Question

A system has a random number of flaws that we will suppose is Poisson distributed with mean \(c\). Each of these flaws will, independently, cause the system to fail at a random time having distribution \(G\). When a system failure occurs, suppose that the flaw causing the failure is immediately located and fixed. (a) What is the distribution of the number of failures by time \(t\) ? (b) What is the distribution of the number of flaws that remain in the system at time \(t ?\) (c) Are the random variables in parts (a) and (b) dependent or independent?

Short answer

In summary, the distribution of the number of system failures by time t, N(t), is given by a Poisson distribution with parameter \(ctG\): \[P(N(t) = n) = \frac{(ctG)^n}{n!} e^{-ctG}\] The distribution of the number of flaws remaining in the system at time t, R(t), is given by a Poisson distribution with parameter \(c(1 - e^{-tG})\): \[P(R(t) = r) = e^{-c(1 - e^{-tG})} \frac{(c(1 - e^{-tG}))^r}{r!}\] Furthermore, the random variables N(t) and R(t) are independent.

Step-by-step solution

Answer (a)

Let N(t) be the number of system failures by time t. Each failure occurs at a random time with distribution G, and fixing the flaw causing the failure is immediate. Let X_i be the time until the i-th flaw causes failure, we have the total time as the sum of these random variables: \[T = \sum_{i=1}^{N} X_i\] Since each failure occurs independently, we can say that the system failures process by time t follows a Poisson process with rate cG, where c is the mean number of flaws and G is the distribution of failures due to the flaws. Thus, the distribution of the number of failures N(t) by time t is given by a Poisson distribution with parameter \(ctG\): \[P(N(t) = n) = \frac{(ctG)^n}{n!} e^{-ctG}\]

Answer (b)

Let R(t) be the number of flaws remaining in the system at time t. Since the total number of flaws in the system follows a Poisson distribution with mean c, we can use the law of total probability to find the distribution of the remaining flaws: \[P(R(t) = r) = \sum_{n=0}^{\infty} P(R(t) = r | N(t) = n) P(N(t) = n)\] Using the identity that the sum of a Poisson random variable and a binomial random variable is a Poisson random variable, we can find the distribution of R(t): \[P(R(t) = r) = e^{-c(1 - e^{-tG})} \frac{(c(1 - e^{-tG}))^r}{r!}\] The remaining flaws in the system at time t follow a Poisson distribution with parameter \(c(1 - e^{-tG})\).

Answer (c)

Now, we will examine if the random variables N(t) and R(t) are dependent or independent. To check for independence, we need to determine if the joint probability distribution of N(t) and R(t) can be factored into the product of their marginal probability distributions. \[P(N(t) = n, R(t) = r) = P(N(t) = n) P(R(t) = r)\] Both N(t) and R(t) are related to the same Poisson process of system failures due to flaws. When a failure occurs, a flaw is fixed, and the remaining number of flaws changes. However, fixing a flaw causing failure doesn't affect the failure distribution, as stated in the problem that fixing the flaw causing the failure is immediate. So, the number of failures and the remaining number of flaws can be seen as independent random processes. Hence, the random variables N(t) and R(t) are independent.