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Chapter 5: Problem 89

Some components of a two-component system fail after receiving a shock. Shocks of three types arrive independently and in accordance with Poisson processes. Shocks of the first type arrive at a Poisson rate \(\lambda_{1}\) and cause the first component to fail. Those of the second type arrive at a Poisson rate \(\lambda_{2}\) and cause the second component to fail. The third type of shock arrives at a Poisson rate \(\lambda_{3}\) and causes both components to fail. Let \(X_{1}\) and \(X_{2}\) denote the survival times for the two components. Show that the joint distribution of \(X_{1}\) and \(X_{2}\) is given by $$ P\left\\{X_{1}>s, X_{1}>t\right\\}=\exp \left\\{-\lambda_{1} s-\lambda_{2} t-\lambda_{3} \max (s, t)\right\\} $$ This distribution is known as the bivariate exponential distribution.

Short Answer

Expert verified
To find the joint distribution of the survival times \(X_1\) and \(X_2\) for the two-component system, we consider cases when both components survive beyond times \(s\) and \(t\). By taking into account the independent Poisson processes for the three shock types, we derive that the joint distribution is given by: $$ P\left\\{X_{1}>s, X_{1}>t\right\\}=\exp \left\\{-\lambda_{1} s-\lambda_{2} t-\lambda_{3} \max (s, t)\right\\} $$ This represents the bivariate exponential distribution.

Step by step solution

01

Let us first consider the probability of survival of Component 1 and Component 2 individually. Since type 1 shock only affects component 1, and type 3 shocks affect both components, we can express the probability of survival of component 1 as \(P\{X_1 > s\} = \exp\{- (\lambda_1 + \lambda_3)s\}\). Similarly, since type 2 shock only affects component 2, we can express the probability of survival of component 2 as \(P\{X_2 > t\} = \exp\{- (\lambda_2 + \lambda_3)t\}\). #Step 2: Finding the joint survival probability#

Now we want to find the probability of both components surviving beyond time points \(s\) and \(t\). We can consider three cases: 1. Component 1 fails after time \(s\) and Component 2 fails after time \(t\). 2. Component 1 fails after time \(s\) and Component 2 fails before time \(t\). 3. Component 1 fails before time \(s\) and Component 2 fails after time \(t\). We want to calculate the probability for case 1 as it represents both components surviving past their respective times \(s\) and \(t\). The other two cases do not fulfill this condition and thus, we should subtract their probabilities when calculating the desired probability. #Step 3: Considering the three cases#
02

To find the probability for the first case, note that the Poisson processes for the three shock types are independent. Thus, we can multiply their respective survival probabilities: $$ P\{X_1>s,X_2>t\text{ (case 1)}\} = \exp\{-\lambda_1 s-\lambda_2 t-\lambda_3\max(s,t)\} $$ For cases 2 and 3, we need to determine what happens when at least one component fails. So we can first calculate the probability of both components failing before time \(s\) and time \(t\). This can be expressed as: $$ P\{X_1t\} + P\{X_1>s,X_2>t\text{ (case 1)}\} $$ Upon substituting the probabilities from Step 1, we get: $$ P\{X_1

Adding the probabilities for cases 2 and 3, and subtracting this sum from 1, we can write the final probability as: \begin{align*} P\{X_1>s,X_2>t\} &= 1 - P\{X_1s, X_{1}>t\right\\}=\exp \left\\{-\lambda_{1} s-\lambda_{2} t-\lambda_{3} \max (s, t)\right\\} $$ This is the bivariate exponential distribution.

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Key Concepts

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

Poisson Process
The Poisson process is a statistical concept that describes a type of event occurring randomly over a given time interval. It is used to model scenarios where events happen independently of each other at a constant average rate. One real-world example of a Poisson process is the arrival of customers at a store, where customers come in at random times, but on average, there is a consistent rate throughout the day.

When applied to the exercise, the Poisson process explains how shocks arrive to possibly cause failure in system components. Each type of shock has its own Poisson rate, \(\lambda_{1}\), \(\lambda_{2}\), and \(\lambda_{3}\), which respectively represent the average number of shocks that will occur in a unit time interval. The key property that we use in modeling the system's failures is that the shocks occur independently. This independence is crucial when we try to determine the joint probability of both components surviving past a certain time.
Joint Distribution
Joint distribution involves understanding the probability of two or more random variables occurring simultaneously. It's like looking at the likelihood of multiple things happening at the same time and figuring out how they are connected. In our exercise, we are interested in the joint distribution of the survival times \(X_{1}\) and \(X_{2}\) of the two system components, which describe the time until failure due to the different shock types.

The survival probabilities calculated individually for each component can be combined to find the overall joint survival probability. Because of the Poisson process鈥檚 independence property, the joint probability is the product of the individual probabilities, factoring in the rate at which both components fail together (shock type three). The resulting joint distribution defines a bivariate exponential distribution, which is a powerful tool in understanding how these two random variables (the survival times of the components) interact.
Survival Probability
Survival probability refers to the chance that a given component, system, or individual continues to function beyond a specified time without failure. It's similar to predicting how long something will last before it stops working. In reliability engineering, which is the field of study related to our exercise, survival probability is crucial for assessing product life and durability.

