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Random Walk
Sarah Days-Merrill
April 27, 2018
1 The Ballot Theorem
Suppose Candidate A gets m votes and Candidate B gets n votes with m > n. We want to see how many ways we can count the votes such that A is always in the lead. Votes will be shown with a lattice path from (0, 0) to (m + n, m − n).
Theorem 1.1. Let n and m be positive integers. There are exactly ( n n+m 2
paths ((0, 0), (1, y 1 ), (2, y 2 ),... , (n − 1 , yn− 1 ), (n, m)) from the origin to the point (n, m), n, m > 0 , |yk − yk+1| = 1, such that y 1 , y 2 ,... , yn > 0.
Lemma 1.2 (The Reflection Principle). For n, m, y > 0 , the number of paths from (0, y) to (n, m) which touch the t-axis equals the number of paths from (0, −y) to (n, m).
Proof. Consider a path ((0, y), (1, y 1 ),... , (n− 1 , yn− 1 ), (n, m)) from (0, y) to (n, m) which has at least one vertex on the t-axis. Let r be the time of the first visit to zero; that is r satisfies yi > 0 ,... , yr− 1 > 0 , yr = 0. Then ((0, −y), (1, −y 1 ),... , (r − 1 , −yr− 1 ), (r, yr),... , (n, m)) is a path leading from (0, −y) to (n, m). This map is a bijection and the Reflection Principle follows.
Lemma 1.3. For m, n ∈ N, the number of paths from (0, 0) to (n, m) is ( n n+m 2
Proof. Denote u as the number of up steps and d as the number of down steps. The total number of steps is u + d = n
and the net number of up steps is u − d = m.
Solving for u gives
u = n + m 2
and the lemma follows, since paths are determined by specifying which of the n steps are up steps.
1.1 Proof of The Ballot Theorem
Proof of 1.1. The Reflection Principle shows that for k, n, m > 0, the number of lattice paths from (0, k) to (n, m) which touch the t-axis (horizontal axis) is equal to the number of paths from (0, −k) to (n, m). Suppose Candidate A gets m votes and Candidate B gets n votes. We want to see how many ways we can count the votes such that A is always in the lead. Votes will be shown with a lattice path from (0, 0) to (m + n, m − n). Then the number of lattice paths from (0, 0) to (m + n, m − n) is (^) ( m + n n
Now, consider the number of ways with Candidate A always in the lead. Candidate A must always get the first vote. Then we need to find the number of ways of counting the votes where A is always in the lead while getting the first vote and the number of ways of counting the votes where A gets the first vote, but B is tied or ahead of A at least one time. In other words, we need to determine the number of lattice paths from (1, 1) to (m+n, m−n) and subtract from it the number of lattice paths from (1, 1) to (m + n, m − n) which touch the horizontal t-axis. We can change one of the starting points (1, 1) to (1, −1) since, according to the Reflection Principle, the number of lattice paths will remain the same. Then we can compare the number of lattice paths from (1, 1) to (m + n, m − n) and subtract from it the number of lattice paths from (1, −1) to (m + n, m − n). Now, suppose we change the starting position to the origin, then we will be subtracting the number of lattice paths from (0, 0) to (m + n − 1 , m − n − 1) from the number of lattice paths from (0, 0) to (m + n − 1 , m − n + 1). Then we have
#paths =
m + n − 1 (m+n−1)+(m−n−1) 2
m + n − 1 (m+n−1)+(m−n+1) 2
m + n − 1 m − 1
m + n − 1 m
(m + n − 1)! (m − 1)!n!
(m + n − 1)! m!(n − 1)! = (m − n)(m + n − 1)! m!n! =
m − n m + n
m + n m
Therefore, the probability that Candidate A is always in the lead is
P (A is always in the lead) =
m−n m+n
(m+n m
(m+n m
) (^) = m^ −^ n m + n
Since
P (S 1 = 1, S 2 ,... , S 2 n− 1 > 0 , S 2 n = 2r)
=
number of paths from (1, 1) to (2n, 2 r) which never touch t-axis 22 n−^1 = number of paths from (1, 1) to (2n, 2 r) − number of paths from (0, 0) to (2n − 1 , 2 r + 1) 22 n =
( (^2) n− 1 n+r− 1
( 2 n− 1 n+r
22 n^
we have
P (S 1 ,... , S 2 n > 0) =
∑^ n
r=
P (S 1 = 1, S 2 ,... , S 2 n− 1 > 0 , S 2 n = 2r)
22 n
∑n
r=
[(
2 n − 1 n + r − 1
2 n − 1 n + r
)]
4 n
[(
2 n − 1 n
2 n − 1 2 n
)]
4 n
2 n − 1 n
so
P (S 1 ,... , S 2 n 6 = 0) =
4 n^
2 n − 1 n
4 n^
(2n − 1)! n!(n − 1)!
