/*! 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 82 In Exercise 5.38 , we determined... [FREE SOLUTION] | 91Ó°ÊÓ

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In Exercise 5.38 , we determined that the joint density function for \(Y_{1}\), the weight in tons of a bulk item stocked by a supplier, and \(Y_{2}\), the weight of the item sold by the supplier, has joint density $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 1 / y_{1}, & 0 \leq y_{2} \leq y_{1} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ In this case, the random variable \(Y_{1}-Y_{2}\) measures the amount of stock remaining at the end of the week, a quantity of great importance to the supplier. Find \(E\left(Y_{1}-Y_{2}\right)\) In this case, the random variable \(Y_{1}-Y_{2}\) measures the amount of stock remaining at the end of the week, a quantity of great importance to the supplier. Find \(E\left(Y_{1}-Y_{2}\right)\)

Short Answer

Expert verified
The expected amount of stock remaining is 0.25 tons.

Step by step solution

01

Define Expected Value

The expected value of a function of two random variables, say \( g(Y_1, Y_2) \), where \( g \) is a function, is calculated by the double integral: \( E[g(Y_1, Y_2)] = \int \int g(y_1, y_2) f(y_1, y_2) \, dy_2 \, dy_1 \) over the region where the joint density \( f(y_1, y_2) \) is non-zero.
02

Set up the Integral for E(Y1 - Y2)

We need to find \( E(Y_1 - Y_2) = \int_0^1 \int_0^{y_1} (y_1 - y_2) \frac{1}{y_1} \, dy_2 \, dy_1 \). This integral is set up using the given joint density function over the valid range \( 0 \leq y_2 \leq y_1 \leq 1 \).
03

Evaluate Inner Integral

Evaluate the inner integral over \( y_2 \) first: \( \int_0^{y_1} (y_1 - y_2) \frac{1}{y_1} \, dy_2 = \int_0^{y_1} \left(1 - \frac{y_2}{y_1}\right) \, dy_2 \). This solves to \( \left[ y_2 - \frac{y_2^2}{2y_1} \right]_0^{y_1} = y_1 - \frac{y_1}{2} = \frac{y_1}{2} \).
04

Evaluate Outer Integral

Evaluate the outer integral over \( y_1 \): \( \int_0^1 \frac{y_1}{2} \, dy_1 = \left[ \frac{y_1^2}{4} \right]_0^1 = \frac{1}{4} \).
05

Conclude the Expected Value

Thus, the expected value \( E(Y_1 - Y_2) \) is \( \frac{1}{4} \), meaning the average amount of stock remaining at the end of the week is 0.25 tons.

