(Y\leq 2\leq X\leq 3) & = \mathbb{P}((Y\leq 2)\cap(2\leq X\leq 3))\\\\ & = \mathbb{P}(Y\leq 2)\mathbb{P}(2\leq X\leq 3)\\\\ & = F_{Y}(2)[F_{X}(3) - F_{X}(2^{-})] \end{align*}

If $X$ is continuous, then you can replace $F_{X}(2^{-})$ by $F_{X}(2)$.

Hopefully this helps!