# Finding the conditional distribution

We are given a joint distribution of $(X,Y)$ as follows : $$f_{X,Y}(x,y)=2I_{(0,y)}(x)I_{(0,1)}(y)$$,where $I$ is an indicator function, i.e $f=2$ if $0<x<y$ and $0<y<1$, otherwise $f=0$.

We want the distribution of $Y|X=x$.

I tried as follows :

$f_{Y|X=x}=\dfrac{f_{(X,Y)}(x,y)}{f_X(x)}$

So, we calculate $f_X(x)$ first.

$f_X(x)=\int_y f_{X,Y}(x,y)\:dy=\int_x^1 f_{X,Y}(x,y)\:dy=\int_x^1 2I_{(0,y)}(x)I_{(0,1)}(y)\:dy$

$I_{(0,1)}(y)$ equals $one$ if $0<y<1$, which is true since we are integrating over $x<y<1$ i.e. $0<x<y<1$, and hence $I_{(0,1)}(y)=1$.

Similarly, $I_{(0,y)}(x)=1$ since we are integrating over $x<y<1$, so, we have $f_X(x)=\int_x^12\:dy=2(1-x)=2(1-x)I_{(0,1)}(x)$.

Thus, $f_{Y|X=x}=\dfrac{2I_{(0,y)}(x)I_{(0,1)}(y)}{2(1-x)I_{(0,1)}(x)}=\dfrac{I_{(0,y)}(x)I_{(0,1)}(y)}{(1-x)I_{(0,1)}(x)}$.

I don't think it can be simplified further, but the solution is given as $\dfrac{I_{(x,1)}(y)}{(1-x)}$, what am I doing wrong ?

Can anyone help ?

It may be the case indeed that we can write $$f_X(x)=\int_x^12\:dy=2(1-x)=2(1-x)I_{(0,1)}(x)$$

i.e. to indicate the support of $x$ through an indicator function.

But because in our case we want $f_X(x)$ in the denominator, we cannot allow it to be zero. So the restriction on the domain of $x$ does not appear in the density expression as an indicator function, for the conditional distribution. The domain restriction that appears in the numerator can stay. So we write

$$f_{Y|X=x}=\dfrac{2I_{(0,y)}(x)I_{(0,1)}(y)}{2(1-x)},\;\;\; x\in (0,1)$$

Moreover, we have that

$$I_{(0,y)}(x)I_{(0,1)}(y) = I_{(x,1)}(y)$$

since

$$\{0<x <y\} \cap \{0<y<1\} = \{x<y<1\}$$