n n
=
4 n
(2n)! n!n! = 4−n
2 n n
3 Recurrence
One can also consider simple random walks with asymmetric increment distributions (so that P (X = 1) 6 = P (X = −1)), as well as simple random walks in higher dimensions. On Zd, simple symmetric random walk is the process defined by Sn =
∑n k=1 Xk^ where the^ Xk’s are independent and equally likely to be any of the 2d vectors
∈^ Zd.
There are also random walks which are not simple, meaning that the walker is not constrained to move only to neighboring sites.
Definition 3.1. A random walk Sn is recurrent if P (Sn = 0 i.o.) = 1. That is, Sn is recurrent iff the lattice path touches the t-axis infinitely often. Otherwise, the random walk is transient.
To state our next result, we define τ 0 = 0 and τn = inf{k > τn− 1 : Sk = 0} for n ≥ 1. In other words, τn is the time of the nth^ return to 0.
Lemma 3.1. For any random walk, the following are equivalent:
- P (τ 1 < ∞) = 1
- P (Sn = 0 i.o.) = 1
∑^ ∞
n=
P (Sn = 0) = ∞
Proof. First note that P (τn < ∞) = P (τ 1 < ∞)n^ for all n ∈ N. Indeed this holds trivially when n = 1, and if P (τn < ∞) = P (τ 1 < ∞)n, then
P (τn+1 < ∞) = P (τn+1 < ∞|τn < ∞)P (τn < ∞) = P (τ 1 < ∞)P (τn < ∞) = P (τ 1 < ∞)n+1.
If P (τ 1 < ∞) = 1, then P (τn < ∞) = 1n^ = 1 for all n, so P (Sn = 0 i.o.) = 1. Using the converse of Borel-Cantelli 1, we see that if P (Sn = 0 i.o.) = 1, then
n=1 P^ (Sn^ = 0) =^ ∞. Finally,
n=1 P^ (Sn^ = 0) =^ ∞^ implies^ P^ (τ^1 <^ ∞) = 1 since considering^ N^ as the number of times the simple random walk touches the t-axis gives
N =
∑^ ∞
k=
1 {Sk = 0}
∑^ ∞
k=
1 {τk < ∞},
so, taking p = P (τ 1 < ∞),
∑^ ∞
k=
P (Sk = 0) = E[N ] =
∑^ ∞
k=
P (τk < ∞)
∑^ ∞
k=
pk^ = p 1 − p if p < 1.
Therefore, all the above statements are equivalent and can be derived from each other.
It follows from the previous result that when d = 1 every site in Z is visited infinitely often with probability one. However, one can show that the expected time to travel between any two sites (or return to the present site) is infinite!
Theorem 3.3. Simple random walk is transient in three or more dimensions.
Proof. When d = 3,
P (S 2 n = 0) = 6−^2 n^
n^ n^1 ,n^2 ,n^3 ≥0: 1 +n 2 +n 3 =n
2 n
n 1 !n 2 !n 3!
= 2−^2 n
2 n n
n^ n^1 ,n^2 ,n^3 ≥0: 1 +n 2 +n 3 =n
3 −n^ n! n 1 !n 2 !n 3!
Now 3−n^ n 1 !nn 2 !!n 3! ≥ 0 for each choice of n 1 , n 2 , n 3 , n, and the multinomial theorem gives
∑
n 1 ,n 2 ,n 3 ≥0: n 1 +n 2 +n 3 =n
3 −n^ n! n 1 !n 2 !n 3!
n 1 ,n 2 ,n 3 ≥0: n 1 +n 2 +n 3 =n
n n 1 , n 2 , n 3
)n 1 ( 1 3
)n 2 ( 1 3
)n 3
so
∑
n 1 ,n 2 ,n 3 ≥0: n 1 +n 2 +n 3 =n
3 −n^ n! n 1 !n 2 !n 3!
max (^0) n≤n 1 ≤n 2 ≤n 3 : 1 +n 2 +n 3 =n
3 −n^ n! n 1 !n 2 !n 3!
n 1 ,n 2 ,n 3 ≥0: n 1 +n 2 +n 3 =n
3 −n^ n! n 1 !n 2 !n 3!
= 3−n^ max (^0) n≤n 1 ≤n 2 ≤n 3 : 1 +n 2 +n 3 =n
n! n 1 !n 2 !n 3!
