Suppose we draw two values $x_1,x_2$ according to a CDF $F$. Independently, we draw another two values $y_1,y_2$ according to another CDF $G$. Both $F$ and $G$ has support $[0,1]$.

Among those four values, I first observe $x_1$ only. And then, I get to observe one of $y_1$ and $y_2$, depending on whether $x_1\leq x_2$ or $x_1>x_2$ following the rule below:

--- If $x_1\leq x_2$, then I observe $y_1$ or $y_2$, whichever is smaller (or equal to) than the other.

--- If $x_1>x_2$, I observe $y_1$ or $y_2$, whichever is greater than the other.

So, if my $x_1$ is smaller than the other $x$, my observation of $y$ equals the smaller value of $y_1$ and $y_2$. If my $x_1$ is larger than the other $x$, the $y$ I observe is the larger one between $y_1$ and $y_2$.

In this case, if I observe $x_1$ and some value $y$, what is the conditional distribution of the other value of $y$? (so, if $y=y_1$, then what is the distribution of $Y_2$?)


1 Answer 1


We assume that $F$ is continuous, so that the probability of $x_1=x_2$ is 0.

Let $y_u$ be the unobserved value of $y$, with $y_o$ the observed value. Then the cdf for $y_u$ is $$\begin{cases} F(x_1)\dfrac{G(y_u)}{G(y_o)} &\text{ if }y_u \le y_o\\ \\ 1 - \big(1-F(x_1))\dfrac{1-G(y_u)}{1-G(y_o)} &\text{ if }y_u \ge y_o \end{cases} $$ Note that this properly gives:

  • a cdf of $0$ when $y_u=0$
  • a cdf of $1$ when $y_u=1$
  • a cdf of $F(x_1)$ when $y_u=y_o$ (calculated from either side).

The last statement reflects that the probability of $y_u<y_o$ is exactly the probability of $x_2<x_1$.

  • $\begingroup$ Thank you for your comment @Matt F. When $y_0=1$, i.e., the upper case, the CDF is not well defined. For example, if $F$ is uniform and $x_1=1/2$, the CDF for $y_u$ becomes $\frac{y_u}{2}$.. $\endgroup$
    – Andeanlll
    Mar 3, 2021 at 8:22
  • $\begingroup$ I agree this isn’t well-behaved for $y_o=1$; I think the reasonable area of concern is $0<y_o<1$. $\endgroup$
    – Matt F.
    Mar 3, 2021 at 9:15

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