Problem 70 - A First Course in Probability

Question

At time \(0,\) a coin that comes up heads with probability \(p\) is flipped and falls to the ground. Suppose it lands on heads. At times chosen according to a Poisson process with rate \(\lambda,\) the coin is picked up and flipped. (Between these times, the coin remains on the ground.) What is the probability that the coin is on its head side at time \(t ?\) Hint: What would be the conditional probability if there were no additional flips by time \(t,\) and what would it be if there were additional flips by time \(t ?\)

Step-by-Step Solution

Answer

The probability that the coin is on its head side at time \(t\) can be found using the law of total probability, considering the cases when there are no additional flips by time \(t\) and when there are additional flips by time \(t\). The final expression is: \[ P(\text{coin on head side}) = p \sum_{n=0, n \text{ even}}^{\infty} \frac{(\lambda t)^n e^{-\lambda t}}{n!} + (1-p) \sum_{n=1, n \text{ odd}}^{\infty} \frac{(\lambda t)^n e^{-\lambda t}}{n!} \]

Step 1: Find the Probability of 'n' Flips by Time 't'

The number of flips by time \(t\) is a Poisson random variable with mean \(\lambda t\). Thus, the probability that 'n' flips have occurred by time \(t\) is given by: \[ P(N=n) = \frac{(\lambda t)^n e^{-\lambda t}}{n!} \] where N is the Poisson random variable.

Step 2: Calculate Conditional Probability Based on 'n' Flips

Now we will calculate the conditional probability that the coin is on its head side at time \(t\), given that 'n' flips have occurred by time \(t\). - If 'n' is even, the probability that the coin is still on its head side is \(p\), since there will be an even number of flips and the coin was initially on its head side. - If 'n' is odd, the probability that the coin is on its head side is \(1-p\), since there will be an odd number of flips and the coin was initially on its head side. So we have: \[ P(\text{coin on head side} | N=n) = \begin{cases} p \quad \text{if } n \text{ is even} \\ 1-p \quad \text{if } n \text{ is odd} \end{cases} \]

Step 3: Using the Law of Total Probability

Now we will combine Steps 1 and 2 using the law of total probability to find the overall probability that the coin is on its head side at time \(t\): \[ P(\text{coin on head side}) = \sum_{n=0}^{\infty} P(\text{coin on head side}|N=n) P(N=n) \] Substituting the expressions from Step 1 and Step 2, we get: \[ P(\text{coin on head side}) = p \sum_{n=0, n \text{ even}}^{\infty} \frac{(\lambda t)^n e^{-\lambda t}}{n!} + (1-p) \sum_{n=1, n \text{ odd}}^{\infty} \frac{(\lambda t)^n e^{-\lambda t}}{n!} \] This is the final expression for the probability that the coin is on its head side at time \(t\).

Key Concepts

Poisson Process

A Poisson process is a mathematical concept used to model randomly occurring events within a certain time frame or area. This process is named after the French mathematician Siméon Denis Poisson and it's quite handy in various fields like physics, finance, and telecommunications. The main idea is to understand how often events happen and how they are spaced in time.
The key feature of a Poisson process is its rate of occurrence, denoted by \( \lambda \). This rate \( \lambda \) is the average number of events in a given period. For instance, if we're looking at phone calls at a call center, \( \lambda \) might be 5 calls per hour. This parameter is crucial because it determines the overall behavior of the number of events.
In our exercise, the Poisson process is used to determine how many times the coin is flipped by a certain time \( t \). The number of these flips follows what's known as a Poisson distribution with mean \( \lambda t \). The probability of having exactly \( n \) flips by time \( t \) is expressed with the formula:

• \( P(N=n) = \frac{(\lambda t)^n e^{-\lambda t}}{n!} \)

This formula helps us calculate the likelihood of the different scenarios of flipping the coin, which is essential for solving the problem at hand.

Conditional Probability

Conditional probability helps us understand the likelihood of an event happening, given that another event has already occurred. In our exercise, we're interested in knowing the probability that the coin is on its head side at time \( t \), considering that a specific number of flips \( n \) have happened by then.
To calculate this, we have to examine two distinct cases:

• When \( n \) is even: The probability that the coin is on its head side remains \( p \), because, if the number of flips is even, the coin ends up back on its original side.
• When \( n \) is odd: The probability alters to \( 1-p \), since an odd number of flips means the coin will land on the opposite side from where it began.

Thus, the conditional probability of the coin being on the head side is defined as:

• \( P(\text{coin on head side} | N=n) = \begin{cases} p & \text{if } n \text{ is even} \ 1-p & \text{if } n \text{ is odd} \end{cases} \)

Understanding this concept is fundamental because it allows us to incorporate specific conditions or events that have an impact on the overall probability outcome.

Law of Total Probability

The law of total probability is a powerful tool in the field of probability theory. It helps us find the probability of an event by considering all possible ways that it can happen. We do this using a partition of the sample space.
In this exercise, we are trying to find the overall probability that the coin is on its head side by a certain time \( t \). This is done by aggregating the conditional probabilities of the coin's state over the different possible numbers of flips (even or odd).
Using the steps from the solution, we understand that:

• For each possible number of flips, we calculate the probability that the coin is on the head side given that number of flips (conditional probability), multiplied by the probability of that number of flips occurring (from the Poisson distribution).

To get the total probability, we sum up these calculated probabilities:

• \( P(\text{coin on head side}) = p \sum_{n=0, n \text{ even}}^{\infty} \frac{(\lambda t)^n e^{-\lambda t}}{n!} + (1-p) \sum_{n=1, n \text{ odd}}^{\infty} \frac{(\lambda t)^n e^{-\lambda t}}{n!} \)

This law simplifies complex calculations by breaking them into manageable parts, enabling a comprehensive understanding of combined probabilities.