I came across a note that says that the joint density of two dependent random variables $x$ and $y$ conditional on the event $\{y\geq z\}$ for a given $z\in\mathbb{R}$ is given as $$f(x,y|y\geq z)=\dfrac{f(x,y)\,\mathbb{1}_{\{y\ge z\}}}{1-\int_{-\infty}^{z}\int_{-\infty}^{\infty}f(x,y)\,dx\,dy}.$$ I might be at a loss here, but is it entirely obvious that this is true?
Thanks a lot.
If $x,y$ are the random variables, they should not also be used as arguments for the pdf , as doing so is a lazy notation that causes unnecessary confusion.
Notably: for the condition to be an event it should be made clear that it involves the random variable, rather than the argument.
So by instead using $X, Y$ as the random variables, and $x,y,z$ as scalar terms, we have the much more obvious:
$$f_{X,Y}(x,y\mid Y\geqslant z) = \dfrac{f_{X,Y}(x,y)\,\mathbf 1_{y\geqslant z}}{\mathsf P(Y\geqslant z)}$$
By way of $$\begin{align}f_{X,Y}(x,y\mid\mathrm Y\geqslant z) \tag 1 &=\dfrac{\mathrm d^2 \mathsf P(X\leqslant x,Y\leqslant y\mid Y\geqslant z) }{\mathrm d x\,\mathrm d y}\\[1ex]\tag 2&=\dfrac{\mathrm d^2~~}{\mathrm d x\,\mathrm d y}\left(\dfrac{\mathsf P(X\leqslant x,z\leqslant Y\leqslant y)}{\mathsf P(z\leqslant Y)}\right)\\[1ex]\tag 3&=\dfrac{f_{X,Y}(x,y)\,\mathbf 1_{z\leqslant y}}{\mathsf P(z\leqslant Y)}\end{align}$$
And, of course, $\mathsf P(z\leqslant Y)=1 - {\small\displaystyle\int_{-\infty}^z \int_{-\infty}^\infty} f_{X,Y}(x,y)\,\mathrm d x\,\mathrm d y$