Charlie joins a new reading club, from which he receives books to read. Suppose that books arrive as a Poisson process at a rate λ of books per week. For each book, the time it takes for Charlie to read it is exponentially distributed with parameter μ; i.e., on average it takes Charlie 1/μ weeks to finish one book. Assume that the reading times for different books are independent, and also independent from the book arrival process.
The problem with Charlie is that he is easily distracted. If he is reading a book when a new book arrives, he immediately turns to read the new one, and only comes back to the older book when he finishes the new book.
For all of the parts below, give your answers in terms of μ and λ, using standard notation. Enter μ as "mu" and λ as "lambda".
Hint: When Charlie starts reading a book, the total time he spends on reading it can be viewed as the first arrival from a Poisson process of rate μ, and you can then think about merging or splitting of Poisson processes.
Question 1 --> When Charlie starts a new book, what is the probability that he can finish this book without being interrupted by a new book?
Question 2 --> Given that Charlie receives a new book while reading a book, what is the probability that he can finish both books, the new one and the interrupted one, without further interruption?
Question 3 --> What is the expected reading time of a book given that it is not interrupted? Hint: The answer is not 1/μ.
To answer these questions, we need to understand the concept of merging or splitting of Poisson processes and the properties of exponential distribution.
1. Question 1: When Charlie starts a new book, what is the probability that he can finish this book without being interrupted by a new book?
To answer this question, we can view the interruptions as a Poisson process with rate λ (arrival rate of books). The time Charlie spends on a book before being interrupted is exponentially distributed with parameter μ (time taken to finish a book). The probability that Charlie finishes the book without being interrupted is the probability that there are no interruptions until he finishes the book.
We can use the complementary probability approach. The probability of no interruptions until Charlie finishes the book can be written as the complement of the probability that at least one interruption occurs. The probability of at least one interruption in an exponential distribution with rate λ is given by the formula:
P(at least one interruption) = 1 - P(no interruptions)
P(at least one interruption) = 1 - e^(-λμ)
Therefore, the probability that Charlie can finish a book without being interrupted is:
P(finishing without interruption) = e^(-λμ)
2. Question 2: Given that Charlie receives a new book while reading a book, what is the probability that he can finish both books, the new one and the interrupted one, without further interruption?
To answer this question, we need to consider the time it takes Charlie to finish the interrupted book and the time it takes him to finish the new book. We know that the time to finish each book is exponentially distributed with parameter μ. Since the exponential distribution has memorylessness property, the fact that Charlie has already spent some time on the interrupted book does not affect the remaining time to finish it.
Therefore, the probability that Charlie can finish both books without further interruption is:
P(finishing both books without further interruption) = P(finishing without interruption) * P(finishing the new book without interruption)
P(finishing both books without further interruption) = e^(-λμ) * e^(-λμ) = e^(-2λμ)
3. Question 3: What is the expected reading time of a book given that it is not interrupted?
When a book is not interrupted, Charlie spends an average of 1/μ weeks to finish it. This is because the exponential distribution has a mean of 1/parameter.
Therefore, the expected reading time of a book given that it is not interrupted is:
Expected reading time = 1/μ weeks