In the context of the exercise, we calculate the survival probability of each component by considering the interval until the first shock that could cause failure. For component 1, any shock of the first or third type could cause failure, and for component 2, shocks of the second or third type. To improve understanding, students could be encouraged to visualize this concept with a timeline where shocks occur at random points, and survival probability is the chance that no shock has occurred up to the time of interest. This visualization aids in comprehending how the survival probability decreases as time goes on, and higher shock rates shorten expected survival times.

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Most popular questions from this chapter

A flashlight needs two batteries to be operational. Consider such a flashlight along with a set of \(n\) functional batteries-battery 1 , battery \(2, \ldots\), battery \(n .\) Initially, battery 1 and 2 are installed. Whenever a battery fails, it is immediately replaced by the lowest numbered functional battery that has not yet been put in use. Suppose that the lifetimes of the different batteries are independent exponential random variables each having rate \(\mu .\) At a random time, call it \(T\), a battery will fail and our stockpile will be empty. At that moment exactly one of the batteries-which we call battery \(X\) -will not yet have failed. (a) What is \(P[X=n\\}\) ? (b) What is \(P[X=1\\} ?\) (c) What is \(P[X=i\\} ?\) (d) Find \(E[T]\). (e) What is the distribution of \(T ?\)

A set of \(n\) cities is to be connected via communication links. The cost to construct a link between cities \(i\) and \(j\) is \(C_{i j}, i \neq j .\) Enough links should be constructed so that for each pair of cities there is a path of links that connects them. As a result, only \(n-1\) links need be constructed. A minimal cost algorithm for solving this problem (known as the minimal spanning tree problem) first constructs the cheapest of all the (in) links. Then, at each additional stage it chooses the cheapest link that connects a city without any links to one with links. That is, if the first link is between cities 1 and 2, then the second link will either be between 1 and one of the links \(3, \ldots, n\) or between 2 and one of the links \(3, \ldots, n .\) Suppose that all of the \(\left(\begin{array}{c}n \\\ 2\end{array}\right)\) costs \(C_{i j}\) are independent exponential random variables with mean \(1 .\) Find the expected cost of the preceding algorithm if (a) \(n=3\), (b) \(n=4\).

Each entering customer must be served first by server 1 , then by server 2 , and finally by server \(3 .\) The amount of time it takes to be served by server \(i\) is an exponential random variable with rate \(\mu_{i}, i=1,2,3 .\) Suppose you enter the system when it contains a single customer who is being served by server \(3 .\) (a) Find the probability that server 3 will still be busy when you move over to server 2 . (b) Find the probability that server 3 will still be busy when you move over to server 3 . (c) Find the expected amount of time that you spend in the system. (Whenever you encounter a busy server, you must wait for the service in progress to end before you can enter service.) (d) Suppose that you enter the system when it contains a single customer who is being served by server \(2 .\) Find the expected amount of time that you spend in the system.

Suppose that the number of typographical errors in a new text is Poisson distributed with mean \(\lambda\). Two proofreaders independently read the text. Suppose that each error is independently found by proofreader \(i\) with probability \(p_{i}, i=1,2 .\) Let \(X_{1}\) denote the number of errors that are found by proofreader 1 but not by proofreader \(2 .\) Let \(X_{2}\) denote the number of errors that are found by proofreader 2 but not by proofreader \(1 .\) Let \(X_{3}\) denote the number of errors that are found by both proofreaders. Finally, let \(X_{4}\) denote the number of errors found by neither proofreader. (a) Describe the joint probability distribution of \(X_{1}, X_{2}, X_{3}, X_{4}\). (b) Show that $$ \frac{E\left[X_{1}\right]}{E\left[X_{3}\right]}=\frac{1-p_{2}}{p_{2}} \text { and } \frac{E\left[X_{2}\right]}{E\left[X_{3}\right]}=\frac{1-p_{1}}{p_{1}} $$ Suppose now that \(\lambda, p_{1}\), and \(p_{2}\) are all unknown. (c) By using \(X_{i}\) as an estimator of \(E\left[X_{i}\right], i=1,2,3\), present estimators of \(p_{1}, p_{2}\) and \(\lambda\). (d) Give an estimator of \(X_{4}\), the number of errors not found by either proofreader.

A two-dimensional Poisson process is a process of randomly occurring events in the plane such that (i) for any region of area \(A\) the number of events in that region has a Poisson distribution with mean \(\lambda A\), and (ii) the number of events in nonoverlapping regions are independent. For such a process, consider an arbitrary point in the plane and let \(X\) denote its distance from its nearest event (where distance is measured in the usual Euclidean manner). Show that (a) \(P[X>t\\}=e^{-\lambda \pi t^{2}}\), (b) \(E[X]=\frac{1}{2 \sqrt{2}}\).
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