Conditional probability (on event) as conditional expectation

Question

Related to the question here, how do we write the following conditional probability for two random variables $X$ and $Y$ conditioned on an event occurring (such as $\mathrm{P}[X > x \mid X>Y]$) as a conditional expectation? It seems that conditioning on the event $\{X>Y\}$ seems to be an abuse of notation per the linked question.

So I'm wondering if this is correct: $$ \mathrm{P}[X > x \mid X>Y] = \mathrm{E}[ 1_{\{X>x\}} \mid X > Y] = \int 1_{\{X>x\}} \frac{f_{X,Y}(x,y)}{\mathrm{P}[X>Y]} dx dy $$ or do I need to integrate against the conditional density $f_{X \mid Y}(x,y)$ instead? If so, I don't see how to get $f_{X \mid Y}(x,y)$.

Answer 1

By definition: $$\mathsf P(X>x\mid X>Y)=\dfrac{\mathsf P(X>\max\{x,Y\})}{\mathsf P(X>Y)}\\\mathsf E(\mathbf 1_{\small X>x}\mid X>Y)=\dfrac{\mathsf E(\mathbf 1_{\small X>\max\{x,Y\}})}{\mathsf E(\mathbf 1_{\small X>Y})}\quad$$

So, when $X$ and $Y$ are continuous random variables with joint density function $f_{\small X,Y}$, then :$$\begin{align}\mathsf P(X>x\mid X>Y)&=\dfrac{\iint_{s>\max\{x,t\}}f_{\small X,Y}(s,t)\,\mathrm d \langle s,t\rangle}{\iint_{s>t}f_{\small X,Y}(s,t)\,\mathrm d \langle s,t\rangle}\\[3ex]&=\dfrac{\displaystyle\int_{-\infty}^x\int_x^\infty f_{\small X,Y}(s,t)\,\mathrm d s\,\mathrm d t+\int_{x}^\infty\int_{t}^\infty f_{\small X,Y}(s,t)\,\mathrm d s\,\mathrm d t}{\displaystyle\int_{-\infty}^\infty\int_t^\infty f_{\small X,Y}(s,t)\,\mathrm d s\,\mathrm d t}\end{align}$$

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Note: Do not use the same token, $x$, as both a free variable and the integral's bound variable.   Always use fresh tokens as an integrals bound variable.

Answer 2

According to the general definition of conditional expectation $P(X >x|X>Y)$ is a random variable which takes the value $\frac {P(X >x, X>Y)} {P(X>Y)}$ on the set $\{X>Y\}$ and the value $\frac {P(X >x, X\leq Y)} {P(X\leq Y)}$ on the set $\{ X \leq Y\}$.