MathIsimple

Probability Theory – Problem 33: Find the probability that these particles meet infinitely often

Question

Let k2k \geq 2. Suppose kk particles independently start simultaneously from the origin and perform one-dimensional simple random walks. We say the kk particles meet whenever all particles occupy the same position. Find the probability that these kk particles meet infinitely often.

Step-by-step solution

Denote the position of the ii-th particle by Xn(i)X_n^{(i)}, n=0,1,2,n=0,1,2,\dots. Consider the relative coordinates: Yn(j)=Xn(j)Xn(1),j=2,,kY_n^{(j)} = X_n^{(j)} - X_n^{(1)}, \quad j=2,\dots,k Then the meeting event is: Yn(2)=Yn(3)==Yn(k)=0Y_n^{(2)} = Y_n^{(3)} = \dots = Y_n^{(k)} = 0 i.e., the vector Yn=(Yn(2),,Yn(k))\mathbf{Y}_n = (Y_n^{(2)},\dots,Y_n^{(k)}) equals the zero vector.

Initial condition: Y0=0\mathbf{Y}_0 = \mathbf{0}. X(1)X^{(1)} and X(j)X^{(j)} are independent, and the increments ΔXn(1),ΔXn(j)\Delta X^{(1)}_n, \Delta X^{(j)}_n are i.i.d., each taking values ±1\pm 1. Thus ΔYn(j)=ΔXn(j)ΔXn(1)\Delta Y^{(j)}_n = \Delta X^{(j)}_n - \Delta X^{(1)}_n with possible values: - 22 (probability 1/41/4); 2-2 (probability 1/41/4); 00 (probability 1/21/2) Hence Y(j)Y^{(j)} is a zero-mean, finite-variance random walk (the increment equals 00 with probability 1/21/2, but it remains a Markov chain). Compute the covariance (for jmj \ne m): Cov(ΔY(j),ΔY(m))=E[(ΔX(j)ΔX(1))(ΔX(m)ΔX(1))]\mathrm{Cov}(\Delta Y^{(j)}, \Delta Y^{(m)}) = E[(\Delta X^{(j)} - \Delta X^{(1)})(\Delta X^{(m)} - \Delta X^{(1)})] Since ΔX(j),ΔX(m),ΔX(1)\Delta X^{(j)},\Delta X^{(m)},\Delta X^{(1)} are pairwise independent with mean 00 and variance 11: =000+1=1= 0 - 0 - 0 + 1 = 1 Covariance matrix: Σ=(211121112)\Sigma = \begin{pmatrix} 2 & 1 & \dots & 1 \\ 1 & 2 & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \dots & 2 \end{pmatrix} Therefore Σ\Sigma is positive definite, and Yn\mathbf{Y}_n can be regarded as a nondegenerate d=k1d=k-1 dimensional random walk (which, after a linear transformation, becomes a standard random walk). By Pólya's theorem, a dd-dimensional simple random walk (nondegenerate, zero-mean, finite variance) is recurrent if and only if d2d \le 2. Here Yn\mathbf{Y}_n is a nondegenerate d=k1d=k-1 dimensional random walk, so: - When k12k-1 \le 2, i.e., k3k \le 3: recurrent \Rightarrow starting from the origin, the walk returns to the origin infinitely often with probability 11 \Rightarrow the probability of meeting infinitely often is 11. - When k13k-1 \ge 3, i.e., k4k \ge 4: transient \Rightarrow the walk returns to the origin only finitely many times \Rightarrow the probability of meeting infinitely often is 00. P(meeting infinitely often)={1,k=2,3,0,k4.P(\text{meeting infinitely often}) = \begin{cases} 1, & k = 2, 3, \\ 0, & k \ge 4. \end{cases} In conclusion, when k=2k=2 or k=3k=3 the probability is 11, and when k4k \ge 4 the probability is 00.

Final answer

When k=2k=2 or k=3k=3 the probability is 11; when k4k \ge 4 the probability is 00.

Marking scheme

The following is a complete marking rubric based on the official solution and undergraduate mathematics grading standards.


Grading Instructions

  • Total: 7 points.
  • The student's approach may differ from the official solution (e.g., using asymptotic estimation). Grade according to [Path A] or [Path B] below.
  • Mutual exclusion rule: Choose the path that best matches the student's main line of reasoning and take the higher score of the two paths. Do not combine scores across paths.

