Steven R. Dunbar
Department of Mathematics
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University of Nebraska-Lincoln
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Topics in
Probability Theory and Stochastic Processes
Steven R. Dunbar

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Recurrence in Higher Dimensions

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Rating

Rating

Mathematicians Only: prolonged scenes of intense rigor.

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Section Starter Question

Section Starter Question

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Key Concepts

Key Concepts

  1. Every centered nearest neighbor random walk on 2 is recurrent.
  2. Every centered nearest neighbor random walk on N for N 3 is transient.

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Vocabulary

Vocabulary

  1. A random walk on the integer lattice N is
    Tn =k=1nY i.

    where Yk(ω) ∈e1,…,emN with a corresponding finite probability distribution (p1,…,pm) where p1 + ⋯ + pm = 1.

  2. The dimension of the random walk is the dimension of the subspace of N generated by the set of directions (e1,…,em) that are the possible steps.
  3. Let b1,…,bN be the standard unit basis vectors for N. The 2N possible steps (e1,…,e2N) = (b1,−b1,…,bN,−bN) with a corresponding finite probability distribution (p1,…,p2N) defines the nearest neighbor random walk on N.
  4. In the special case that pi = 1 2N for i = 1,…, 2N, the nearest neighbor random walk is the simple random walk in N.
  5. For each positive integer n the Rodrigues formula for the (non-normalized) Legendre polynomial of degree n is
    Ln(x) = d n dxn (x2 − 1)n.

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Mathematical Ideas

Mathematical Ideas

Random Walks on Integer Lattices

Let e1,…,em be a set of vectors in N with a corresponding finite probability distribution (p1,…,pm) where p1 + ⋯ + pm = 1. The probability space associated to an elementary trial is Ω1 = 1,…,m. The sample space for a sequence of independent trials of the experiment is Ωn = Ω1n with the point probability

ℙn ω(n) =k=1np ωk =i=1mp i|{k:ωk=i,1≤k≤n}|

where ω(n) = (ω 1,…,ωn) ∈ Ωn. Then the random variable is Yi(ω) = eωi for each ω ∈ Ωn. Then a random walk on the integer lattice is the sequence of random variables Tn defined by

Tn =k=1nY i.

A random walk is recurrent if the walk almost surely returns to the origin infinitely many times. That is, the walk is recurrent if the event [Tn = 0] i.o. is an almost sure event. A random walk is transient if [Tn = 0] i.o. is a negligible event, alternatively, if the walk almost surely returns to the origin only finitely many times.

A random walk is centered if 𝔼 Yi = 0.

The dimension of the random walk is the dimension of the subspace of N generated by the set of directions (e1,…,em) that are the possible steps.

Definition. Let S be the semigroup generated under addition on the set E = e1,…,em of allowed steps in the random walk, where all probabilities in the associated distribution (p1,…,pm) are positive. That is, S is the set of all elements in N that are a sum of integer multiples of elements of E. Alternatively, S is the set of points attainable in the random walk.

General Theorems about Recurrence

The lemmas and theorems in the previous section about recurrence and transience carry over to higher dimensions with only minimal changes. This is because the proofs do not depend on the dimensionality of the state space. The proofs do depend on the probabilities of return to a specific value.

Definition. For m,s,t ∈ ℕ with s ≤ t, let As,tm be the event consisting of the random walks that return to the starting point at least m times between steps s and t. That is,

ω ∈ As,tm if and only if |{n : s ≤ n ≤ t,T n(ω) = 0}|≥ m.

Lemma 1. For every m and t

A1,tm(ℙ A 1,t1)m.

Lemma 2. For every m and t

(ℙ A1,t1)m ≤ ℙ A 1,mtm .

Theorem 3. The walk is recurrent, that is,

[Tn = 0] i.o.  = 1,

if and only if

lim n→∞there exists k ≤ n such that Tk = 0  = 1.

Theorem 4 (Recurrence Criteria).

  1. If
    n=1T n = 0 < +∞.

    then

    [Tn = 0] i.o.  = 0,

    that is, the walk is transient.

