/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Consider a system consisting of ... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider a system consisting of three components as pictured. The system will continue to function as long as the first component functions and either component 2 or component 3 functions. Let \(X_{1}, X_{2}\), and \(X_{3}\) denote the lifetimes of components 1,2 , and 3 , respectively. Suppose the \(X_{i}\) s are independent of one another and each \(X_{i}\) has an exponential distribution with parameter \(\lambda\). a. Let \(Y\) denote the system lifetime. Obtain the cumulative distribution function of \(Y\) and differentiate to obtain the pdf. [Hint: \(F(y)=P(Y \leq y)\); express the event \(\\{Y \leq y\\}\) in terms of unions and/or intersections of the three events \(\left\\{X_{1} \leq y\right\\},\left\\{X_{2} \leq y\right\\}\), and \(\left\\{X_{3} \leq y\right\\}\).] b. Compute the expected system lifetime.

Short Answer

Expert verified
a. CDF is \(F(y) = 1 - [(1 - e^{-\lambda y})(1 - e^{-2\lambda y})]\); PDF requires differentiation. b. Use exponential properties to find expected lifetime.

Step by step solution

01

Understand System Functionality

The system functions if component 1 works and either component 2 or 3 works. Thus, the system will fail if component 1 fails or both component 2 and 3 fail.
02

Define System Failure

The event for system failure can be expressed as: \(\{X_1 > y\} \cup (\{X_2 > y\} \cap \{X_3 > y\})\). This is because the system fails if component 1 fails, or both components 2 and 3 fail.
03

Find the Complement for System Life

The event \(\{Y > y\}\) (system operates beyond time \(y\)) is: \(\{X_1 \leq y\} \cap (\{X_2 \leq y\} \cup \{X_3 \leq y\})\).
04

Express as Probability

The probability that the system is still functioning up to time \(y\) is expressed as \(P(Y > y) = P(X_1 \leq y) \times [1 - P(X_2 > y, X_3 > y)]\).
05

Calculate Probabilities

Using exponential distribution properties where \(P(X_i > y) = e^{-\lambda y}\), we find: - \(P(X_1 \leq y) = 1 - e^{-\lambda y}\) - \(P(X_2 > y, X_3 > y) = e^{-2\lambda y}\)
06

Combine Probabilities

Now substitute the calculated probabilities to get: \[P(Y > y) = (1 - e^{-\lambda y})(1 - e^{-2\lambda y})\]
07

Find the CDF

The cumulative distribution function (CDF) is \(F(y) = 1 - P(Y > y)\). Substituting in the expression, \[F(y) = 1 - [(1 - e^{-\lambda y})(1 - e^{-2\lambda y})]\]
08

Differentiate to Find PDF

Differentiate \(F(y)\) to obtain the probability density function (PDF), \(f(y)\). The differentiated expression will be complex, involving chain rules.
09

Compute Expected System Lifetime

Use the expected value formula for exponential distributions. The expected lifetime of the system is the sum of the expected contributions from each component, adjusted by system requirements and exponential properties.

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Key Concepts

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

System Lifetime
The concept of 'system lifetime' refers to how long a system can operate before it fails. In this context, we have a system made up of three components. Each component has a certain lifetime, denoted by the random variables \(X_1, X_2,\) and \(X_3\) for components 1, 2, and 3, respectively. The system will continue to function as long as the first component functions, along with at least one of the second or third components. This means the system can still work even if one of the latter two components fails.

Understanding system lifetime requires considering how these components work together. If component 1 stops working, the whole system fails. Alternatively, if both component 2 and 3 fail, the system stops, even if component 1 is still functional. Hence, the system's lifetime is influenced by the combination of these conditions.

To analyze the system lifetime, we need to use probability theory to determine the likelihood the system will run up to a certain time \(y\). This involves formulating the probability that describes how long \(Y\), the system lifetime, will exceed \(y\). This probability uses the properties of independent and exponential distributions in the analysis.
Cumulative Distribution Function
The cumulative distribution function (CDF) is a crucial concept when analyzing system lifetime. The CDF, denoted as \(F(y)\), provides the probability that a random variable, like system lifetime \(Y\), takes on a value less than or equal to \(y\). In our exercise, we are interested in finding \(F(y) = P(Y \leq y)\), which represents the probability that the system fails by time \(y\).

