The joint probability density function for two random variables X and Y is given by $f_{XY}(x,y)=1 $ $(0<x<1), (0<y<1)$.

and the conditional joint PDF is $f_{XY}(x,y | X>Y)=2 $ $(0<x<y<1))$ by bayes' theorem.

Thus, the conditional marginal PDF is $f_{X}(x | X>Y)= \int_{0}^y 2dy=2x $ $(0<x<y<1))$

I'd like to calculate the conditional expectation and I have a little confusion here : whether I should use $E(X|X>Y)=\int_{0}^12xdx$ or $E(X|X>Y)=\int_{y}^12xdx$.

  • $\begingroup$ So the joint distribution of $X$ and $Y$ is given by $f_{X,Y}(x,y)= \mathbb{1}_{0<x<1}\mathbb{1}_{0<y<1}$ and $f_X(x) = \int f_{X,Y}(x,y)dy$. We know that $E[X|X>Y]=\frac{E[X\mathbb{1}_{X>Y}]}{P(X>Y)}$. I think you have all the ingredients to compute the expectation. $\endgroup$ – Sesame May 30 at 10:41
  • $\begingroup$ How does E[X|X>Y]=E[X|X>Y]P(X>Y) hold? It doesn't make sense. $\endgroup$ – i9100 May 30 at 12:18
  • $\begingroup$ Look at this link $\endgroup$ – Sesame May 30 at 12:58
  • $\begingroup$ You are overthinking. What you have is two independent $U(0,1)$ variables. See math.stackexchange.com/a/3006539/321264, which is in the same spirit as the answer below. $\endgroup$ – StubbornAtom May 30 at 16:11

We know that $E[X|X>Y]=\frac{E[X\mathbb{1}_{X>Y}]}{P(X>Y)}$. Therefore, \begin{align} E[X|X>Y] &= \frac{E[X\mathbb{1}_{X>Y}]}{P(X>Y)} \\ &=\frac{\int_0^1\int_0^1x\mathbb{1}_{x\geq y}dxdy}{\int_0^1\int_0^1\mathbb{1}_{x\geq y}dxdy} \\ &= \frac23 \end{align}

  • 1
    $\begingroup$ (+1) It is enough to directly say $P(X>Y)=1/2$ using symmetry. $\endgroup$ – StubbornAtom May 30 at 16:14

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