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Consider randomly selecting a point \(\left(X_{1}, X_{2}, X_{3}\right)\) in the unit cube \(\left\\{\left(x_{1}, x_{2}, x_{3}\right): 0

Short Answer

Expert verified
The pdf of \(Y_3\) is \(4y_3\) for \(0 < y_3 < 1\).

Step by step solution

01

Understand the Problem

We are given a probability density function (pdf) of a joint distribution for random variables \(X_1, X_2, X_3\), and are asked to find the pdf of the volume \(Y_3 = X_1 X_2 X_3\) of a rectangular solid formed by these variables.
02

Define Intermediate Variables

Define intermediate transformed variables: Let \(Y_1 = X_1\) and \(Y_2 = X_1 X_2\). Ultimately, \(Y_3 = X_1 X_2 X_3\). We will find the pdf of \(Y_3\) using these transformations.
03

Find Joint pdf of Intermediate Variables

Since \(X_1, X_2, X_3\) are independent, their joint pdf is \(f(x_1, x_2, x_3) = 8x_1 x_2 x_3\). For \(Y_1, Y_2, Y_3\), we use the transformation of variables to find their joint pdf.
04

Use Transformation Technique

Using the transformations: \(X_1 = Y_1\), \(X_2 = \frac{Y_2}{Y_1}\), and \(X_3 = \frac{Y_3}{Y_2}\). Substitute these in the joint pdf and adjust for the Jacobian of transformation.
05

Calculate the Jacobian

The Jacobian of the transformation is the determinant of the matrix of partial derivatives: \[ J = \begin{vmatrix} 1 & 0 & 0 \ 0 & \frac{1}{X_1} & 0 \ 0 & -\frac{X_3}{X_1X_2} & \frac{1}{X_2} \end{vmatrix} = \frac{1}{X_1 X_2} \]
06

Find Transformed pdf

Consider \(f(y_1, y_2, y_3) = f(x_1, x_2, x_3) \times |J|\). Simplifying, \(f(y_1, y_2, y_3) = 8 y_1 \cdot \frac{y_2}{y_1} \cdot \frac{y_3}{y_2} \times \frac{1}{y_1 \cdot \frac{y_2}{y_1}} = 8 \cdot \frac{y_3}{y_1} = 8 y_3\) within the limits.
07

Marginalize Out Intermediate Variables

To obtain the pdf of \(Y_3\), integrate out \(Y_1\) and \(Y_2\). Calculate \[ f_{Y_3}(y_3) = \int_{0}^{1} \int_{0}^{y_1} 8 y_3 \, dy_2 \, dy_1 \].
08

Perform Integration

Perform the integration: 1. Integrate with respect to \(y_2\): \(\int_{0}^{y_1} 8 y_3 \, dy_2 = 8 y_3 y_1\)2. Integrate with respect to \(y_1\): \(\int_{0}^{1} 8 y_3 y_1 \, dy_1 = 4 y_3 \) over the unit cube.

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Key Concepts

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

Joint Distribution
Joint distribution is a fundamental concept in probability theory that deals with multiple random variables considered together. When we talk about the joint distribution of random variables, we're interested in understanding how these variables behave in relation to one another. In the original exercise, we have three variables:
  • \(X_1\)
  • \(X_2\)
  • \(X_3\)
These variables are all uniformly distributed across a unit cube defined as \((0,0,0) \) to \( (1,1,1)\). The joint probability density function (pdf) of these three independent variables is:\[f(x_1, x_2, x_3) = 8 x_1 x_2 x_3 \]for the case where all variables lie within the unit interval \((0,1)\) and zero otherwise.Since these variables are independent, the joint pdf is the product of the individual pdfs. This independence simplifies calculations greatly, especially when transforming variables. If they were not independent, we would need to consider their covariance or correlation; but in this case, it is not necessary.
Transformation Technique
The transformation technique is essential when dealing with scenarios where a relationship between variables isn’t direct. It is used to map one set of random variables into another. In this exercise, the transformation of variables is key to solving for the pdf of volume \(Y_3 = X_1 X_2 X_3\).First, we introduce intermediate variables:
  • \(Y_1 = X_1\)
  • \(Y_2 = X_1 X_2\)
This helps to form simplified expressions for further calculations. The core idea is transforming variables from \( (X_1, X_2, X_3) \) to \( (Y_1, Y_2, Y_3) \). The transformation equations become:
  • \(X_1 = Y_1\)
  • \(X_2 = \frac{Y_2}{Y_1}\)
  • \(X_3 = \frac{Y_3}{Y_2}\)
By substituting these transformations into the joint pdf formula, we can express the problem in terms of the new variables. Thus, the original multidimensional problem becomes more manageable. This highlights how transformation can simplify complex functions into more obvious solutions.
Jacobian Calculation
The Jacobian calculation is a crucial step whenever we use transformation techniques, especially when shifting among different sets of variables. The Jacobian is a determinant that accounts for how volume scales under the transformation.For the transformation from \( (X_1, X_2, X_3) \) to \( (Y_1, Y_2, Y_3) \), we calculate the Jacobian matrix, composed of the partial derivatives of the transformations:\[J = \begin{vmatrix} \frac{\partial x_1}{\partial y_1} & \frac{\partial x_1}{\partial y_2} & \frac{\partial x_1}{\partial y_3} \ \frac{\partial x_2}{\partial y_1} & \frac{\partial x_2}{\partial y_2} & \frac{\partial x_2}{\partial y_3} \ \frac{\partial x_3}{\partial y_1} & \frac{\partial x_3}{\partial y_2} & \frac{\partial x_3}{\partial y_3}\end{vmatrix} = \begin{vmatrix} 1 & 0 & 0 \ 0 & \frac{1}{y_1} & 0 \ 0 & -\frac{y_3}{y_1y_2} & \frac{1}{y_2} \end{vmatrix}\]By evaluating this determinant, we find:\[\text{Jacobian} = \frac{1}{Y_1 Y_2}\]Thus, when transforming the joint pdf to the new variables, we have to multiply by the absolute value of this Jacobian determinant. This adjustment accommodates the change in volume introduced by transforming coordinate systems. Ultimately, calculating the Jacobian is pivotal as it ensures that the transformed density maintains the integral properties, allowing us to derive the correct pdf for \(Y_3\).

