A short proof of Rayleighs Theorem Olivier Bernardi (MIT) Microsoft - - PowerPoint PPT Presentation

Microsoft Research, January 2011 Olivier Bernardi (MIT)

A short proof of Rayleigh’s Theorem

Spitzer’s walks in the plane Walks made of n unit steps. Direction of each step is uniformly random.

Spitzer’s walks in the plane Theorem [Rayleigh] The probability for the walk to end at distance less than 1 from the origin is 1 n + 1. Walks made of n unit steps. Direction of each step is uniformly random.

Spitzer’s walks in the plane Theorem [Rayleigh] The probability for the walk to end at distance less than 1 from the origin is 1 n + 1. Walks made of n steps of random lengths distributed as X. Direction of each step is uniformly random. X∗6 X∗4 Theorem [B.] For any positive random variable X, P(X∗i > X∗j) = i i + j .

• Lemma. For any positive random variables A, B, C,

P(A > B ∗ C) + P(B > A ∗ C) + P(C > A ∗ B) = 1 Proof.

• Lemma. For any positive random variables A, B, C,

P(A > B ∗ C) + P(B > A ∗ C) + P(C > A ∗ B) = 1 Proof.

• Proof. We condition on A = a, B = b, C = c and prove:

P(a > b ∗ c) + P(b > a ∗ c) + P(c > a ∗ b) = 1.

• If a ≥ b + c or b ≥ a + c or c ≥ a + b, obvious.
• Otherwise, consider the angles α, β, γ of the triangle.

One has P(a > b ∗ c) = 2α 2π etc. Hence P(a > b ∗ c)+P(b > a ∗ c)+P(c > a ∗ b) = α + β + γ π = 1. a b c α γ β

• Lemma. For any positive random variables A, B, C,

P(A > B ∗ C) + P(B > A ∗ C) + P(C > A ∗ B) = 1 Proof. Equivalently, P(A > B ∗ C) + P(B > A ∗ C) = P(A ∗ B > C). Let P(i, n) = P(X∗i > X∗(n−i)). We want to prove P(i, n) = i n.

• Lemma. For any positive random variables A, B, C,

P(A > B ∗ C) + P(B > A ∗ C) + P(C > A ∗ B) = 1 Proof. Equivalently, P(A > B ∗ C) + P(B > A ∗ C) = P(A ∗ B > C). Let P(i, n) = P(X∗i > X∗(n−i)). We want to prove P(i, n) = i n. Apply the lemma to A = X∗i, B = X∗j, C = X∗(n−i−j). This gives P(i, n) + P(j, n) = P(i + j, n) whenever i + j ≤ n. Thus n P(1, n) = P(n, n) = 1. This gives P(1, n) = 1 n, and P(i, n) = i P(1, n) = i n.

The end.