/*! 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 37 In Exercise \(5.18, Y_{1}\) and ... [FREE SOLUTION] | 91Ó°ÊÓ

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In Exercise \(5.18, Y_{1}\) and \(Y_{2}\) denoted the lengths of life, in hundreds of hours, for components of types I and II, respectively, in an electronic system. The joint density of \(Y_{1}\) and \(Y_{2}\) is given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} (1 / 8) y_{1} e^{-\left(y_{1}+y_{2}\right) / 2}, & y_{1} > 0, y_{2} > 0 \\ 0, & \text { elsewhere } \end{array}\right.$$

Short Answer

Expert verified
The joint density function is valid and integrates to 1 over its domain.

Step by step solution

01

Understand the Joint Density Function

The joint density function is given by \( f(y_1, y_2) = \frac{1}{8} y_1 e^{-(y_1 + y_2)/2} \) for \( y_1 > 0 \) and \( y_2 > 0 \), and 0 elsewhere. The expression represents how the probabilities are distributed between the random variables \( Y_1 \) and \( Y_2 \) over their domains.
02

Verify the Joint Density Function

To ensure \( f(y_1, y_2) \) is a valid probability density function, it must integrate to 1 over the entire positive domain. Check by evaluating \( \int_0^\infty \int_0^\infty \frac{1}{8} y_1 e^{-(y_1+y_2)/2} \, dy_2 \, dy_1 \).
03

Integrate with respect to \( y_2 \)

First, integrate \( \frac{1}{8} y_1 e^{-(y_1+y_2)/2} \) with respect to \( y_2 \) from 0 to \( \infty \). This results in \( \int_0^\infty e^{-y_2/2} \, dy_2 = 2 \). The integration yields \( \frac{1}{8} y_1 \times 2 = \frac{1}{4} y_1 e^{-y_1/2} \).
04

Integrate with respect to \( y_1 \)

Next, integrate \( \frac{1}{4} y_1 e^{-y_1/2} \) with respect to \( y_1 \) from 0 to \( \infty \). Use integration by parts or a substitution \( u = -y_1/2 \), resulting in \( \int_0^\infty y_1 e^{-y_1/2} \, dy_1 = 4 \).
05

Verify normalization

Multiply the result of integrating over \( y_1 \) by the \( \frac{1}{4} \) factor, yielding \( 1 \). This confirms that \( f(y_1, y_2) \) is correctly normalized over its domain.

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Key Concepts

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

Probability Density Function
In probability theory, a probability density function (PDF) describes the likelihood of a continuous random variable falling within a particular range of values. The PDF is a fundamental concept because it explains how probabilities are distributed across possible values of a random variable.
In the given exercise, the joint density function is expressed as \[ f(y_1, y_2) = \frac{1}{8} y_1 e^{-(y_1 + y_2)/2}, \quad y_1 > 0,\, y_2 > 0 \] This expression tells us how two continuous random variables, \( Y_1 \) and \( Y_2 \) (representing the lifespan of two types of components in hundreds of hours), behave together. When we say 'joint density,' we're referring to the combined behavior of both variables.
  • The joint PDF should be non-negative over its domain.
  • The total probability over the domain must equal one for it to be a valid PDF.
Understanding the behavior of these variables through the joint PDF allows us to draw conclusions about the likelihood of component lifespan combinations in the system.
Integration
Integration is a mathematical process used to calculate totals, areas, or volumes. In the context of probability density functions, integration helps to determine probabilities. When dealing with a continuous random variable, integration is used to compute the area under the curve of the PDF.
### Computing Total ProbabilityThe integration performed in the exercise ensures that the total probability is equal to one, thus confirming the function's validity.
1. **Integrate with Respect to \( y_2 \):** This step involves integrating the function \[ \frac{1}{8} y_1 e^{-(y_1+y_2)/2} \] from \( y_2 = 0 \) to \( y_2 = \infty \). The result here is \[ \int_0^\infty e^{-y_2/2} \, dy_2 = 2 \].
2. **Integrate with Respect to \( y_1 \):** After obtaining the result of the first integration, the remaining expression \[ \frac{1}{4} y_1 e^{-y_1/2} \] is integrated over \( y_1 \) from 0 to \( \infty \), yielding:\[ \int_0^\infty y_1 e^{-y_1/2} \, dy_1 = 4 \].