  2. If
    n=1T n = 0 = +∞.

    then

    [Tn = 0] i.o.  = 1,

    that is, the walk is recurrent.

Corollary 1. Every random walk is either recurrent or transient.

Definition. For xN, the norm in the following theorem can be any norm, but for convenience, the proof uses the L-norm, x∥ = max{|x1|,…,|xN|}.

Lemma 5.

  1. The random walk Tn on N satisfies
    n=1T n = x ≤ 1 +n=1T n = 0

    for every xN

  2. n=1T n∥≤ K(2K + 1)N 1 +n=1T n = 0

    for every K > 0.

Theorem 6.

  1. If the random walk is transient, then every point of S is almost surely reached only finitely many times.
  2. If the random walk is transient, then almost surely lim n→∞Tn∥ = +∞.
  3. If the random walk is recurrent, then S is a group.
  4. If the random walk is recurrent, then the random walk almost surely reaches every point in S infinitely many times.

Random Walks in Higher Dimensions

Let Yn = (Y n1,Y n2,…,Y nN) be the sequence of ordered N-tuples of the steps of the random walk, and Tn = (Tn1,T n2,…,T nN) be the sequence of ordered N-tuples of the random walk itself.

If the random walk is not centered, then there exists at least one index i with 1 ≤ i ≤ N such that 𝔼 Y ni ≠0. Then the Strong Law of Large Numbers implies that lim n→∞|Tni| = ∞ almost surely. Thus the full walk in N is transient.

Whether a centered random walk is recurrent depends on its dimension. The general result is that every centered walk in dimension at most 2 is recurrent. On the other hand, every random walk in dimension 3 or greater is transient. The proof of these results in full generality requires Fourier analysis, so the results here are only for nearest neighbor random walks.

In the integer lattice N each point has 2N nearest neighbors. Let b1,…,bN be the standard unit basis vectors for N. The 2N possible steps (e1,…,e2N) = (b1,−b1,…,bN,−bN) define the nearest neighbor random walk on N. The random walk defined by these possible steps is centered if and only if p2i−1 = p2i for i = 1,…,n. Without loss of generality, assume that p2i−1 and p2i are non-zero for each i, so that the dimension of the random walk is N. In the special case that pi = 1 2N we say that this walk is the simple random walk in N.

Proposition 7. Suppose (Tn)n≥1 be a nearest neighbor random walk in N.

  1. T2n−1 = 0 = 0

  2. T2n = 0 =k1+k2+⋯+kN=n (2n)! (k1!k2!kN!)2i=1N(p 2i−1p2i)ki

    where each ki is a positive integer.

Proof.

  1. If ω is an element of Ω that satisfies Tm(ω) = 0, then there exist integers k1, k2kN such that the finite sequence of steps (Y1(ω),Y2(ω),…Ym(ω)) takes the value bi exactly ki times and the value bi exactly ki times for 1 ≤ i ≤ N.
  2. For such an ω to exist m must be even, setting m = 2n, then k1 + k2 + ⋯ + kN = n. This proves part 1.
  3. Let the set of sets {E1,F1,E2,F2,…,EN,FN} be a partition of the set {1,…, 2n} such that |Ei| = |Fi| = ki for 1 ≤ i ≤ N.
  4. The event that the indices j in Ei are those satisfying Yj = bj and the indices j in Fi are those satisfying Yj = bj has probability i=1N(p 2k−1p2k)k1, by independence.
  5. To be absolutely explicit, for fixed (k1,k2,…,kN) there are exactly 2n k1 2n − k1 k1 2n − 2k1 k2 2n − 2k1k2 k2 2n − 2k1 − 2k2 −⋯ − 2kN−1 kN

    ways of choosing the partition {E1,F1,E2,F2,…,EN,FN}. This product is the multinomial coefficient

    (2n)! (k1!k2!kN!)2

  6. Then summing over all possible disjoint partitions, it follows that
    T2n = 0 =k1+k2+⋯+kN=n (2n)! (k1!k2!kN!)2i=1N(p 2i−1p2i)ki

The following proofs use classical properties of Legendre polynomials. For each positive integer n the Rodrigues formula for the (non-normalized) Legendre polynomial of degree n is

Ln(x) = d n dxn (x2 − 1)n.

