The intersection exponent for two simple random walks on the plane is 5/8

You need to know: Integer lattice ${\mathbb Z}^2$ , vertices of ${\mathbb Z}^2$ , basic probability theory, simple random walk on ${\mathbb Z}^2$ .

Background: For a simple random walk S on ${\mathbb Z}^2$ and integer $k\geq 0$ , denote $S[0,k]$ to be the (random) set of vertices S visited after k steps.

The Theorem: On 27th March 2000, Gregory Lawler, Oded Schramm and Wendelin Werner submitted to Acta Mathematica a paper in which they proved, among other results, that there exists a constant $c>0$ such that inequality $c^{-1}k^{-5/8} \leq P[S[0,k]\cap S'[0,k] = \emptyset] \leq ck^{-5/8}$ holds for all $k\geq 1$ , where S and S’ are two independent simple random walks that start from neighbouring vertices in ${\mathbb Z}^2$ .

Short context: If $P[S[0,k]\cap S'[0,k] = \emptyset]$ decays proportional to $k^{-\gamma}$ for some constant $\gamma>0$ , then $\gamma$ is called the intersection exponent. In 1988 Duplantier and Kwon used ideas from theoretical physics to predicts that, for two simple random walks on the plane, $\gamma=5/8$ . The Theorem provides a rigorous mathematical proof of this prediction.