Question

Suppose 2 distinguishable fair coins are tossed simultaneously and repeatedly. An account of the tallies of heads and tails are recorded. Consider the event \(E_{n}\) that at the \(n\)th toss the cumulative number of heads on both tallies are equal. Relate the event \(E_{n}\) to the recurrence time of a given state for a symmetric random walk on the integers.

Short answer

In this problem, we have a symmetric random walk on the integers, where the walker moves one step forward when two coins show the same result and one step backward when the coins show different results. The event \(E_n\) is the event that the cumulative number of heads in both tallies being equal at the \(n\)th toss, which can be interpreted as the random walker being back at the starting position after \(n\) steps. The probability of event \(E_n\) can be written as \(P(E_n) = \sum_{k=0}^{\infty} (\frac{1}{2})^n\), where \(n\) is the number of steps taken. Thus, event \(E_n\) is related to the recurrence time of a given state for a symmetric random walk on the integers.

Step-by-step solution

Identify the random walk properties

In this problem, the random walker can either move one step forward (when the two coins show the same result, either heads-heads or tails-tails) or one step backward (when the two coins show different results, either heads-tails or tails-heads). This situation creates a symmetric random walk on the integers, with a 50% chance of moving forward and a 50% chance of moving backward.

Define event \(E_n\)

Event \(E_n\) is the event that at the \(n\)th toss, the cumulative number of heads in both tallies are equal. This can be interpreted as the random walker being back at the starting position after \(n\) steps.

Calculate the probability of event \(E_n\)

Whenever the coins show the same result, we move one step forward, and whenever the coins show a different result, we move one step backward. The probability of moving one step forward, \(P_H\) (heads-heads or tails-tails), is \(P_H = \frac{1}{2}\) and the probability of moving one step backward, \(P_T\) (heads-tails or tails-heads), is \(P_T = \frac{1}{2}\).

Relate event \(E_n\) to recurrence time

Recurrence time is the expected time for a random walker to return to its starting point. Since we are considering a symmetric random walk, the random walker will eventually return to its starting point with a probability of 1. Now, let's consider the number of steps needed to return to the starting point, which is the event \(E_n\), i.e., the cumulative number of heads in both tallies being equal at the \(n\)th toss. Since the random walker has an equal probability (50%) of moving forward or backward, the recurrence time can be related to the event \(E_n\). At each step, the walker has a 50% chance of returning to the starting point. Therefore, the probability of event \(E_n\) occurring can be written as \(P(E_n) = \sum_{k=0}^{\infty} (\frac{1}{2})^n\), where \(n\) is the number of steps taken. In conclusion, the event \(E_n\) can be interpreted as the recurrence time of a given state for a symmetric random walk on the integers. The probability of this event can be calculated by considering the 50% chance of moving forward or backward and by relating it to the expected recurrence time of the random walk.