The first few of the (non-normalized) Legendre polynomials are L0(x) = 1, L1(x) = 2x, L2(x) = 12x2 − 4, L3(x) = 120x3 − 72x. The standard normalized definition of the Legendre polynomials is

Pn(x) = 1 2nn! d n dxn (x2 − 1)n.

Also note that the standard notation for the normalized Legendre polynomials is Pn(x).

For each positive integer n define

gn = (2n)! (2nn!)2.

Note that gn is the probability of exactly an equal number of heads and tails in 2n flips of a fair coin. This fact is used later in the proofs, but here it shows that gn is a familiar quantity.

Lemma 8. For every positive integer n and every nonzero real number y,

Ln 1 2 y + 1 y = 2nn!k=0ng kgn−ky2k−n.

Proof.

  1. The binomial expansion is
    (x2 − 1)n =k=0n(−1)kn kx2(n−k).

  2. Differentiating n times term by term gives
    Ln(x) = n!k=0⌊n∕2⌋(−1)k (2(n − k))! k!(n − k)!(n − 2k)!xn−2k.

  3. Use the binomial theorem for the series expansion of (1 − u)−1∕2
    (1 − u)−1∕2 = 1 + 1 2u + 1 2!1 23 2u2 + 1 3!1 23 25 2u3 + ….

    With mathematical induction on the coefficients this is a way to show that the Taylor series for f1(u) = (1 − u)−1∕2 about 0 is

    1 1 − u =n=0g nun.

  4. Set u = 2tx − t2 to see that (1 − 2xt + t2)−1∕2 = 1 + 1 2(2xt − t2) + 1 2!1 23 2(2xt − t2)2+ 1 3!1 23 25 2(2xt − t2)3 + ….

  5. Expand powers and collect in powers of t (1 − 2xt + t2)−1∕2 = 1 + xt + 1 2(3x2 − 1)t2 + (5x33x 2 )t3+ (3 815x2 4 )t4 + 15x 8 t55 16t6 + ⋯.

    Since L0(x) = 1, L1(x) = 2x, L2(x) = 12x2 − 4 and L3(x) = 120x3 − 72x, it becomes apparent that

    1 1 − 2xt + t2 =n=0 1 2nn!Ln(x)tn.

  6. Let y≠0 and x = 1 2(y + 1 y). Then
    1 − 2xt + t2 = (1 − yt) 1 −1 yt

    so

    1 1 − 2xt + t2 = 1 1 − yt 1 1 − 1 yt = n=0g nyntn n=0g ny−ntn .

  7. Then multiplying the series, gathering powers of t, and comparing coefficients
    2nn! Ln(x) =k=0ng kgn−ky2k−n.

Remark.


Vector diagram

Figure 1: Vector diagram for the potential at point P from an electric charge q.

Here is another view of the generating function for Legendre polynomials in step 5. The origin of the Legendre polynomials is in the treatment of potential problems from gravity and electrostatics. For example, the potential at r due to a point charge q located at the point r, assuming r > r, is given by

Φ(r) = 1 𝜖0 q |rr| = 1 𝜖0 q (r2 − 2rr + r2)1∕2 = q 𝜖0r 1 (1 − 2(r∕r) cos 𝜃 + (r∕r)2)1∕2

Let cos 𝜃 = x and r∕r = t to express the potential as

Φ(r) = q 𝜖0r 1 − 2xt + t2 −1∕2.

The expansion of the electrostatic potential in spherical harmonics written with Legendre polynomials is

Φ(r) = q 𝜖0rn=0P n(x)t2.

In physics, the spherical harmonics are also called the monopole moment, dipole moment, quadrupole moment, and so on, leading to the multipole expansion. Comparing these expressions for the potential,

1 1 − 2xt + t2 =n=0 1 2nn!Ln(x)tn.

This is the generating function in step 5 in the proof of Lemma 8.

Lemma 9. The two-dimensional centered random walk (Tn)n=1 satisfies

T2n = 0 = gnk=0ng kgn−k(4p − 1)2k

for every n.

