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This is a simple question.

Suppose $V_1$ and $V_2$ are distributed independently and uniformly on the unit interval. I want to find the expected difference $V_2 - V_1$ conditional on $V_2 > V_1$.

I know the answer is $\int_0^1\int_0^{v_2}2(v_2-v_1)dv_1dv_2 = 1/3$. I have also verified it by computation. Why is there a "2" in the integrand? I understand that conditioning on the difference being positive changes the bounds of integration, and obviously understand that $f_1 = f_2 = 1$, but I am confused about the 2.

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  • $\begingroup$ If you were to remove the $(v_2-v_1)$ term, you would simply be computing the probability. If the $2$ weren't there, what would the total probability work out to? $\endgroup$
    – whuber ♦
    Commented Apr 7, 2017 at 21:33
  • $\begingroup$ Okay, so the probability would be $1/2$ without the 2. Does the 2 come from the density of $V_2 - V_1$? $\endgroup$
    – TCW
    Commented Apr 7, 2017 at 21:38
  • $\begingroup$ The density remains uniform in $(V_1,V_2)$. The $2$ arises because there are two distinct ways to order the $V_i$ and the region of integration is limited to just one of them; namely, where $V_2\gt V_1$. The analog of this problem with $n$ iid random variables will involve a factor of $n!$. $\endgroup$
    – whuber ♦
    Commented Apr 7, 2017 at 21:39
  • $\begingroup$ I deleted the answer because Bill Huber is right about the inner integral. $\endgroup$
    – Michael R. Chernick
    Commented Apr 7, 2017 at 22:02

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You want to compute $E[V_2-V_1|V_2>V_1]$.Writing out the integrals:

$$E[V_2-V_1|V_2>V_1]=\int_0^1\int_0^1(v_2-v_1)f(v_1,v_2|v_2>v_1)dv_1dv_2.$$

The crux is computing the density correctly. $f(v_1,v_2|v_2>v_1)=0$ when $v_2 \leq v_1$ and $c$ otherwise. The question is, what is $c$? Well total probability states that $\int_0^1\int_0^1f(v_1,v_2|v_2>v_1)dv_1dv_2=1$. So:

$\int_0^1\int_0^{v_2}cdv_1dv_2=1.$

You'll get $c=2$ because the region of integration is a triangle.

So an even easier way of seeing this is to visualize the region. You have that $f(v_1,v_2)$ lives on the unit rectangle $[0,1]\times [0,1]$. Conditioning $v_2>v_1$ puts you in a triangular region, which has half the area, hence when conditioning you'll need to multiply by 2.

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