Let $X,Y$ be iid such that $X\sim F>0$ and $Y \sim F>0$ ($X$ and $Y$ have the same probability distribution). Find $\mathbb{P}(X \le x | \max(X,Y)) $.

I know that $\max(X,Y) \sim F^2$.

I would use $$\mathbb{P} (X \in B | Y) = \frac{\int_B g(x,Y)dx}{\int_{\mathbb{R}}g(x,Y)dx}$$

Where $Y = \max(X,Y)$ and $B=(0, x)$.

But the problem is how to find density function $g$ ?

  • $\begingroup$ Got something from the answer below? $\endgroup$ – Did Oct 21 '14 at 19:25

The final formula is $$P(X\leqslant x\mid\max(X,Y))=u(x,\max(X,Y)),$$ where the function $u$ is defined for every $(x,z)$ by $$u(x,z)=\left\{\begin{array}{ccc}\frac{F(x)}{2F(z)}&\mathrm{if}&x\lt z,\\ 1&\mathrm{if}&x\geqslant z.\end{array}\right.$$ To explain why this identity holds, one needs to know the definition of conditional expectation you are acquainted with and the tools you know to compute these.

First, you could explain how you define $P(A\mid Z)$, for every event $A$ and random variable $Z$...

  • $\begingroup$ I was surprised to find that the answer is a discontinuous function, but on further consideration it seems it must be. $\endgroup$ – David K Oct 16 '14 at 2:45
  • $\begingroup$ Related: math.stackexchange.com/q/102938 $\endgroup$ – Did Oct 17 '14 at 5:49
  • $\begingroup$ In a more general case ($X$ and $Y$ arbitrary distributions, not necessarily iid), I believe $P(X\leq x \mid \max(X,Y))$ is discontinuous if $P(Y > X) > 0.$ $\endgroup$ – David K Oct 17 '14 at 12:54

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