Remark. Note that if p = 1∕4 and q = 1∕4, the probability of return to the origin in n steps is the just the first term of the sum. Specifically T2n = 0 = gn(g0gn) = gn2. For small values of n, this is easy to confirm. For example, let L = (1, 0), R = (−1, 0), U = (0, 1), and D = (0,−1). Return to the origin in 2 steps can occur in 4 ways, RL, LR, UD and DU, each with probability 1∕16. The total probability of return to the origin in 2 steps is 4∕16 = 1∕4. On the other hand, g1 = 1∕2 so g12 = 1∕4.

Return to the origin in 4 steps can occur in several ways. The walk can take steps L, R, L and R in 4! 2!2! = 6 ways. The walk can take steps U, D, U and D in 6 ways. The walk can take steps R, L, U, D in 4! = 24. This makes a total of 36 paths with return to the origin, each with probability (1∕4)4 = 1∕256 for a total probability 9∕64. On the other hand, g2 = 3∕8 so g22 = 9∕64.

The proof below uses polynomial expansions to keep track of the possible combinations in longer collections of steps returning to the origin. Then the Rodrigues definition of the Legendre polynomials arises naturally to express the probability of return to the origin.

Proof.

  1. Since the two-dimensional random walk is centered, p1 = p2 is the probability that the walk steps in one direction and p3 = p4 = 1 2p1 is the probability that the walk steps in the complementary direction. Set p1 = p and p3 = q. (Note that this is not the frequent use of p and q for the complementary probabilities of heads and tails in a coin flip.)
  2. Lemma 8 shows that T2n = 0 =k=0n (2n)! (k!(n − k)!)2p2kq2(n−k) = 22ng nk=0n n k2p2kq2(n−k).
  3. From binomial expansions
    (p2 + x)n =k=0nn kp2kxn−k

    and

    (x + q2)n =k=0nn kxkq2(n−k).

  4. Comparing these expansions, see that T2n = 0 is the coefficient of xn in the polynomial
    Q(x) = 22ng n(p2 + x)n(x + q2)n.

  5. In the case that p = q = 1 4
    T2n = 0 = 22ng n2n n 1 4 2n = g n2

    using the binomial identity

    k=0n n k2 = 2n n .

  6. In the case that p≠q, Q(x) = 22ng n(p2q2 + (p2 + q2)x + x2)n and using step 4 that T2n = 0 is the coefficient of xn in the polynomial
    Q(x) = 22ng n(p2 + x)n(x + q2)n

    then

    T2n = 0 = 1 n!Q(n)(0).

  7. Completing the square on (p2q2 + (p2 + q2)x + x2), rewrite Q(x) as
    Q(x) = 22ng n p2q2 2 2n 2x + p2 + q2 p2q2 2 − 1 n.

  8. Using the variable x = 2x+p2+q2 p2q2 , and the Rodrigues formula definition of Legendre polynomials, then
    Q(n)(0) = 22ng n p2q2 2 2L n p2 + q2 p2q2 .

  9. Since q = 1 2 − p, write this as
    Q(n)(0) = 22ng n 4p − 1 8 nL n 1 2 4p − 1 + 1 4p − 1 .

  10. Finally, using Lemma 8
    Q(n)(0) = n!g nk=0ng kgn−k(4p − 1)2k.

    Thus

    T2n = 0 = gnk=0ng kgn−k(4p − 1)2k

    for every n.

Theorem 10. Every centered nearest neighbor random walk on 2 is recurrent.

Remark. The strategy of the proof is to use the formula for the probability of the uniform centered random walk with p = 1∕4 returning to the origin to show that the uniform centered random walk is recurrent. Since the probability of returning when p≠1∕4 is less than when p = 1∕4, this implies that all centered random walks in 2 are recurrent.

Proof.

  1. Take p = 1∕4. By Lemma 9
    T2n = 0 = gn2 = ((2n)!)2 (2nn!)4 .

  2. Note that gn is the probability that a simple one-dimensional random walk returns to the origin after 2n steps.
  3. Estimate gn with Stirling’s Formula to obtain
    gn1 .