The latter quantity is maximized when n 1 !n 2 !n 3! is minimized. This happens when n 1 , n 2 , n 3 are as close as possible: If ni < nj − 1 for i < j, then ni!nj! > ni n+1j ni!nj! = (ni + 1)!(nj − 1)!. It follows that
max (^0) n≤n 1 ≤n 2 ≤n 3 : 1 +n 2 +n 3 =n
n! n 1 !n 2 !n 3!
n! ([n 3
]
2 πn
(n e
)n (√ 2 πn 3
( (^) n 3 e
)ne^ )^3
32 (n e
)n
2 πn
( (^) n 3 e
)n ≤ 3 n n
Putting all this together and recalling that (^212) n
( 2 n n
≈ √^1 πn shows that
P (S 2 n = 0) = 2−^2 n
2 n n
n 1 ,n 2 ,n 3 ≥0: n 1 +n 2 +n 3 =n
3 −n^
n! n 1 !n 2 !n 3!
≤ 2 −^2 n
2 n n
n
c n
where c is a constant.
Hence
n=1 P^ (Sn^ = 0)^ <^ ∞^ and we conclude that Simple Random Walk is transient in 3-dimensions.
Transience in higher dimensions follows by letting Tn = (S n^1 , S n^2 , S n^3 ) be the projection onto the first three coordinates and letting N (n) = inf{m > N (n − 1) : Tm 6 = TN (n−1)} to be the nth^ time that the random walker moves in any of the first three coordinates (with the convention that N (0) = 0). Then TN (n) is a simple random walk in three dimensions and the probability that TN (n) = 0 infinitely often is 0. Since the first three coordinates of Sn are constant between N (n) and N (n + 1) and N (n + 1) − N (n) is almost surely finite, this implies that Sn is transient.
4 Arcsine Laws
In this section, we focus on simple random walk on Z. Define
Ln = max{ 0 ≤ k ≤ n : Sk = 0}
to be the time of the last visit to zero by time n.
Lemma 4.1. Let u 2 m = P (S 2 m = 0). Then P (L 2 n = 2k) = u 2 ku 2 n− 2 k for k = 0, 1 ,... n.
Proof.
P (L 2 n = 2k) = P (S 2 k = 0, S 2 k+1 6 = 0,... , S 2 n 6 = 0) = P (S 2 k = 0, X 2 k+1 6 = 0,... , X 2 k+1 +... + X 2 n 6 = 0) = P (S 2 k = 0)P (X 2 k+1 6 = 0,... , X 2 k+1 +... + X 2 n 6 = 0) = P (S 2 k = 0)P (S 1 6 = 0,... , S 2 n− 2 k 6 = 0) = u 2 ku 2 n− 2 k.
The preceding observation allows us to prove the second arcsine law.
Theorem 4.2. For 0 < a < b < 1 ,
P
a ≤ L 2 n 2 n ≤ b
∫ (^) b
a
π
x(1 − x)
dx.
Proof. We first note that
nP (L 2 n = 2k) = nu 2 ku2(n−k) ≈ n √ πk
π(n − k)
π
k n (1^ −^
k n )
so if (^) nk → x, then
nP (L 2 n = 2k) =
(nP (L 2 n = 2k) 1 π √ (^) k n (1−^ kn )
π
k n (1^ −^
k n )
π
x(1 − x)
Proposition 3 (Continuity Above). If A 1 ⊇ A 2 ⊇ A 3 ⊇.. ., then
P
i=
Ai
= lim n→∞ P (An).
Proof. If A 1 ⊇ A 2 ⊇ A 3 ⊇.. ., then AC 1 ⊆ AC 2 ⊆.. ., so
P
i=
Ai
= 1 − P
i=
Ai
)C )
= 1 − P
i=
ACi
= 1 − (^) nlim→∞ P (ACn ) = 1 − (^) nlim→∞(1 − P (An)) = lim n→∞ P (An)
Proposition 4 (Borel-Cantelli I). If A 1 , A 2 ,... are events with
n=1 P^ (An)^ <^ ∞, then
P (An i.o.) := P
n=
⋃^ ∞
m=n
Am
Proof. If Bn =
m=n Am, then^ B^1 ⊇^ B^2 ⊇^ B^3 ⊇^.. ., so
P (An i.o.) = P
n=
⋃^ ∞
m=n
Am
= P
n=
Bn
= lim n→∞ P (Bn).
The result follows since
n=1 P^ (An)^ <^ ∞^ implies
P (Bn) = P
m=n
Am
∑^ ∞
m=n
P (Am) → 0
as n → ∞.
