# Finding the correlation between the maximum and minimum of two uniform random variables

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.

• 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? Aug 14 '21 at 3:44
• Yes, that was a typo. Sorry! I have corrected it now. Aug 14 '21 at 3:51
• All good! Also, may I know why $\mathbb{E}[YZ] = \mathbb{E}[X_1X_2]$? I am a bit lost here! Aug 14 '21 at 3:57
• 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 $X_2$. Aug 14 '21 at 4:09
• Yes, it is. We can check that it holds when $X_1 \le X_2$ and also when $X_1 > X_2$. Aug 14 '21 at 6:08