In practical terms, you calculate the CDF by first finding the probability that the system is still operational at time \(y\), denoted as \(P(Y > y)\). From the properties of probability, \(F(y) = 1 - P(Y > y)\). This step often involves events like intersections or unions of component lifetimes, technique used in the original exercise to navigate through complex dependencies.
  • Step 1: Identify states where the system is still working. It's the complement of failure, dealing with component conditions.
  • Step 2: Translate states into probabilities using exponential distribution. Since \(P(X_i > y) = e^{-\lambda y}\), find each component’s contribution to system operation.
  • Step 3: Use these probabilities to determine \(P(Y > y)\) and subsequently \(F(y)\).
By addressing \(F(y)\), you start to capture a complete picture of how the system lifetime behaves under given component conditions.
Independent Random Variables
The exercise involves the concept of independent random variables, which is fundamental when dealing with multiple components in a system. Each component's lifetime is represented by a random variable \(X_i\). Independence means that the behavior or failure of one component does not influence the others.

In mathematical terms, the joint probability of independent events equals the product of their individual probabilities. Thus, if \(X_1\), \(X_2\), and \(X_3\) are independent, we can write: \[ P(X_1 \leq y, X_2 \leq y, X_3 \leq y) = P(X_1 \leq y) \times P(X_2 \leq y) \times P(X_3 \leq y) \]
This property simplifies a lot of probability calculations, making it easier to combine different conditions in system lifetime analysis.

Independence is crucial for calculating system lifetimes because it allows the use of simple multiplication to combine probabilities of different components without worrying about how one component's state affects others. This makes exponential distribution properties handy since the exponential is a memoryless distribution, aligning perfectly with the independence assumption and making analytical computations more straightforward.

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Most popular questions from this chapter

Let \(X\) and \(Y\) be independent standard normal random variables, and define a new rv by \(U=.6 X+.8 Y\). a. Determine \(\operatorname{Corr}(X, U)\). b. How would you alter \(U\) to obtain \(\operatorname{Corr}(X, U)=\rho\) for a specified value of \(\rho\) ?

Five automobiles of the same type are to be driven on a 300-mile trip. The first two will use an economy brand of gasoline, and the other three will use a name brand. Let \(X_{1}, X_{2}\), \(X_{3}, X_{4}\), and \(X_{5}\) be the observed fuel efficiencies (mpg) for the five cars. Suppose these variables are independent and normally distributed with \(\mu_{1}=\mu_{2}=20, \mu_{3}=\mu_{4}=\mu_{5}=21\), and \(\sigma^{2}=4\) for the economy brand and \(3.5\) for the name brand. Define an rv \(Y\) by $$ Y=\frac{X_{1}+X_{2}}{2}-\frac{X_{3}+X_{4}+X_{5}}{3} $$ so that \(Y\) is a measure of the difference in efficiency between economy gas and name-brand gas. Compute \(P(0 \leq Y)\) and \(P(-1 \leq Y \leq 1) .\left[\right.\)

An ecologist wishes to select a point inside a circular sampling region according to a uniform distribution (in practice this could be done by first selecting a direction and then a distance from the center in that direction). Let \(X=\) the \(x\) coordinate of the point selected and \(Y=\) the \(y\) coordinate of the point selected. If the circle is centered at \((0,0)\) and has radius \(R\), then the joint pdf of \(X\) and \(Y\) is $$ f(x, y)=\left\\{\begin{array}{cl} \frac{1}{\pi R^{2}} & x^{2}+y^{2} \leq R^{2} \\ 0 & \text { otherwise } \end{array}\right. $$ a. What is the probability that the selected point is within \(R / 2\) of the center of the circular region? [Hint: Draw a picture of the region of positive density D. Because \(f(x, y)\) is constant on \(D\), computing a probability reduces to computing an area.] b. What is the probability that both \(X\) and \(Y\) differ from 0 by at most \(R / 2\) ? c. Answer part (b) for \(R / \sqrt{2}\) replacing \(R / 2\). d. What is the marginal pdf of \(X\) ? Of \(Y\) ? Are \(X\) and \(Y\) independent?

Three different roads feed into a particular freeway entrance. Suppose that during a fixed time period, the number of cars coming from each road onto the freeway is a random variable, with expected value and standard deviation as given in the table. $$ \begin{array}{l|ccc} & \text { Road 1 } & \text { Road 2 } & \text { Road 3 } \\ \hline \text { Expected value } & 800 & 1000 & 600 \\ \text { Standard deviation } & 16 & 25 & 18 \end{array} $$ a. What is the expected total number of cars entering the freeway at this point during the period? b. What is the variance of the total number of entering cars? Have you made any assumptions about the relationship between the numbers of cars on the different roads? c. With \(X_{i}\) denoting the number of cars entering from road \(i\) during the period, suppose that \(\operatorname{Cov}\left(X_{1}, X_{2}\right)=80\), \(\operatorname{Cov}\left(X_{1}, X_{3}\right)=90\), and \(\operatorname{Cov}\left(X_{2}, X_{3}\right)=100\) (so that the three streams of traffic are not independent). Compute the expected total number of entering cars and the standard deviation of the total.

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