Math Problem Statement

  1. With a simple random walk (Sn) show that P (S1 S2 . . . S2n 6 = 0) = P (S2n = 0).

Solution

To demonstrate the equality P(S1,S2,,S2n0)=P(S2n=0)P(S_1, S_2, \ldots, S_{2n} \neq 0) = P(S_{2n} = 0), let's consider the properties of a simple symmetric random walk. Here, SnS_n represents the position of the walk after nn steps, starting at the origin S0=0S_0 = 0.

Key Concepts:

  1. Simple Random Walk Definition: A simple random walk SnS_n on the integers is defined by Sn=X1+X2++Xn,S_n = X_1 + X_2 + \cdots + X_n, where XiX_i are independent and identically distributed random variables taking values +1+1 or 1-1 with equal probability 12\frac{1}{2}.

  2. Event Descriptions:

    • P(S2n=0)P(S_{2n} = 0) is the probability that the random walk returns to the origin after 2n2n steps.
    • P(S1,S2,,S2n0)P(S_1, S_2, \ldots, S_{2n} \neq 0) represents the probability that none of the intermediate positions S1,S2,,S2n1S_1, S_2, \ldots, S_{2n-1} are equal to zero, while S2n=0S_{2n} = 0.

Explanation:

To solve the problem, we need to establish a relationship between these two probabilities.

Step 1: Reflection Principle

The reflection principle for random walks states that for a symmetric random walk, if a walk reaches zero at any intermediate step kk, it is equally likely that the walk could reach any point symmetric to the origin (such as the point where it started).

Step 2: Counting Paths

  • Paths returning to zero after 2n2n steps without touching zero at any intermediate step: We can count the number of such paths by considering all possible paths of length 2n2n that start and end at zero, excluding paths that touch zero before reaching 2n2n. For such paths, the number is given by the Catalan number Cn=1n+1(2nn)C_n = \frac{1}{n+1} \binom{2n}{n}.
  • Paths that reach zero exactly at step 2n2n: The total number of such paths is given by (2nn)\binom{2n}{n}.

Step 3: Probability Computation

The probability that the walk returns to zero at step 2n2n, having not touched zero at any intermediate step, is given by the ratio: P(S1,S2,,S2n0)=Cn(2nn)=P(S2n=0).P(S_1, S_2, \ldots, S_{2n} \neq 0) = \frac{C_n}{\binom{2n}{n}} = P(S_{2n} = 0).

Hence, we have shown that: P(S1,S2,,S2n0)=P(S2n=0).P(S_1, S_2, \ldots, S_{2n} \neq 0) = P(S_{2n} = 0).

Would you like more details on any step, or do you have any questions?

Here are five related questions to deepen your understanding:

  1. What is the reflection principle in the context of random walks, and how is it used in probability?
  2. How are Catalan numbers related to random walks and lattice paths?
  3. Why is it necessary to take an even number of steps for the probability calculations in this context?
  4. Can this result be generalized for a random walk in higher dimensions?
  5. How would the probability change if the steps had unequal probabilities for +1+1 and 1-1?

Tip: In symmetric random walks, understanding how paths behave over time is essential for calculating probabilities, especially when using combinatorial methods like the reflection principle.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability Theory
Combinatorics
Random Walks

Formulas

S_n = X_1 + X_2 + ... + X_n (for simple random walk)
P(S_1, S_2, ..., S_{2n} ≠ 0) = C_n / binom(2n, n)
Catalan number formula: C_n = (1 / (n+1)) * binom(2n, n)

Theorems

Reflection Principle for Random Walks
Catalan Numbers and Lattice Paths

Suitable Grade Level

Undergraduate (Probability Theory, Combinatorics)