A short proof of Rayleigh’s Theorem
Olivier Bernardi (MIT)
Microsoft Research, January 2011
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
Walks made of n unit steps.
Direction of each step is uniformly random.
Theorem [Rayleigh] The probability for the walk to end at distance
1
less than 1 from the origin is
.
n+1
Spitzer’s walks in the plane
Walks made of n steps of random lengths distributed as X.
Direction of each step is uniformly random.
X ∗6
X ∗4
Theorem [Rayleigh] The probability for the walk to end at distance
1
less than 1 from the origin is
.
n+1
Theorem [B.] For any positive random variable X,
i
∗i
∗j
P(X > X ) =
.
i+j
Proof.
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. 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.
2α
One has P(a > b ∗ c) =
etc.
2π
α+β+γ
Hence P(a > b ∗ c)+P(b > a ∗ c)+P(c > a ∗ b) =
= 1.
π
a
γ
β
b
α
c
Proof.
Lemma. For any positive random variables A, B, C,
P(A > B ∗ C) + P(B > A ∗ C) + P(C > A ∗ B) = 1
Equivalently, P(A > B ∗ C) + P(B > A ∗ C) = P(A ∗ B > C).
Let P (i, n) = P(X
∗i
>X
∗(n−i)
i
). We want to prove P (i, n) = .
n
Proof.
Lemma. For any positive random variables A, B, C,
P(A > B ∗ C) + P(B > A ∗ C) + P(C > A ∗ B) = 1
Equivalently, P(A > B ∗ C) + P(B > A ∗ C) = P(A ∗ B > C).
Let P (i, n) = P(X
∗i
>X
∗(n−i)
i
). We want to prove P (i, n) = .
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.
1
i
This gives P (1, n) = , and P (i, n) = i P (1, n) = .
n
n
The end.