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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 \(t ?\)

Short Answer

Expert verified
The probability that the coin is on its head side at time \(t\) is given by: \[P(\text{Head at time } t) = \sum_{k=0}^{\infty} \left[ \sum_{i=0}^{\lfloor k/2 \rfloor} \binom{k}{2i} p^{k-2i}(1-p)^{2i} \right] \frac{e^{-\lambda t} (\lambda t)^k}{k!}\]

Step by step solution

01

Define the Random Variables

Let \(N_t\) be the number of flips up to time \(t\). We observe that given k flips, the coin is on the head side exactly when an even number of flips occured. So, we need to find the probability that given k flips up to time \(t\), there is an even number of them. To find this, we will first find the probability that there are k flips up to the time \(t\), that is, \(P(N_t = k)\).
02

Find the Probability of k Flips by Time t

We are given that the flipping events follow a Poisson process with rate \(\lambda\). So, we can find the probability of \(k\) flips by time \(t\) using the Poisson probability mass function: \[P(N_t = k) = \frac{e^{-\lambda t} (\lambda t)^k}{k!}\]
03

Calculate the Conditional Probability

We need to find the conditional probability that the coin is on its head side at time \(t\) given that there were \(k\) flips up to time \(t\). As mentioned before, the coin will be on its head side exactly when an even number of flips occured. So, we need to find the probability of an even number of tails occurring among the k flips. Let \(B_k\) be the number of tails among the \(k\) flips. Then, the probability of getting an even number of tails, given \(k\) flips can be written as: \[P(B_k \text{ is even} | N_t=k) = \sum_{i=0}^{\lfloor k/2 \rfloor} \binom{k}{2i} p^{k-2i}(1-p)^{2i}\]
04

Use Total Probability Theorem

Now, we can use the total probability theorem to find the probability that the coin is on its head side at time \(t\): \[P(\text{Head at time } t) = \sum_{k=0}^{\infty} P(B_k \text{ is even} | N_t=k) P(N_t = k)\]
05

Plug in the probabilities found in Steps 2 and 3

Plugging in the probabilities found in steps 2 and 3, we have: \[P(\text{Head at time } t) = \sum_{k=0}^{\infty} \left[ \sum_{i=0}^{\lfloor k/2 \rfloor} \binom{k}{2i} p^{k-2i}(1-p)^{2i} \right] \frac{e^{-\lambda t} (\lambda t)^k}{k!}\] This is the probability that the coin is on its head side at time \(t\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conditional Probability
In probability theory, conditional probability is a crucial concept that helps us understand the likelihood of an event occurring given that another event has already happened. To grasp this better, think about flipping a coin multiple times. If we know that a certain number of flips occurred, we might want to determine how this influences whether the coin lands on heads or tails.
The conditional probability of an event A given event B is expressed as \( P(A | B) \), which can be understood as "the probability of A occurring under the condition that B has occurred."
In our scenario, we are interested in the conditional probability that the coin lands heads-up under the condition that some flips have already occurred by time \( t \). We consider each possible number of flips, \( k \), and find the probability that the coin is heads-up, given those \( k \) flips. This involves calculating the probability of having an even number of these flips result in tails, as an even number of tails means the last flip keeps the coin on heads.
Random Variables
Random variables are used in probability and statistics to quantify uncertain outcomes. In simple terms, a random variable assigns a numerical value to each outcome of a random event. There are two types of random variables: discrete and continuous.
In our exercise involving the Poisson process, we define a discrete random variable \( N_t \), which represents the number of flips up to time \( t \). The role of \( N_t \) is crucial—it tells us how many times the coin has potentially changed its state due to flips.
A key aspect of working with random variables is understanding their probability distributions. In this case, \( N_t \) follows a Poisson distribution because the flips occur according to a Poisson process with rate \( \lambda \). This means we can calculate the probability of having exactly \( k \) flips by time \( t \) using a special formula based on \( \lambda \) and \( t \). This formula is vital in determining the likelihood of different outcomes and is used to further calculate the probability of ultimately landing on heads.
Total Probability Theorem
The total probability theorem plays an integral role when dealing with complex problems that involve multiple scenarios. This theorem allows us to find the probability of an event by considering all the different ways that event can happen and summing their probabilities.
This is pivotal in our problem. We need to find the overall probability of the coin being on its head side at time \( t \). To accomplish this, we look at all possible scenarios (different numbers of flips \( k \)), calculate each scenario's probability (the coin being heads), and then combine them.
First, we calculate the conditional probability for each \( k \) flips. Then, we multiply this by the likelihood of having \( k \) flips, given by the Poisson distribution. Finally, we sum up all these contributions using the total probability theorem. This comprehensive approach ensures that we account for every way the coin might end up heads-up, leading to the final probability expression.

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Most popular questions from this chapter

The National Basketball Association (NBA) draft lottery involves the 11 teams that had the worst won-lost records during the year. A total of 66 balls are placed in an urn. Each of these balls is inscribed with the name of a team: Eleven have the name of the team with the worst record, 10 have the name of the team with the second-worst record, 9 have the name of the team with the thirdworst record, and so on (with 1 ball having the name of the team with the 11 th-worst record). A ball is then chosen at random, and the team whose name is on the ball is given the first pick in the draft of players about to enter the league. Another ball is then chosen, and if it "belongs" to a team different from the one that received the first draft pick, then the team to which it belongs receives the second draft pick. (If the ball belongs to the team receiving the first pick, then it is discarded and another one is chosen; this continues until the ball of another team is chosen.) Finally, another ball is chosen, and the team named on the ball (provided that it is different from the previous two teams) receives the third draft pick. The remaining draft picks 4 through 11 are then awarded to the 8 teams that did not "win the lottery," in inverse order of their won-lost records. For instance, if the team with the worst record did not receive any of the 3 lottery picks, then that team would receive the fourth draft pick. Let \(X\) denote the draft pick of the team with the worst record. Find the probability mass function of \(X\)

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