Average distance (maximum norm) from origin to the nearer of two random points in a unit square

Question

I got a unit square, the origin (0,0) and two random points which are uniformly distributed on the 1x1 unit square. What is the average distance in maximum norm from origin to the nearer of the two random points?

So for example we've got (0,0) and (0.4,0.3) and (0.2,0,9) as random points => distance to origin in maximum norm is 0.4 and 0.9 => so the minimum would be 0.4.

I already know that the average distance between origin and ONE random point in a unit square is 2/3 (maximum norm)...

Thanks for your help.

Answer 1

For two independent random variables $X, Y$ you can calculate the CDF of $\min(X, Y)$ and $\max(X, Y)$ via the formulae

$$ P(\max(X, Y) \le x) = P(X \le x \wedge Y \le x) = P(X \le x) P(Y \le x)$$ $$P(\min(X, Y) \le x) = 1 - P(\min(X, Y) > x) = 1 - P(X > x \wedge Y > x)\\ = 1 - P(X > x) P(Y > x) = 1 - (1 - P(X \le x))(1 - P(Y \le x))$$

Now apply this to determine the CDF of $\min(\max(X_1, X_2), \max(X_3, X_4))$, where the $X_i$ are i.i.d uniformly on $[0, 1]$ distributed. From the CDF you can then calculate the mean via the well-known formulae.