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So two independent random variables $X$ and $Y$ are uniformly distributed on $[0,1]$

I want to find the conditional densities of $X$ and $Y$ given that $X>Y$.

How can I find these conditional densities and their expected values? i.e $\mathbb E(X|X>Y)$ ?

Thanks

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  • $\begingroup$ Are $X$ and $Y$ defined on the same probability space? If so then what is their joint distribution? Are they independent maybe? $\endgroup$ – drhab Aug 12 '19 at 9:50
  • $\begingroup$ Yes they are independent $\endgroup$ – StatsHelp Aug 12 '19 at 10:13
  • $\begingroup$ You should add to your question in an edit, because it is essential information. Without it your question cannot be answered. $\endgroup$ – drhab Aug 12 '19 at 10:14
  • $\begingroup$ I have edited the question. Are you able to provide an answer now by any chance? $\endgroup$ – StatsHelp Aug 12 '19 at 10:21
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Assuming that $X$ and $Y$ are independent (since there is no way to proceed with assuming some joint distribution, and this is the most natural one) then the question has already been answered here Conditional density question.

In particular, $\mathbb E(X\mid X>Y)=\int _0^1 2x^2\ dx=\frac{2}{3}$.

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  • $\begingroup$ I tried to understand that comment of yours but I could not make sense out of the first part (the triangle part) $\endgroup$ – StatsHelp Aug 12 '19 at 10:17
  • $\begingroup$ Consider the pair $(X,Y)$. It is a random point in the plane. Before the conditioning, it belongs to the unit square $[0,1]\times [0,1]$. You can think of the conditioning as cutting the square in half along the line $x=y$. The bottom right half is all that remains after conditioning. Since the area is half, the density must double so as to maintain a total integral of $1$. That is the entire concept... $\endgroup$ – pre-kidney Aug 12 '19 at 10:29
  • $\begingroup$ Oh right. That makes sense. Thank you that was very helful $\endgroup$ – StatsHelp Aug 12 '19 at 10:31
  • $\begingroup$ If the questions are answered, you can mark them as such using the green checkmark. $\endgroup$ – pre-kidney Aug 12 '19 at 10:32
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Under condition $X>Y$ we are dealing with uniform distribution on the triangle: $$\Delta:=\{(x,y)\in[0,1]^2\mid x>y\}$$

The PDF of this joint distribution is constant on that triangle and takes value $0$ outside the triangle.

As always its integral must take value $1$ and this together leads to PDF:$$f_{X,Y\mid X>Y}(x,y)=2\mathbf1_{\Delta}(x,y)$$where $\mathbf1_{\Delta}$ denotes the indicator function of set $\Delta$.

Then we find: $$\mathbb E[X\mid X>Y]=\int xf_{X,Y\mid X>Y}(x,y)dydx=2\int_0^1\int_0^xxdydx=2\int_0^1x\int_0^xdydx=$$$$2\int_0^1x^2dx=\frac23$$

You can find the marginal distributions of the conditional distribution on the usual way by finding for $x\in[0,1]$:$$f_{X\mid X>Y}(x)=\int f_{X,Y\mid X>Y}(x,y)dy=2\int_0^xdy=2x\tag1$$and for $y\in[0,1]$ similarly:$$f_{Y\mid X>Y}(y)=\int f_{X,Y\mid X>Y}(x,y)dx=2\int_y^1dy=2(1-y)$$

Outside interval $[0,1]$ these marginals evidently both take value $0$.

Also note that $(1)$ reveals another way to find $\mathbb E[X\mid X>Y]$:$$\mathbb E[X\mid X>Y]=\int_0^1xf_{X\mid X>Y}(x)dx=\int_0^1x\cdot2xdx=\frac23$$

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