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As the title says, I'm looking for the distribution of $X$ given that $X>Y$.

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Assume that $X$ and $Y$ are independent $\mathrm{N}(0,1)$ rv's. Denote by $\phi$ and $\Phi$ the corresponding pdf and df, respectively. By definition, $$ P(X\leq x\mid X>Y) = \frac{P(X\leq x,X>Y)}{P(X>Y)} \ \, . $$ Since $X$ and $Y$ are iid, by symmetry we have $P(X>Y)=P(X<Y)$. But we know that $P(X>Y)+P(X<Y)=1$ (the probability of a tie is zero). Hence, $P(X>Y)=1/2$.

Now, $$ \begin{align*} P(X\leq x,X>Y) &= \mathrm{E}[P(X\leq x,X>Y\mid Y)] \quad \text{(total expectation)} \\ &= \int_{-\infty}^\infty P(X\leq x,X>Y\mid Y=y)\,\phi(y)\,dy \\ &= \int_{-\infty}^\infty P(X\leq x,X>y\mid Y=y)\,\phi(y)\,dy \quad \text{(use what you know)} \\ &= \int_{-\infty}^\infty P(y<X\leq x)\,\phi(y)\,dy \quad \text{(independence)} \\ &= \int_{-\infty}^\infty (\Phi(x)-\Phi(y)) I_{(-\infty,x]}(y)\,\phi(y)\,dy \\ &= \Phi^2(x) - \int_{-\infty}^x \Phi(y)\,\phi(y)\,dy \, . \\ \end{align*} $$ Therefore, $$ P(X\leq x\mid X>Y) = 2\left(\Phi^2(x) - \int_{-\infty}^x \Phi(y)\,\phi(y)\,dy\right) \, . $$ Integrating by parts, we have $$ \int_{-\infty}^x \Phi(y)\,\phi(y)\,dy = \left[\Phi(y)\Phi(y)\right]_{-\infty}^x - \int_{-\infty}^x \Phi(y)\,\phi(y)\,dy \, , $$ yielding $$ \int_{-\infty}^x \Phi(y)\,\phi(y)\,dy = \frac{\Phi^2(x)}{2} \, , $$ and finally $$ P(X\leq x\mid X>Y) = \Phi^2(x) \, . $$

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    $\begingroup$ Note that $\Pr(X\le x|X\gt Y)$ = $\Pr(\max(X,Y)\le x)$ = $\Pr(X\le x\text{ and }Y\le x)$ = $\Phi(x)^2$. This agrees with your result after you perform the final integration using the substitution $z=\Phi(y)$ (no Monte-Carlo needed). $\endgroup$ – whuber Apr 8 '14 at 18:54
  • $\begingroup$ Thanks! This is much simpler. How would you prove the (intuitively clear) first equality? $\endgroup$ – Zen Apr 8 '14 at 20:01
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    $\begingroup$ You need (a) exchangeability of $X$ and $Y$ and (b) that $\Pr(X=Y)=0$. It follows immediately that $\Pr(X\gt Y)=\Pr(Y\gt X)=1/2$. A total probability law then gives $\Pr(\max(X,Y)\le x) \\= \Pr(X\le x|X\gt Y)\Pr(X\gt Y)+\Pr(Y\le x|Y\gt X)\Pr(Y\gt X)+\Pr(X\le x|X=Y)\Pr(X=Y)\\ = \Pr(X\le x|X\gt Y)(1/2+1/2)+0\\ = \Pr(X\le x|X\gt Y).\\$ All of this is essentially already in the first few lines of your answer. The remainder (about eight lines) therefore seems to be just a particularly cumbersome way to derive the distribution of the maximum of two continuous iid variables. $\endgroup$ – whuber Apr 8 '14 at 20:11

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