I have been considering the question of probabilities conditioned on inequalities. My intuition says that I should end up with a double integral, but I'm worried I've missed something in my solution.
Notation: $F_X(x)=\mathbb{P}(X\le x)$ denotes the CDF of RV X and p(x) the corresponding pdf.
I know the joint distribution $\mathbb{P}(X\le \alpha, Y\le \alpha)=\int_{-\infty}^{\alpha}\int_{-\infty}^{\alpha}p(x,y) \;dy\;dx$
Condsider $\mathbb{P}(X\le \alpha|Y\le \alpha)$ where X and Y are RV's, for some constant $\alpha>0$. \begin{eqnarray} \mathbb{P}(X\le \alpha|Y\le \alpha) &=& \int_{-\infty}^{\alpha}p(x|Y\le \alpha) \; dx \\ &=& \int_{-\infty}^{\infty}\int_{-\infty}^{\alpha}p(x,y|Y\le \alpha) \; dx \; dy \;(*)\\ &=& \int_{-\infty}^{\alpha}\int_{-\infty}^{\alpha}p(x,y) \;dx \;dy \\ &=& \int_{-\infty}^{\alpha}\int_{-\infty}^{\alpha}p(x,y) \;dx \;dy \\ \end{eqnarray} This has turned into the joint distribution. I think there is something wrong with step (*) but I can't seem to figure out what. Can anyone help me?
I could apply Bayes to $p(x|Y\le \alpha)$ but I can't find a way to make the inequality into something that doesn't involve an inequality.