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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.~ 0<x<1;~0 <y<x \\ 0 &x <0~;~x>1 \\ \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

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1 Answer 1

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$f_Y(y)=\int_y^1 f(x,y)dx=2\int_y^1 dx=2(1-y), 0<y<1$
$f_{X|Y}(x|y)= \dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\dfrac{2}{2(1-y)}=\dfrac{1}{1-y}, 0<y<1$

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  • $\begingroup$ 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). $\endgroup$ Oct 18, 2019 at 19:27

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