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Consider a random walk on the integers such that \(P_{i, l+1}=p, P_{1, i-1}=q\) for all integer i \((0

Short Answer

Expert verified
\(P_{00}^{2 m}=\left(\begin{array}{c} 2 m \\\ m \end{array}\right) p^{m} q^{m}, \quad P_{00}^{2 m+1}=0\)

Step by step solution

01

Understanding Random Walks

A random walk is a mathematical object, known as a stochastic or random process, which describes a path consisting of a succession of random steps. In this case, the walk is taking place on a set of integers, and at each integer 'i', there's a probability 'p' of moving to the integer i+1 and 'q' of moving to the integer i-1.
02

Probabilities

Here 'p' and 'q' denote probabilities, so they are both between 0 and 1 (0<p<1). Also, for each integer 'i', the sum of the probabilities of all possible next steps must equals to 1, which explains the equation 'p + q = 1'.
03

Representing the Probability of Return

\(P_{00}^{n}\) represents the probability of returning to the starting point (0 in this case) after 'n' steps. This value is what the exercise asks to calculate.
04

Solutions to the Problem

It can be determined that for an even number of steps '2m', the probability of return is: \(P_{00}^{2m}=\binom{2m}{m} p^{m} q^{m}\). Where \(\binom{2m}{m}\) is the binomial coefficient, a key element of probability theory which helps calculate the number of possible combinations for 'm' successes in '2m' trials. On the other hand, for an odd number of steps '2m+1', the probability of return is zero: \(P_{00}^{2m+1}=0\). This is simply because one cannot return to the starting point in an odd number of steps, when the move can only go one step to the left or right at each step. One needs an equal number of leftward and rightward step to return to the starting point, which is only possible with an even number of steps. Therefore, the probability is zero for an odd number of steps.

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