Min and Max of Two Uniform Random Variables
Let
\(X_1,X_2\sim\mathrm{Uniform}(0,1)\)
independently, with
\(Y=\min(X_1,X_2)\)
and
\(Z=\max(X_1,X_2)\).
Part 1: Conditional Probability
$$
P(Y\ge y\mid Z\le z),
\qquad 0\le y\le z\le1
$$
A useful way to think about the problem is geometrically.
The pair
\((X_1,X_2)\)
is uniformly distributed over the unit square.
Step 1: Interpret \(Z\le z\)
Since
$$
Z=\max(X_1,X_2),
$$
the event \(Z\le z\) means
$$
X_1\le z
\quad\text{and}\quad
X_2\le z.
$$
Geometrically, this is a square of side length \(z\).
$$
P(Z\le z)=z^2.
$$
Step 2: Also require \(Y\ge y\)
Since
$$
Y=\min(X_1,X_2),
$$
\(Y\ge y\) means that both variables must be at least \(y\).
$$
X_1\ge y,
\qquad
X_2\ge y.
$$
Combining this with \(Z\le z\),
$$
y\le X_1\le z,
\qquad
y\le X_2\le z.
$$
This is another square, now with side length
\(z-y\).
$$
P(Y\ge y,\ Z\le z)
=(z-y)^2.
$$
Step 3: Divide by the conditioning region
$$
P(Y\ge y\mid Z\le z)
=
\frac{
P(Y\ge y,\ Z\le z)
}{
P(Z\le z)
}
=
\frac{(z-y)^2}{z^2}.
$$
Geometric interpretation:
Once we condition on \(Z\le z\), our new sample space is
the \(z\times z\) square.
The successful region is the smaller
\((z-y)\times(z-y)\) square.
$$
\text{Conditional probability}
=
\frac{\text{successful area}}
{\text{conditioning area}}.
$$
Part 2: Correlation between \(Y\) and \(Z\)
We want
$$
\operatorname{Corr}(Y,Z)
=
\frac{
\operatorname{Cov}(Y,Z)
}{
\sqrt{
\operatorname{Var}(Y)
\operatorname{Var}(Z)
}
}.
$$
Known moments of the order statistics
| Quantity |
Value |
| \(E[Y]\) |
\(\frac13\) |
| \(E[Z]\) |
\(\frac23\) |
| \(\operatorname{Var}(Y)\) |
\(\frac1{18}\) |
| \(\operatorname{Var}(Z)\) |
\(\frac1{18}\) |
The key shortcut: \(YZ=X_1X_2\)
Because \(Y\) is the smaller of \(X_1,X_2\)
and \(Z\) is the larger,
multiplication does not care about their order:
$$
YZ
=
\min(X_1,X_2)
\max(X_1,X_2)
=
X_1X_2.
$$
Since \(X_1\) and \(X_2\) are independent,
$$
E[YZ]
=
E[X_1X_2]
=
E[X_1]E[X_2]
=
\frac12\cdot\frac12
=
\frac14.
$$
Compute the covariance
$$
\begin{aligned}
\operatorname{Cov}(Y,Z)
&=
E[YZ]-E[Y]E[Z]
\\[4pt]
&=
\frac14
-
\frac13\frac23
\\[4pt]
&=
\frac14-\frac29
\\[4pt]
&=
\frac{9-8}{36}
\\[4pt]
&=
\frac1{36}.
\end{aligned}
$$
Compute the correlation
$$
\begin{aligned}
\operatorname{Corr}(Y,Z)
&=
\frac{
\frac1{36}
}{
\sqrt{
\frac1{18}\cdot\frac1{18}
}
}
\\[6pt]
&=
\frac{
\frac1{36}
}{
\frac1{18}
}
\\[6pt]
&=
\boxed{\frac12}.
\end{aligned}
$$
Intuition:
\(Y\) and \(Z\) are positively correlated.
If the minimum of the two observations is large,
then both observations must be relatively large.
That also tends to make their maximum large.
However, the relationship is not perfect, so the correlation
is \(1/2\), rather than \(1\).
Final Answers
$$
\boxed{
P(Y\ge y\mid Z\le z)
=
\frac{(z-y)^2}{z^2}
}
$$
for
$$
0\le y\le z\le1.
$$
$$
\boxed{
\operatorname{Corr}(Y,Z)=\frac12
}
$$