  4. Thus
    n=1T 2n = 0 = ∞

    and by part 2 of Theorem 4 the walk is recurrent.

Theorem 11. Every centered nearest neighbor random walk on N for N ≥ 3 is transient.

Remark. The overall strategy is to show that the centered random walk in N dimensions, where N ≥ 3, has

T2n = 0 = O(n−N∕2).

This implies that nTn = 0 converges so by part 1 of Theorem 4 the walk is transient. To be specific, this proof is just for N = 3. Since higher dimensional lattices contain 3, random walks in N are also transient. The steps in the overall strategy are to first prove some inequalities about binomial coefficients and factorials. Then break the sum expressing the probability of return to the origin into 4 parts. The first 3 parts express the contributions where the number of steps in each of the three coordinate directions is “far” from np, nq,and nr respectively. Using the Large Deviations Estimate, these three parts are each exponentially small. The fourth sum is the probability that the number of steps in each of the three coordinate directions is “approximately” np, nq, and nr. Using the binomial estimates, this probability is of order n−3∕2. Then the sum of probabilities of returns to the origin is a convergent series and the walk is transient.

Proof.

  1. Begin with the observation that 1 (n!)222n (2n)!. This follows from the binomial expansion estimate (1 + 1)2n 2n n .
  2. Also use the estimate that there is a c > 0 such that
    1 (n!)2 ≤ c 22n n(2n)! (1)

    for every n ≥ 0 This follows from gn1 already used above in step 3 in the proof of recurrence for the random walk in 2.

  3. Let the step probabilities in the centered random walk in 3 be p = p1 = p2, q = p3 = p4, r = p5 = p6, with p,q,r > 0 and p + q + r = 1∕2. Using Proposition 7
    T2n = 0 =i+j+k=n (2n)! (i!j!k!)2p2iq2jr2k.

  4. Make the following definitions:
    1. I(p,n) = i : 0 ≤ i ≤ n, p − i 2n < p 2.

      An equivalent definition is

      I(p,n) = i : 0 ≤ i ≤ n,pn < i < 3pn. (2)

      See Figure 2 for an illustration of the first definition.


      Example of definition of I(p,n)

      Figure 2: Example of the definition of I(p,n) with p = 1∕8 and n = 16.

    2. Ap = i+j+k=n, i∉I(p,n) (2n)! (i!j!k!)2p2iq2jr2k

    3. B = i+j+k=n, i∈I(p,n),j∈I(q,n),k∈I(r,n) (2n)! (i!j!k!)2p2iq2jr2k (3)
  5. By step 1 Ap i+j+k=n i∉I(p,n) (2n)!22n (2i)!(2j)!(2k)!p2iq2jr2k.
  6. Rewriting the sum in the inequality
    Ap 0≤i≤n i∉I(p,n) (2n)!22n (2i)! p2ij=0n 1 (2j)!((2(n − i − j))!)q2jr2(n−i−j).

  7. By the binomial expansion,
    j=0n 1 (2j)!((2(n − i − j))!)q2jr2(n−i−j) = 1 (2(n − i))!(q + r)2(n−i).

  8. Combine the last two expressions, express the factorials as a binomial coefficient, and factor the powers of 2 through to rewrite the inequality as
    Ap 0≤i≤n i∉I(p,n) 2n 2i (2p)2i(2(q+r))2(n−i). (4)
  9. When i∉I(p,n) we have 2i 2n − 2p ≥ p Therefore, the upper bound for Ap is a probability controlled by the large deviations estimate: Let Sn be a binomial random variable with parameters n and 2p. Then (4) becomes
    Ap ≤ ℙ 2p −Sn 2n ≥ p.

  10. By the Large Deviations Estimate,
    Ape−nd

    for a constant d > 0. Likewise the sums Aq and Ar of

    (2n)! (i!j!k!)2p2iq2jr2k

    for j∉I(q,n) and k∉I(r,n) are also bounded above by exponentially small quantities.

