We are given two independent and identically distributed (i.i.d.) random variables, $X$ and $Y$, both following a uniform distribution on $(0,1)$. We define $Z = \max(X,Y)$ and $W = \min(X,Y)$. We need to find the probability $P((Z-W) > 1/2)$.

Understanding the Difference Z-W

First, let's simplify the expression $Z-W$.

• If $X \ge Y$, then $Z=X$ and $W=Y$. So, $Z-W = X-Y$.
• If $Y > X$, then $Z=Y$ and $W=X$. So, $Z-W = Y-X$.

In both cases, $Z-W = |X-Y|$. Therefore, the probability we need to calculate is $P(|X-Y| > 1/2)$.

Geometric Interpretation

Since $X$ and $Y$ are i.i.d. uniform $(0,1)$ variables, their joint distribution is uniform over the unit square $[0,1] \times [0,1]$. The total area of this sample space is $1 \times 1 = 1$. The probability $P(|X-Y| > 1/2)$ is equivalent to the area of the region within the unit square where $|X-Y| > 1/2$.

The condition $|X-Y| > 1/2$ splits into two inequalities:

1. $X-Y > 1/2 \implies Y < X - 1/2$
2. $Y-X > 1/2 \implies Y > X + 1/2$

Calculating the Area

Let's find the area corresponding to these inequalities within the unit square:

• Region 1 ($Y < X - 1/2$): This inequality defines the area below the line $y = x - 1/2$. Within the unit square, this region is a right-angled triangle with vertices at $(1/2, 0)$, $(1, 0)$, and $(1, 1/2)$. The base of this triangle is $1 - 1/2 = 1/2$, and the height is $1/2$. The area is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$.
• Region 2 ($Y > X + 1/2$): This inequality defines the area above the line $y = x + 1/2$. Within the unit square, this region is a right-angled triangle with vertices at $(0, 1/2)$, $(0, 1)$, and $(1/2, 1)$. The base of this triangle is $1 - 1/2 = 1/2$, and the height is $1/2$. The area is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$.

These two regions are disjoint.

Final Probability Calculation

The total probability $P(|X-Y| > 1/2)$ is the sum of the areas of these two regions:

$ P((Z-W) > 1/2) = P(|X-Y| > 1/2) = \text{Area(Region 1)} + \text{Area(Region 2)} $

$ P((Z-W) > 1/2) = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4} $

Thus, the probability is $1/4$.