3
$\begingroup$

Here is the problem from the book:

Let $X = \min(U,V)$ and $Y = \max(U,V)$ for independent $\text{uniform}(0,1)$ variables $U$ and $V$. Find the distributions of a) $X$; b) $1-Y$; c) $Y-X$.

I know that each has density $2(1-x)$ for $0<x<1$. What I'm hoping for is an intuitive explanation of how to understand this problem and arrive at that solution without knowing the beta distribution.

$\endgroup$
2
$\begingroup$

An explanation for the result that $X$, $1-Y$, and $Y-X$ have the same distribution in this case is as follows.

First, consider a plane with coordinate axes $u$ and $v$ and let $(U,V)$ be a random point in the plane chosen according to some joint density function $f_{U,V}(u,v)$. $U$ and $V$ need not be independent random variables. Then, $$\begin{align*} P\{X > \alpha\} &= P\{\min(U,V) > \alpha\} = P\{U > \alpha, V > \alpha\},\\ P\{1-Y > \alpha\} &= P\{1-\max(U,V) > \alpha\} = P\{\max(U,V) < 1 - \alpha\}\\ &= P\{U < 1- \alpha, V < 1- \alpha\},\\ P\{Y-X > \alpha\} &= P\{\max(U,V)-\min(U,V) > \alpha\}\\ &= P\{U-V > \alpha\} + P\{V-U > \alpha\}. \end{align*}$$ These three probabilities can be found in the general case by integrating $f_{U,V}(u,v)$ over the appropriate region which can be described in the three cases respectively as

  • the northeast quadrant of the plane with southwest corner $(\alpha, \alpha)$

  • the southwest quadrant of the plane with northeast corner $(1-\alpha, 1-\alpha)$

  • the half-plane below the line $v < u - \alpha$ and the half-plane above the line $v > u + \alpha$

So much for generalities. If the random point $(U,V)$ is uniformly distributed on a region $A$ of the plane (that is, $f_{U,V}(u,v)$ is nonzero and constant for $(u,v) \in A$, $f_{U,V}(u,v) = 0$ for $(u,v) \notin A$) and $B$ is any region of the plane, then $$P\{(U,V) \in B\} = P\{(U,V) \in A\cap B\} = \frac{\mathrm{Area}(A\cap B)}{\mathrm{Area}(A)}.$$ In particular, if we can compute areas via mensuration formulas learned in school, we do not need to integrate formally.

Finally, in the special case when $A$ is the unit-area square with opposite corners $(0,0)$ and $(1,1)$, and $\alpha$ is a number between $0$ and $1$, $$\begin{align*} P\{X > \alpha\} &= P\{U > \alpha, V > \alpha\}\\ &= P\{(U,V) \in ~\mathrm{square~with~opposite~corners}~ (\alpha,\alpha) ~ \mathrm{and}~ (1,1)\\ &= (1-\alpha)^2,\\ P\{1-Y > \alpha\} &= P\{U < 1- \alpha, V < 1- \alpha\}\\ &= P\{(U,V) \in ~\mathrm{square~with~opposite~corners}~ (0,0) ~ \mathrm{and}(1-\alpha,1-\alpha)~ \\ &= (1-\alpha)^2,\\ P\{Y-X > \alpha\} &= P\{U-V > \alpha\} + P\{V-U > \alpha\}\\ &= P\{(U,V) \in ~\mathrm{triangle~with~corners}~ (\alpha,0), (1,1-\alpha) ~\mathrm{and}~(1,0)\}\\ &\quad \quad + P\{(U,V) \in ~\mathrm{triangle~with~corners}~ (0,\alpha), (1-\alpha,1) ~\mathrm{and}~(0,1)\}\\ &= \frac{1}{2}(1-\alpha)^2 + \frac{1}{2}(1-\alpha)^2 = (1-\alpha)^2.\\ \end{align*}$$ So the complementary cumulative distribution of the three random variables $X$, $1-Y$ and $Y-X$ is the same $(1-\alpha)^2$ in this case, and so the three random variables have the same density function $2(1-\alpha)$, $0 \leq \alpha \leq 1$.

$\endgroup$
2
$\begingroup$

I'll work out the first one and leave the other two to you. First, find $\Pr(X \leq x)$. We have $$\begin{align*}\Pr(X \leq x) &= 1 - \Pr(X > x) \\ &= 1 - \Pr(U > x, V > x) \\ &= 1 - \Pr(U > x)\Pr(V > x) \\ &= 1 - (1-x)^2 \end{align*}$$ The first step follows that, for $x$ to be a minimum $U$ and $V$ must be greater than that value. Independence gives us the second line. Use of the CDF of the uniform gives us the third.

To get the density, take the derivative to get $2(1-x)$.

$\endgroup$
  • $\begingroup$ Thanks, Charlie. You were actually the head GSI for my Econ 140 class. I've always found your explanations to be very clear and precise. $\endgroup$ – Glendon Trullinger Dec 11 '11 at 6:43
  • $\begingroup$ Are you sure it's not $P(X \geq x)$ in the first line? $\endgroup$ – mark999 Dec 11 '11 at 6:53
  • $\begingroup$ @mark999, you're right, that was a typo. I actually rewrote it to be clearer. Thanks for the catch. $\endgroup$ – Charlie Dec 11 '11 at 7:38
  • $\begingroup$ @GlendonTrullinger, Glad that I could be of help. Sorry for the typo in the first draft. $\endgroup$ – Charlie Dec 11 '11 at 7:45
  • $\begingroup$ For the third case on $Y-X$, you have to get the joint distribution of $(X,Y)$ first, as $X$ and $Y$ are not independent. $\endgroup$ – Xi'an Dec 11 '11 at 10:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.