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Key Concepts

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

Joint Density Function
A joint density function is a fundamental concept in probability and statistics. It describes the likelihood of two continuous random variables taking on specific values simultaneously. In this problem, we have the joint density function \( f(y_1, y_2) \) which is defined as \( \frac{1}{y_1} \) for the range \( 0 \leq y_2 \leq y_1 \leq 1 \), and zero elsewhere. This function effectively describes the relationship between the weight of a bulk item stocked \( Y_1 \) and the weight sold \( Y_2 \).
Understanding joint density functions is essential as they provide the foundation for calculating probabilities and expected values involving multiple random variables. By understanding this function, we can assess probabilities over specific ranges and compute derived metrics, like the remaining stock in this scenario.
Double Integral
The use of a double integral is a powerful technique to evaluate expected values involving two random variables. A double integral extends the concept of an integral to two dimensions, allowing us to compute sums over areas rather than lines.
For the expected value \( E(Y_1 - Y_2) \), we set up the double integral \( \int_0^1 \int_0^{y_1} (y_1 - y_2) \frac{1}{y_1} \, dy_2 \, dy_1 \). The outer integral considers \( y_1 \), and the inner integral considers \( y_2 \).
  • The process involves evaluating the inner integral first, resulting in a function of \( y_1 \).
  • Next, the outer integral is evaluated to obtain the expected value over the entire region.
Using a double integral helps us compute the probabilistic expectation, taking account all possible combinations of \( Y_1 \) and \( Y_2 \) as described by the joint density function.
Random Variables
Random variables are quantities whose values depend on the outcomes of a random phenomenon. In this problem, \( Y_1 \) and \( Y_2 \) are random variables representing the item weight in tons stocked and sold, respectively.
Because these quantities can vary from one observation to the next, they require statistical measures like expected values to summarize their behavior.
  • \( Y_1 \) defines the supply component and is constrained such that it lies between 0 and 1 ton.
  • \( Y_2 \) reports the sale amount, subject to being less than or equal to \( Y_1 \).
These constraints define their joint density function, illustrating the relationship between them, allowing us to analyze further outcomes such as the amount of stock remaining.
Mathematical Statistics
Mathematical statistics provides tools for rigorous analysis of data, including methods for dealing with joint distributions and computing expectations. In this exercise, statistical methods help determine the average remaining stock given the joint probabilities involved.
The expected value, \( E(Y_1 - Y_2) \), is calculated through the framework of mathematical statistics using double integrals and joint density functions.
  • It measures the average outcome of random variables under uncertainty.
  • The computation involves integrating over the valid probability space, as defined by the problem constraints.
Through these methods, we gain insights into expected quantities, which are crucial for planning and decision-making, like estimating stock levels in supply chain management. Mathematical statistics precisely defines these complex relationships and provides robust conclusions supported by mathematical reasoning.

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Most popular questions from this chapter

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.

A retail grocery merchant figures that her daily gain \(X\) from sales is a normally distributed random variable with \(\mu=50\) and \(\sigma=3\) (measurements in dollars). \(X\) can be negative if she is forced to dispose of enough perishable goods. Also, she figures daily overhead costs \(Y\) to have gamma distribution with \(\alpha=4\) and \(\beta=2\). If \(X\) and \(Y\) are independent, find the expected value and variance of her net daily gain. Would you expect her net gain for tomorrow to rise above \(\$ 70 ?\)

Suppose that \(Y_{1}\) and \(Y_{2}\) are independent binomial distributed random variables based on samples of sizes \(n_{1}\) and \(n_{2},\) respectively. Suppose that \(p_{1}=p_{2}=p .\) That is, the probability of "success" is the same for the two random variables. Let \(W=Y_{1}+Y_{2} .\) In Chapter 6 you will prove that \(W\) has a binomial distribution with success probability \(p\) and sample size \(n_{1}+n_{2}\). Use this result to show that the conditional distribution of \(Y_{1}\), given that \(W=w\), is a hypergeometric distribution with \(N=n_{1}+n_{2},\) and \(n=w,\) and \(r=n_{1}\)

In the production of a certain type of copper, two types of copper powder (types A and B) are mixed together and sintered (heated) for a certain length of time. For a fixed volume of sintered copper, the producer measures the proportion \(Y_{1}\) of the volume due to solid copper (some pores will have to be filled with air) and the proportion \(Y_{2}\) of the solid mass due to type A crystals. Assume that appropriate probability densities for \(Y_{1}\) and \(Y_{2}\) are $$\begin{array}{l} f_{1}\left(y_{1}\right)=\left\\{\begin{array}{ll} 6 y_{1}\left(1-y_{1}\right), & 0 \leq y_{1} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right. \\ f_{2}\left(y_{2}\right)=\left\\{\begin{array}{ll} 3 y_{2}^{2}, & 0 \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right. \end{array}$$ The proportion of the sample volume due to type A crystals is then \(Y_{1} Y_{2} .\) Assuming that \(Y_{1}\) and \(Y_{2}\) are independent, find \(P\left(Y_{1} Y_{2} \leq .5\right)\)

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 \(V\left(Y_{1}+Y_{2}\right)\)

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