Let $X$ and $Y$ be independent random variables, $X,Y \sim unif(0,1)$. Let $U = \min \{X,Y\}$ and $V = \max\{X,Y\}$. Find the correlation coefficient of $U$ and $V$.

I think we can assume that $U = X$ and $V = Y$ because they both have the same distribution. Further, $E[U] = E[V] = 1/2$ and $Var(U) = Var(V) = 1/12$. Next is $E[UV]$ and this is where I am stuck, because I can't solve this $E[UV] = \int_{}^{}\int_{\mathbb{R}^2}^{}uvf(u,v)dudv$. I think I can substitute $f(u,v)$ with $1$, but I'm not sure.

  • 1
    $\begingroup$ $U$ and $X$ do not have the same distribution. $P(U\geq s) = P(X\geq s)P(Y\geq s) = (1-s)^2$. Also you have not properly defined $V$. $\endgroup$
    – Alex R.
    Feb 10, 2015 at 23:00
  • $\begingroup$ Then could you please tell me how to find E(U)? $\endgroup$
    – user44697
    Feb 10, 2015 at 23:22
  • $\begingroup$ Well $F(s)=P(U\leq s)$, the cdf, you could differentiate that and get a density. Or maybe you remember that for a non-negative random variable $U$, $E[U]=\int_0^\infty (1-F(s))ds$ $\endgroup$
    – Alex R.
    Feb 10, 2015 at 23:24
  • $\begingroup$ $U$ and $V$ are not independent, as $U<V$ ! $\endgroup$
    – user65203
    Feb 13, 2015 at 15:03
  • $\begingroup$ math.stackexchange.com/q/2146136/321264 $\endgroup$ May 2, 2021 at 14:35

3 Answers 3


This problem has several nice shortcuts.

For the first term, you can note that $UV=XY$. This is because one or the other of $X,Y$ is the smaller one and the other is the larger, and multiplication doesn't care what order they come in. So since $ X, Y $ are independent, $ E [UV]= E [X] E [Y] = 1/4$.

We can use a similar trick to calculate $E[U]$ and $E[V]$. Let's think about $U$. We first consider $X=x$. Now with probability $1-x$, $Y>x$, in which case $U=x$. With probability $x$, $Y \leq x$, in which case $U$ is uniform on $[0,x]$. I'll write $U_x$ for a uniform variable on $[0,x]$. Now linearity of expectation tells us that:

$$E[U|X=x]=x (1-x) + x E[U_x]=x (1-x) + x^2/2$$

Now the total expectation formula gives:

$$E[U]=\int_0^1 E[U|X=x] f_X(x) dx = \int_0^1 (x (1-x) + x^2/2) dx.$$

The situation is similar to calculate $E[V]$, $E[U^2]$, and $E[V]^2$, then you can just combine everything to get the correlation coefficient.


You can also check whether your answer is correct using a computer algebra system. In this instance, since $X$ and $Y$ are independent standard Uniform random variables, their joint pdf, say $f(x,y)$ is:

f = 1;   domain[f] = {{x, 0, 1}, {y, 0, 1}};

Then, you seek:

Corr[{Min[x, y], Max[x, y]}, f]


where I am using here the Corr function from the mathStatica package for Mathematica. As disclosure, I should perhaps add that I am one of the authors.

The expectation you are having trouble with is $E[U V]$:

Expect[Min[x, y] Max[x, y], f]


Also, your calculation $E[U] = E[V] = 1/2$ is plainly wrong: this should be obvious since $E[X] = \frac12$, so $E[min[X,Y]]$ must be less than $\frac12$.


Let's find the joint distribution. When $u \leq v$ then \begin{align*} P(U<u,V<v)&=P(\min\{X,Y\}<u,\max\{X,Y\}<v)\\ &=P(X<u,Y<u)+P(X<u,u<Y<v)+P(u<X<v,Y<u)\\ &=P(X<u)P(Y<u)+P(X<u)P(u<Y<v)+P(u<X<v)P(Y<u)\\ &=u^2+2u(v-u)=2uv-u^2 \end{align*} When $v<u$ then \begin{align*} P(U<u,V<v)&=P(\min\{X,Y\}<u,\max\{X,Y\}<v)=v^2 \end{align*} Hence $f(u,v)=2$ when $u \leq v$ and $f(u,v)=0$ when $v<u$. Then $f(v)=2v$, $f(u)=2(1-u)$. Then $E[U]=1/3$, $E[U^2]=1/6$, $E[V]=2/3$, $E[V^2]=1/2$ and $$E[UV]=\int_0^1\int_0^{1}uv f(u,v)dudv =2\int_0^1\int_0^{v}uv dudv =\int_0^1v^3 dv=\frac{1}{4}$$

Now using this, I get the correlation is $1/2$.

btw. More generally if $X$ and $Y$ are independent and have distributions $F_x$ and $F_y$. Then when $u \leq v$, \begin{align*} P(U<u,V<v)&=P(\min\{X,Y\}<u,\max\{X,Y\}<v)\\ &=F_x(u)F_y(u)+F_x(u)(F_y(v)-F_y(u)) +F_y(u)(F_x(v)-F_x(u))\\ &=F_x(v)F_y(u)+F_x(u)F_y(v)-F_x(u)F_y(u) \end{align*} and when $v<u$, $$ P(U<u,V<v)=P(\min\{X,Y\}<u,\max\{X,Y\}<v)=F_x(v)F_y(v) $$ so $f(v,u)=f_x(v)f_y(u)+f_x(u)f_y(v)$ when $u \leq v$ and $f(v,u)=0$ when $v<u$. Using this you can calculate $E[g(U,V)]$ for any function $g$.

  • $\begingroup$ The PDF of (U,V) is wrong and not useful. The value of E(UV) is wrong. $\endgroup$
    – Did
    Feb 13, 2015 at 14:37
  • 1
    $\begingroup$ That value of $E[UV]$ doesn't even make sense, $U \in [0,1]$ and $V \in [0,1]$ so $E[UV] \in [0,1]$... $\endgroup$
    – Ian
    Feb 13, 2015 at 16:23
  • $\begingroup$ @Did PDF is corrected. The PDF is useful to calculate $E[UV]$. $\endgroup$
    – user103828
    Feb 14, 2015 at 10:01
  • $\begingroup$ @Ian You are correct. I made a mistake in the PDF calculation. It's been corrected now. $\endgroup$
    – user103828
    Feb 14, 2015 at 10:03
  • $\begingroup$ @user103828 In general yes, of course, but not in this case. Did you read Ian's answer? $\endgroup$
    – Did
    Feb 14, 2015 at 20:06

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