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The lengths of life \(Y\) for a type of fuse has an exponential distribution with a density function given by $$f(y)=\left\\{\begin{array}{ll} (1 / \beta) e^{-y / \beta}, & y \geq 0 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. If two such fuses have independent life lengths \(Y_{1}\) and \(Y_{2}\), find their joint probability density function. b. One fuse from part (a) is in a primary system, and the other is in a backup system that comes into use only if the primary system fails. The total effective life length of the two fuses. therefore, is \(Y_{1}+Y_{2} .\) Find \(P\left(Y_{1}+Y_{2} \leq a\right),\) where \(a>0\)

Short Answer

Expert verified
a. The joint PDF is \(f(y_1, y_2) = \frac{1}{\beta^2} e^{-(y_1+y_2)/\beta}\) for \(y_1, y_2 \geq 0\). b. \(P(Y_1+Y_2 \leq a) = 1 - e^{-a/\beta} - \frac{a}{\beta}e^{-a/\beta}\).

Step by step solution

01

Understanding the Independent Life Lengths

The life lengths of the two fuses, \(Y_1\) and \(Y_2\), are independent and identically distributed exponential random variables with parameter \(\beta\). For independent exponential random variables, the joint probability density function (PDF) is the product of their individual PDFs.
02

Formulate the Joint PDF

The PDF of a single exponential random variable \(Y\) with parameter \(\beta\) is given as \( (1/\beta) e^{-y/\beta} \). For two independent exponential random variables \(Y_1\) and \(Y_2\), their joint PDF is given by:\[f(y_1, y_2) = \left(\frac{1}{\beta}\right) e^{-y_1/\beta} \cdot \left(\frac{1}{\beta}\right) e^{-y_2/\beta} = \frac{1}{\beta^2} e^{-(y_1+y_2)/\beta} \quad \text{for} \quad y_1, y_2 \geq 0\]
03

Define the Total Life Length

The total life length of the two fuses, since the backup system only comes into play after the primary system fails, is given by \(Y_1+Y_2\). We need to find the probability \(P(Y_1+Y_2 \leq a)\), where \(a > 0\). This setup tells us that the sum \(Y = Y_1 + Y_2\) is a gamma distribution with shape parameter \(2\) because it's a sum of two independent exponential variables, each with rate \(\beta\).
04

Compute the Desired Probability

The sum \(Y = Y_1 + Y_2\) takes the form of a gamma distribution with shape \(k=2\) and scale \(\beta\). For a gamma distribution of such parameters, the cumulative distribution function (CDF) is:\[P(Y \leq a) = 1 - e^{-a/\beta} - \frac{a}{\beta}e^{-a/\beta}\]This is derived from the property of the gamma distribution.
05

Conclusion

After computing the cumulative probability expression, the final result is useful to find the probability that the total effective life of the two fuses is less than or equal to a certain time \(a\). The CDF examined above gives us this probability directly.

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Key Concepts

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

Probability Density Function
The Probability Density Function (PDF) is a fundamental concept in probability and statistics, especially when dealing with continuous random variables like the ones described in an exponential distribution. The PDF tells us how the probabilities are distributed over the values of the random variable. For instance, if you have a PDF for a random variable, you can use it to determine the likelihood that the variable takes on a certain value.

In the case of an exponential distribution, the PDF is given by:
  • For non-negative values: \( f(y) = \frac{1}{\beta} e^{-y/\beta} \)
  • For negative values: \( f(y) = 0 \), since the exponential distribution can't take negative values.
The formula is derived based on the rate parameter \( \beta \), which dictates the spread or the rate at which events occur. Understanding the PDF is crucial because it helps us compute probabilities for various intervals of values, which is foundational for performing further analyses such as obtaining the joint distribution or cumulative distribution.
Joint Distribution
The joint distribution involves studying two or more random variables together and is critical when these variables are related or dependent. When we say two random variables, say \( Y_1 \) and \( Y_2 \), have a joint probability density function, we refer to the distribution of their combined outcomes. With exponential distributions, if \( Y_1 \) and \( Y_2 \) are independent, their joint PDF is simply the product of their individual PDFs.

