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So two variables $X$ and $Y$ are uniformly distributed on $[0,1]$. I want to find the conditional densities of $X$ and $Y$ given $X > Y$.

I am approaching this question using the formula $h(x|x>y) = \frac{f(x,y)}{\textrm{marginal pdf of }y}$ What would the $\textrm{pdf} (f(x,y))$ be in this scenario?

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  • $\begingroup$ The answers will generally assume $X$ and $Y$ are independent. No reason to think they aren't, but you should probably state this. $\endgroup$ – Brian Tung Aug 12 '19 at 7:00
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$P(X\leq t|X>Y)=\frac {P(Y<X\leq t)} {P(X>Y)}$. By symmetry $P(X>Y)=P(X<Y)=\frac 1 2$ so $P(X\leq t|X>Y)=2 P(Y<X\leq t)=2\int _0^{t}\int_y^{t}dxdy=t^{2}$ for $0< t <1$. The density if $X$ given $X>Y$ is therefore $2t$ for $0< t <1$.

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Since $(X,Y)$ is uniformly distributed on the unit square, the conditional distribution given $X>Y$ is uniformly distributed on the triangle $$\{(x,y)\colon 1>x>y>0\}.$$

Since the triangle has area $\tfrac12$, we have to multiply by $2$ to get a probability distribution. Thus, the probability density function of the conditional distribution of the pair $(X,Y)$ is given by the function equal to $2$ on this triangle, and $0$ everywhere else, or in other words: $$ \textrm{pdf}(x,y)=2\cdot 1_{x>y},\qquad (x,y)\in [0,1]^2. $$ To find the $x$ and $y$ marginals from this pdf, simply integrate out the other variable: $$ \textrm{pdf}(x)=\int_0^1 2\cdot 1_{x>y}\ dy=2\int_0^x\ dy=2x, $$ and $$ \textrm{pdf}(y)=\int_0^1 2\cdot 1_{x>y}\ dx=2\int_y^1\ dy=2(1-y). $$

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  • $\begingroup$ is pdf(x) = 2x the conditional density of X given X > Y? $\endgroup$ – StatsHelp Aug 12 '19 at 10:19
  • $\begingroup$ Yes, I write pdf$(x,y)$ for the joint conditional density and pdf$(x)$ for the $x$ marginal (and similarly for $y$) $\endgroup$ – pre-kidney Aug 12 '19 at 10:27
  • $\begingroup$ Okay. Thanks a lot for your solution. $\endgroup$ – StatsHelp Aug 12 '19 at 10:30

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