We place a red dot within the unit square randomly and then place two blue dots also within the square, what is the expected distance from the closest blue dot to the red dot?
I can tell that when there is one blue dot the expected distance is around $\int \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} dx dy = 0.521$ when there is just one blue dot, how would we apply this problem to more blue dots?
Thanks.
Let the position of the red dot be $(a,b)$. Let $$c(a,b,r)=\int_0^1 \int_0^1 1_{(x-a)^2+(y-b)^2<r^2}\, dx \, dy.$$ Then the answer for $n$ dots is $$\int_0^1 \int_0^1 \int_0^\sqrt{2} n(1-c(a,b,r))^{n-1} \frac{\partial c(a,b,r)}{\partial r}r \, dr\, da \, db.$$ So there is a closed-form solution, but it seems too inaccessible for computing asymptotics. The problem is that the first integral is a mess of cases, since the circle and the square can intersect 0, 2, 4, 6, or 8 times.
