Suppose $X_1, X_2 \stackrel{\mathrm{iid}}{\sim}$ Uniform$(-1, 2)$. I am interested in finding Corr$(Y, Z)$, where $Y = \min\{X_1, X_2\}$ and $Z = \max\{X_1, X_2\}$.

Now, I can solve the problem following the traditional methods i.e. by first finding the respective probability density functions and so on. However, this process is very long and tedious.

After a bit of searching online, I came across this post, where the question is very similar to mine but I am more interested in the answer, which gives the following three "short-cuts" for $X_1, X_2 \stackrel{\mathrm{iid}}{\sim}$ Uniform$(0, 1)$ and $Y$ and $Z$ defined as above:

  1. $\mathbb{E}[Y] + \mathbb{E}[Z] = 1$
  2. Var$(Y)$ = Var$(Z)$
  3. $\mathbb{E}[YZ] = \frac 1 4$

I am unable to see how these relationships are true. Moreover, do such relationships hold in general (possibly just simply requiring to change the values on the RHS of 1 and 3 accordingly) or are they only true for standard iid uniform random variables?

Any intuitive explanations will be greatly appreciated! :)


We get $\text{Var}[Y] = \text{Var}[Z]$ from the symmetry in the setup: if you reverse the interval $[0,1]$, the distribution of $X_1$ and $X_2$ is unchanged, but $\min$ and $\max$ are swapped. Formally: \begin{align} \text{Var}[Y] &= \text{Var}[1-Y] \\ &= \text{Var}[1-\min\{X_1, X_2\}] \\ &= \text{Var}[\max\{1 - X_1, 1-X_2\}] \\ &= \text{Var}[\max\{X_1, X_2\}] \\ &= \text{Var}[Z]. \end{align} The part that relies on distributions is the step where we replace $\max\{1-X_1, 1-X_2\}$ by $\max\{X_1, X_2\}$: that's because when $X_1, X_2$ are independent $\text{Uniform}(0,1)$, so are $1-X_1$ and $1-X_2$.

Looking at this argument, we see that it continues to work for $X_1, X_2 \sim \text{Uniform}(-1,2)$: in fact, the same transformation $t \to 1-t$ reverses the interval $[-1,2]$. More generally, we can do this for any symmetric distribution (where $X$ and $c-X$ are identically distributed for some $c$).

The other two facts are even more general: all they need is that addition and multiplication are commutative. Therefore $Y+Z = \min\{X_1, X_2\} + \max\{X_1, X_2\} = X_1 + X_2$: sorting $X_1$ and $X_2$ before adding them doesn't do anything. We conclude that $$ \mathbb E[Y] + \mathbb E[Z] = \mathbb E[Y+Z] = \mathbb E[X_1 + X_2] = 2 \mathbb E[X_1] = 1. $$ The value is unchanged, because $\frac{0+1}{2} = \frac{-1+2}{2} = \frac12$ is the value of $\mathbb E[X_1]$ in both problems.

Similarly, $\mathbb E[YZ] = \mathbb E[X_1X_2] = \mathbb E[X_1]^2 = \frac14$ in both cases.

  • $\begingroup$ Hi! Thanks for your answer! May I know if there is a typo though? Where you say "$Y + Z = \min\{X_1, X_2\} + \max\{X_1, X_2\} = X_1X_2$", did you mean "$Y + Z = \min\{X_1, X_2\} + \max\{X_1, X_2\} = X_1 + X_2$" instead? $\endgroup$
    – Ethan Mark
    Aug 14 '21 at 3:44
  • $\begingroup$ Yes, that was a typo. Sorry! I have corrected it now. $\endgroup$ Aug 14 '21 at 3:51
  • $\begingroup$ All good! Also, may I know why $\mathbb{E}[YZ] = \mathbb{E}[X_1X_2]$? I am a bit lost here! $\endgroup$
    – Ethan Mark
    Aug 14 '21 at 3:57
  • $\begingroup$ The same reason: if you multiply the smaller of $X_1, X_2$ by the larger of $X_1, X_2$, you will get the product of $X_1$ and $X_2$. $\endgroup$ Aug 14 '21 at 4:09
  • 1
    $\begingroup$ Yes, it is. We can check that it holds when $X_1 \le X_2$ and also when $X_1 > X_2$. $\endgroup$ Aug 14 '21 at 6:08

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