# what is the conditional density of $X$ given $Y$? [duplicate]

If the joint p.d.f of a two-dimensional random variable $$(X,Y)$$ is given by:

$$f(x,y) = \ \left\{ \begin{array}{ll} 2 &s.t.~ 01 \\ \end{array} \right.$$

Then, what is the conditional density of $$X$$ given $$Y$$?

Attempt: $$f_{X|Y}(x|y)= \dfrac{P_{XY}(x,y)}{P_Y(y)}$$

As per given condition : $$\int^0_1 (\int_0^xP_{XY}(x,y)~dy) dx=2$$

How does one calculate $$P_{XY}(x,y)$$ from here.

Also, I have trouble visualizing a way to find out $$P_Y(y) = \int_0^\infty P_{XY}(x,y)~ dx$$

Thanks a lot for your help

$$f_Y(y)=\int_y^1 f(x,y)dx=2\int_y^1 dx=2(1-y), 0
$$f_{X|Y}(x|y)= \dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\dfrac{2}{2(1-y)}=\dfrac{1}{1-y}, 0
• No, this is wrong, The conditional pdf $f_{X\mid Y}(x\mid y)$ is a function of $x$, not of $y$ ($y$ is just a parameter in the formulas that you might come up with). Oct 18, 2019 at 19:27