For example, if \( Y_1 \) and \( Y_2 \) are independently distributed exponential variables with parameter \( \beta \), the joint PDF is computed as:
  • \( f(y_1, y_2) = \left(\frac{1}{\beta}\right) e^{-y_1/\beta} \times \left(\frac{1}{\beta}\right) e^{-y_2/\beta} = \frac{1}{\beta^2} e^{-(y_1+y_2)/\beta} \)
This expression reveals how the probability density spreads over the plane formed by \( Y_1 \) and \( Y_2 \). It's useful in scenarios where the cumulative effect of several events is studied collectively.
Gamma Distribution
The Gamma Distribution is a two-parameter family of continuous probability distributions. It is widely used to model the sum of exponential random variables, which is precisely what happens in our exercise when considering \( Y = Y_1 + Y_2 \). Naming \( Y \) as a Gamma-distributed variable stems from the sum of two independent exponential variables \( Y_1 \) and \( Y_2 \), each having the same rate parameter \( \beta \). The resulting distribution is gamma with a shape parameter equal to the number of those variables, here 2.

Gamma Distribution is usually specified by:
  • Shape parameter \( k \): Indicates the number of exponential variables included in the sum, here it is \( k = 2 \).
  • Rate or scale parameter \( \beta \): Dictates how quickly the probabilities decay.
Understanding the role of the gamma distribution helps greatly in assessing scenarios involving sums of exponential random variables, which occur often in reliability engineering and queueing theory, where processes are modeled over time.
Cumulative Distribution Function
The Cumulative Distribution Function (CDF) is an essential tool in probability because it tells us the probability that a random variable takes on a value less than or equal to a certain threshold. In our exercise, the CDF helps us determine \( P(Y_1 + Y_2 \leq a) \) with \( Y_1 + Y_2 \) being a gamma-distributed variable.

The CDF for a sum of two exponential variables (equivalently, a Gamma distribution with shape \(k = 2\)) is given by:
  • \( P(Y \leq a) = 1 - e^{-a/\beta} - \frac{a}{\beta}e^{-a/\beta} \)
This formula helps compute the desired probability for any given \( a > 0 \). By using the CDF, we efficiently address questions like how likely the total lifespan of components is within a specific time frame, which is criticial in reliability studies and failure forecasting.

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Most popular questions from this chapter

The number of defects per yard \(Y\) for a certain fabric is known to have a Poisson distribution with parameter \lambda. However, \lambda itself is a random variable with probability density function given by $$f(\lambda)=\left\\{\begin{array}{ll} e^{-\lambda}, & \lambda \geq 0 \\ 0, & \text { elsewhere } \end{array}\right.$$ Find the unconditional probability function for \(Y\)

Refer to Exercises 5.6,5.24 , and \(5.50 .\) Suppose that a radioactive particle is randomly located in a square with sides of unit length. A reasonable model for the joint density function for \(Y_{1}\) and \(Y_{2}\) is $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 1, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. What is \(E\left(Y_{1}-Y_{2}\right) ?\) b. What is \(E\left(Y_{1} Y_{2}\right) ?\) c. What is \(E\left(Y_{1}^{2}+Y_{2}^{2}\right) ?\) d. What is \(V\left(Y_{1} Y_{2}\right) ?\)

Of nine executives in a business firm, four are married, three have never married, and two are divorced. Three of the executives are to be selected for promotion. Let \(Y_{1}\) denote the number of married executives and \(Y_{2}\) denote the number of never-married executives among the three selected for promotion. Assuming that the three are randomly selected from the nine available, find the joint probability function of \(Y_{1}\) and \(Y_{2}\).

Assume that \(Y_{1}, Y_{2},\) and \(Y_{3}\) are random variables, with $$E\left(Y_{1}\right)=2, \quad E\left(Y_{2}\right)=-1, \quad E\left(Y_{3}\right)=4 $$ $$V\left(Y_{1}\right)=4, \quad V\left(Y_{2}\right)=6, \quad V\left(Y_{3}\right)=8$$ $$\operatorname{Cov}\left(Y_{1}, Y_{2}\right)=1, \operatorname{Cov}\left(Y_{1}, Y_{3}\right)=-1, \operatorname{Cov}\left(Y_{2}, Y_{3}\right)=0$$ Find \(E\left(3 Y_{1}+4 Y_{2}-6 Y_{3}\right)\) and \(V\left(3 Y_{1}+4 Y_{2}-6 Y_{3}\right)\)

A box contains four balls, numbered 1 through 4 . One ball is selected at random from this box. Let \(X_{1}=1\) if ball1or ball2is drawn, \(X_{2}=1\) if ball1or ball3is drawn, \(X_{3}=1\) if ball1or ball4is drawn. The \(X_{i}\) values are zero otherwise. Show that any two of the random variables \(X_{1}, X_{2},\) and \(X_{3}\) are independent but that the three together are not.

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