Finding correlation of Max and Min of two IID random variable in U[0,1]

Question

I have a hw problem and can't figure out how to do it. Basically, $X,Y$ are iid $U[0,1]$, we need to find the correlation between max$(X,Y)$ and min$(X,Y)$. My thought is to find the pdf of $U=$max$(X,Y)$ and $V=$min$(X,Y)$ and then find pdf of UV and use the definition of Variance to find the correaltion. But this seems way to substantial to do. I wonder if there is some simpler method.

Answer 1

Let $W=\max\{X,Y\}$ and $Z=\min\{X,Y\}$. We can write the following: $$ F_W(w)=P(\max\{X,Y\}\le w)=P(X\le w \text{ and } Y\le w)=w^2 \\ F_Z(z)=P(\min\{X,Y\}\le z)=1-P(X\ge z \text{ and } Y\ge z)=1-(1-z)^2 $$ Now, you can compute, e.g. $$ E[W]=\int_0^1w(2w)\, \mathrm{d}w = \frac{2}{3}\\ E[W^2]=\int_0^1w^2(2w)\, \mathrm{d}w = \frac{1}{2}\\ \sigma_W=\sqrt{E(W^2)-[E(W)]^2}=\sqrt{\frac{1}{2}-\frac{4}{9}}=\sqrt{\frac{1}{18}} $$ (and similarily for $Z$). The last piece is to figure out $\mathop{cov}(W,Z)$: $$ \mathop{cov}(W,Z)=E[WZ]-E[W]E[Z] $$ where $E[WZ]=\frac{1}{4}$ (why??).

Now you can finish the calculation: $$ \rho(W,Z)=\dfrac{E[WZ]-E[W]E[Z]}{\sigma_W \sigma_Z}=\dfrac{1/4-(1/3)(2/3)}{(1/\sqrt{18})(1/\sqrt{18})}=\dfrac{1}{2}. $$

Here are the useful shortcuts - think a little about why they should hold:

1. $E[W]+E[Z] = 1$
2. $\sigma_W^2=\sigma_Z^2$
3. $E[WZ] = 1/4$

Answer 2

Here is a possible answer that avoids using calculus (as it is always asked in some quant interviews, calculus would introduce heavy calculations).

Following @baudolino's notations, we also define $W=\max\{X,Y\}$ and $Z=\min\{X,Y\}$. Then, by symmetry, we have $E[W] = E[W-Z] = 1-E[W]-E[Z]$. Also note that $E[W] + E[W-Z] + 1-E[W]-E[Z] =1$. Thus, $E[W] = \frac{1}{3}$ and $E[Z]= E[W] + E[W-Z] = \frac{2}{3}$.

In terms of $E[W^2]$, also by symmetry, $E[W^2] = E[1-Z^2]$. Note that, $E[W^2]+ E[W^2-Z^2] + E[1-Z^2]=1$, and $E[W^2-Z^2] = E[(W+Z)(W-Z)]$, implying $E[(W+Z)(W-Z)] = E[(X+Y)(|X-Y|)] = \frac{1}{2}E[X^2-Y^2|X>Y]+\frac{1}{2}E[Y^2-X^2|X<Y] = 0$. The last equation is also by symmetry.

Thus, $2E[W^2]=1$, $E[W^2] = E[Z^2]=\frac{1}{2}$.

Finally, $E[WZ] = E[XY] = E[X][Y]= \frac{1}{4}$. The last equation is by the independent of X and Y.

Putting what we obtained above into the definition of corrlation, we have $\rho(W,Z)=\dfrac{E[WZ]-E[W]E[Z]}{\sqrt{E[W^2]- E[W]^2}\sqrt{E[Z^2]- E[Z]^2}} = \frac{1}{2}$.