Question

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

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

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#

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.

Key concepts

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’s 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.