This integration steps demonstrate that the integral of the joint PDF across its entire domain is equal to one.
Random Variables
In probability and statistics, a random variable is a variable that can take on different values, each with a certain probability. Random variables are fundamental to understanding statistical experiments and real-world probability scenarios.
### Types of Random Variables
  • **Discrete Random Variables:** These take on distinct, separate values, often as counts or listed values.
  • **Continuous Random Variables:** These can take any value within a range, which is often represented by intervals on a number line.
In the exercise, \( Y_1 \) and \( Y_2 \) are examples of continuous random variables.
**Joint Random Variables:** When two or more random variables are considered together, they become joint random variables. Joint distributions illustrate the behavior and relationships between multiple random variables.
In this context, \( Y_1 \) and \( Y_2 \) together form a joint random distribution which helps to understand how the lifetimes of components I and II vary in relation to each other.

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Most popular questions from this chapter

A firm purchases two types of industrial chemicals. Type I chemical costs \(\$ 3\) per gallon, whereas type II costs \(\$ 5\) per gallon. The mean and variance for the number of gallons of type I chemical purchased, \(Y_{1},\) are 40 and \(4,\) respectively. The amount of type II chemical purchased, \(Y_{2}\), has \(E\left(Y_{2}\right)=65\) gallons and \(V\left(Y_{2}\right)=8 .\) Assume that \(Y_{1}\) and \(Y_{2}\) are independent and find the mean and variance of the total amount of money spent per week on the two chemicals.

Let \(Y_{1}\) denote the weight (in tons) of a bulk item stocked by a supplier at the beginning of a week and suppose that \(Y_{1}\) has a uniform distribution over the interval \(0 \leq y_{1} \leq 1\). Let \(Y_{2}\) denote the amount (by weight) of this item sold by the supplier during the week and suppose that \(Y_{2}\) has a uniform distribution over the interval \(0 \leq y_{2} \leq y_{1},\) where \(y_{1}\) is a specific value of \(Y_{1}\) a. Find the joint density function for \(Y_{1}\) and \(Y_{2}\) b. If the supplier stocks a half-ton of the item, what is the probability that she sells more than a quarter-ton? c. If it is known that the supplier sold a quarter-ton of the item, what is the probability that she had stocked more than a half-ton?

In Exercise \(5.12,\) we were given the following joint probability density function for the random variables \(Y_{1}\) and \(Y_{2},\) which were the proportions of two components in a sample from a mixture of insecticide: $$ f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 2, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1,0 \leq y_{1}+y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right. $$ For the two chemicals under consideration, an important quantity is the total proportion \(Y_{1}+Y_{2}\) found in any sample. Find \(E\left(Y_{1}+Y_{2}\right)\) and \(V\left(Y_{1}+Y_{2}\right)\)

A learning experiment requires a rat to run a maze (a network of pathways) until it locates one of three possible exits. Exit 1 presents a reward of food, but exits 2 and 3 do not. (If the rat eventually selects exit 1 almost every time, learning may have taken place.) Let \(Y_{i}\) denote the number of times exit \(i\) is chosen in successive runnings. For the following, assume that the rat chooses an exit at random on each run. a. Find the probability that \(n=6\) runs result in \(Y_{1}=3, Y_{2}=1,\) and \(Y_{3}=2\). b. For general \(n\), find \(E\left(Y_{1}\right)\) and \(V\left(Y_{1}\right)\). c. Find \(\operatorname{Cov}\left(Y_{2}, Y_{3}\right)\) for general \(n\). d. To check for the rat's preference between exits 2 and \(3,\) we may look at \(Y_{2}-Y_{3} .\) Find \(E\left(Y_{2}-Y_{3}\right)\) and \(V\left(Y_{2}-Y_{3}\right)\) for general \(n\).

The weights of a population of mice fed on a certain diet since birth are assumed to be normally distributed with \(\mu=100\) and \(\sigma=20\) (measurement in grams). Suppose that a random sample of \(n=4\) mice is taken from this population. Find the probability that a. exactly two weigh between 80 and 100 grams and exactly one weighs more than 100 grams. b. all four mice weigh more than 100 grams.

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