Expected Value based on condition with range.

Question

Let's say I'm asked $E[X\mid Y<1]$, where, $0<X<2$ and $0<Y<2.$ I know that I need $f_{X\mid Y<1.5}.$ My question is, is $f_{X\mid Y<1.5} = \dfrac{f(x,y)}{\int_0^2\int_0^{1.5}f(x,y)\,dy\,dx}$ OR is it $\dfrac{f(x,y)}{\int_0^{1.5}f_y(y)\,dy}$?

Answer 1

It is neither. It is this:

$$\begin{align}f_{X\mid Y\leqslant 1.5}(x) &=\dfrac{f_X(x)\,\mathsf P(Y\leqslant 1.5\mid X=x)}{\mathsf P(Y\leqslant 1.5)}\\[2ex]&=\dfrac{\int_0^{1.5}f_{X,Y}(x,t)\,\mathrm d t}{\int_0^{1.5} f_{Y}(t)\,\mathrm d t}\\[2ex] &=\dfrac{\int_0^{1.5}f_{X,Y}(x,t)\,\mathrm d t}{\int_0^2\int_0^{1.5} f_{X,Y}(s,t)\,\mathrm d t\,\mathrm d s}\end{align}$$

And thus $$\mathsf E(X\mid Y\leqslant 1.5) = \dfrac{\int_0^2\int_0^{1.5} s\, f_{X,Y}(s,t)\,\mathrm d t\,\mathrm d s}{\int_0^2\int_0^{1.5} f_{X,Y}(s,t)\,\mathrm d t\,\mathrm d s}$$