I. Step-by-Step Grading Criteria (Checkpoints)

Path A: Relative Coordinates and Random Walk Dimension Reduction (Official Solution Approach)

  • Constructing the relative coordinate system and problem reformulation (2 points)
  • Introduce relative coordinate variables (e.g., Yn(j)=Xn(j)Xn(1)Y_n^{(j)} = X_n^{(j)} - X_n^{(1)} or a similar linear transformation) and identify that the meeting of kk particles is equivalent to the (k1)(k-1)-dimensional vector process Yn\mathbf{Y}_n returning to the origin (or the zero vector).
  • [Bonus criterion] As long as the student correctly identifies that the essence of the problem is to examine the recurrence of a d=k1d=k-1 dimensional random walk, this score is awarded.
  • Verifying nondegeneracy / computing the covariance (2 points)
  • [2 points] By computing the covariance matrix (e.g., diagonal entries equal to 2, off-diagonal entries equal to 1), proving the matrix is positive definite, or using linear independence arguments to establish that Yn\mathbf{Y}_n is a nondegenerate random walk on Zk1\mathbb{Z}^{k-1} (i.e., it can traverse the lattice with positive probability rather than being confined to a lower-dimensional subspace).
  • [1 point] Merely asserting that Yn\mathbf{Y}_n constitutes a (k1)(k-1)-dimensional simple random walk without verifying its nondegeneracy or describing the covariance structure.
  • Applying Pólya's theorem / dimension criterion (2 points)
  • Cite the recurrence criterion for dd-dimensional simple random walks (zero-mean, finite variance): recurrent if and only if d2d \le 2 (probability 1); transient for d3d \ge 3 (probability 0).
  • Explicitly state that in this problem the dimension is d=k1d = k-1.
  • Final classification and conclusion (1 point)
  • Correctly combine the value of kk to state the conclusion:
  • When k12k-1 \le 2 (i.e., k=2,3k=2, 3), the probability is 1.
  • When k13k-1 \ge 3 (i.e., k4k \ge 4), the probability is 0.
  • *Note: If only the answer is given without derivation, this point is not awarded.*

Path B: Asymptotic Probability Estimation (Analytic Approach)

  • Establishing the meeting probability model (2 points)
  • Express the probability of meeting at time nn, unu_n, as the sum of probabilities that all particles occupy the same site: un=x[P(Xn(1)=x)]ku_n = \sum_{x} [P(X_n^{(1)}=x)]^k or an equivalent integral form.
  • Deriving the asymptotic order (2 points)
  • Using the local central limit theorem (Local CLT) or Stirling's formula, correctly derive the asymptotic behavior of unu_n:

unCn(k1)/2u_n \sim C \cdot n^{-(k-1)/2}

  • *Note: If an incorrect order is obtained (e.g., nk/2n^{-k/2}, missing the summation over the spatial variable xx), this item receives 0 points.*
  • Connecting series convergence/divergence to recurrence (2 points)
  • State that the probability of meeting infinitely often depends on the convergence or divergence of the series un\sum u_n (via the Borel--Cantelli lemma or Markov chain properties):
  • un=    \sum u_n = \infty \implies probability is 1 (recurrent).
  • un<    \sum u_n < \infty \implies probability is 0 (transient / finitely many meetings).
  • Final classification and conclusion (1 point)
  • Analyze the series n(k1)/2\sum n^{-(k-1)/2}:
  • For k=2,3k=2,3, the exponent is 1\le 1, the series diverges \to probability 1.
  • For k4k \ge 4, the exponent is >1> 1, the series converges \to probability 0.

II. Zero-Credit Items

  • Modeling error: Incorrectly modeling the meeting of kk particles in 1 dimension as a single particle returning to the origin in kk-dimensional space.
  • *Consequence*: This leads to an incorrect dimension (d=kd=k), yielding the erroneous conclusion that the probability is 1 for k=2k=2 and 0 for k3k \ge 3. This typically receives 0 points, unless the solution contains some correct general theorem statements (at most 1--2 points).
  • Guessing the critical value of kk by intuition alone, without mathematical derivation or citing a theorem.
  • Simply restating the given conditions of the problem.

III. Deductions

*In this section, apply the largest single-item deduction; deductions stop at zero (no negative scores).*

  • Imprecise logic/conclusion (-1 point)
  • For the transient case (k4k \ge 4), the student concludes "the probability is less than 1" or "the probability is very small" without explicitly stating that the probability is 0 (neglecting the zero-one law for tail events).
  • Similarly, for the recurrent case, failing to explicitly state that the probability is 1.
  • Dimension/exponent calculation error (-1 point)
  • In Path A, an error in computing the rank of the matrix; or in Path B, an error in the integral/summation, leading to an incorrect critical value (e.g., concluding that k=3k=3 is also transient).
  • Missing necessary conditions (-1 point)
  • When applying the recurrence theorem for random walks, completely failing to mention conditions such as "zero mean" or "step distribution," and directly applying the formula (not enough for zero credit, but insufficiently rigorous).

Total Check

  • Full marks: 7 points
  • Path A and Path B scores are not additive.
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