  11. Combining the exponential inequalities for i∉I(p,n), j∉I(q,n) and k∉I(r,n) leaves only estimating the quantity B where the indices i, j, k satisfy i > np, j > nq, and k > nr respectively. By (1), (??), (??)
    B ≤i+j+k=n(2n)! c22i i(2i)! c22j j(2j)! c22k k(2k)!p2iq2jr2k

    so

    B ≤ c3 pqrn3∕2 i+j+k=n i∉I(p,n),j∉I(q,n),k∉I(r,n) (2n)! (2i)!(2j)!(2k)!(2p)2i(2q)2j(2r)2k.

  12. By the multinomial expansion
    i+j+k=n (2n)! (i)!(j)!(k)!(2p)i (2q)j (2r)k = (2p + 2q + 2r)2n = 1.

  13. Hence
    B ≤ c3 pqrn3∕2

    which is enough to complete the proof of transience using the convergence criterion 1 of the Recurrence Criteria, Theorem 4

Application

Consider N simultaneous fair coin-tossing games carried on independently of each other. In terms of random variables, the games are

Y 1(1),Y 2(1),… Y 1(2),Y 2(2),… Y 1(N),Y 2(N),…

where each Y i(j) = ±1 is the ith Heads or Tails outcome in the jth game. Let Tn(j) = i=1jY i(j) be the cumulative fortune in the jth game. Consider the N-tuple Tn = (Tn(1),…T n(N)) of fortunes at time n. Then Tn = 0 if and only equalization takes place in all N games simultaneously. Then

Tn = 0 i.o.  = 1,N = 1, 20, N ≥ 3.

Sources

This section is adapted from: E. Lesigne, Heads or Tails: An Introduction to Limit Theorems in Probability, Chapter 13, American Mathematical Society, Student Mathematical Library, Volume 28, 2005. The remark on the connection of the generating function for the Legendre polynomials to potential for a charge is adapted from The Legendre Polynomials.. The application is from L. Breiman, Probability, Section 3.7, Problem 18, page 58.

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Algorithms, Scripts, Simulations

Algorithms, Scripts, Simulations

Algorithm

Scripts

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Problems to Work

Problems to Work for Understanding

  1. Prove Lemma 1 in n.
  2. Prove Lemma 2 in n.
  3. Prove Theorem 3 in n.
  4. Prove part 1 of Lemma 5 in n.
  5. Prove part 2 of Lemma 5 in n.
  6. Prove part 1 of Lemma 6 in n.
  7. Prove part 2 of Lemma 6 in n.
  8. Prove part 3 of Lemma 6 in n.
  9. Prove part 4 of Lemma 6 in n.
  10. Show by examples for n = 3 and 4 that differentiating the Rodrigues definition of Legendre polynomials n times term by term gives
    Ln(x) = n!k=0⌊n∕2⌋(−1)k (2(n − k))! k!(n − k)!(n − 2k)!xn−2k.

  11. Show that the leading terms agree in both representations of the Legendre polynomials:
    Ln(x) = d n dxn (x2 − 1)n

    and

    Ln(x) = n!k=0⌊n∕2⌋(−1)k (2(n − k))! k!(n − k)!(n − 2k)!xn−2k.

  12. For n even, show that the final constant terms agree in both representations of the Legendre polynomials:
    Ln(x) = d n dxn (x2 − 1)n

    and

    Ln(x) = n!k=0⌊n∕2⌋(−1)k (2(n − k))! k!(n − k)!(n − 2k)!xn−2k.

  13. Show that the Taylor series for f1(u) = (1 − u)−1∕2 about 0 is
    1 1 − u =n=0g nun.

  14. Prove the binomial identity
    k=0n n k2 = 2n n .

  15. Show that completing the square on (p2q2 + (p2 + q2)x + x2), allows rewriting Q(x) as
    Q(x) = 22ng n p2q2 2 2n 2x + p2 + q2 p2q2 2 − 1 n.

  16. Estimate gn with Stirling’s Formula to obtain
    gn1 .

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Books

Reading Suggestion:

References

[1]   Emmanuel Lesigne. Heads or Tails: An Introduction to Limit Theorems in Probability, volume 28 of Student Mathematical Library. American Mathematical Society, 2005.

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Links

Outside Readings and Links:

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