, 2020

Solution:

Let $\Delta x$ and $\Delta y$ be the positive differences between the $x$ coordinates and $y$ coordinates of the centers of the squares, respectively. Then, the length of the intersection of the squares along the $x$ dimension is $1-\Delta x$, and likewise the length along the $y$ dimension is $1-\Delta y$. In order to find the expectation of $\Delta x$ and $\Delta y$, we can find the volume of the set of points $(a, b, c)$ such that $0 \leq a, b \leq 1$ and $c \leq|a-b|$. This set is composed of the two pyramids of volume $\frac{1}{6}$ shown below:

Since the expected distance between two points on a unit interval is therefore $\frac{1}{3}$, we have that $\mathbb{E}[1-\Delta x]=\mathbb{E}[1-\Delta y]=\frac{2}{3}$. The expectation of the product of independent variables equals the product of their expectations, so the expected area of intersection is $\frac{4}{9}$ and the expected area of union is $2-\frac{4}{9}=\frac{14}{9}$.