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Most popular questions from this chapter

A restaurant serves three fixed-price dinners costing $$\$ 20, \$ 25$$, and $$\$ 30$$. For a randomly selected couple dining at this restaurant, let \(X=\) the cost of the man's dinner and \(Y=\) the cost of the woman's dinner. The joint pmf of \(X\) and \(Y\) is given in the following table: $$ \begin{array}{ll|ccc} & & {l}{y} \\ {p(x, y)} & & 20 & 25 & 30 \\ \hline {x} & 20 & .05 & .05 & .10 \\ & 25 & .05 & .10 & .35 \\ & 30 & 0 & .20 & .10 \end{array} $$ a. Compute the marginal pmf's of \(X\) and \(Y\). b. What is the probability that the man's and the woman's dinner cost at most \(\$ 25\) each? c. Are \(X\) and \(Y\) independent? Justify your answer. d. What is the expected total cost of the dinner for the two people? e. Suppose that when a couple opens fortune cookies at the conclusion of the meal, they find the message "You will receive as a refund the difference between the cost of the more expensive and the less expensive meal that you have chosen." How much does the restaurant expect to refund?

Annie and Alvie have agreed to meet for lunch between noon \((0: 00\) p.m. \()\) and 1:00 p.m. Denote Annie's arrival time by \(X\), Alvie's by \(Y\), and suppose \(X\) and \(Y\) are independent with pdf's $$ \begin{aligned} &f_{X}(x)=\left\\{\begin{array}{cc} 3 x^{2} & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. \\ &f_{Y}(y)=\left\\{\begin{array}{cl} 2 y & 0 \leq y \leq 1 \\ 0 & \text { otherwise } \end{array}\right. \end{aligned} $$ What is the expected amount of time that the one who arrives first must wait for the other person?

Show that if \(X, Y\), and \(Z\) are rv's and \(a\) and \(b\) are constants, then \(\operatorname{Cov}(a X+b Y, Z)=a \operatorname{Cov}(X, Z)+\) \(b \operatorname{Cov}(Y, Z)\)

When an automobile is stopped by a roving safety patrol, each tire is checked for tire wear, and each headlight is checked to see whether it is properly aimed. Let \(X\) denote the number of headlights that need adjustment, and let \(Y\) denote the number of defective tires. a. If \(X\) and \(Y\) are independent with \(p_{X}(0)=.5\), \(p_{X}(1)=.3, p_{X}(2)=.2\), and \(p_{Y}(0)=.6, p_{Y}(1)\) \(=.1, p_{Y}(2)=p_{Y}(3)=.05, p_{Y}(4)=.2\), display the joint pmf of \((X, Y)\) in a joint probability table. b. Compute \(P(X \leq 1\) and \(Y \leq 1)\) from the joint probability table, and verify that it equals the product \(P(X \leq 1) \cdot P(Y \leq 1)\) c. What is \(P(X+Y=0)\) (the probability of no violations)? d. Compute \(P(X+Y \leq 1)\)

A stick is one foot long. You break it at a point \(X\) (measured from the left end) chosen randomly uniformly along its length. Then you break the left part at a point \(Y\) chosen randomly uniformly along its length. In other words, \(X\) is uniformly distributed between 0 and 1 and, given \(X=x, Y\) is uniformly distributed between 0 and \(x\). a. Determine \(E(Y \mid X=x)\) and then \(V(Y \mid X=x)\). Is \(E(Y \mid X=x)\) a linear function of \(x\) ? b. Determine \(f(x, y)\) using \(f_{X}(x)\) and \(f_{Y \mid X}(y \mid x)\). c. Determine \(f_{Y}(y)\). d. Use \(f_{Y}(y)\) from (c) to get \(E(Y)\) and \(V(Y)\). e. Use (a) and the theorem of this section to get \(E(Y)\) and \(V(Y)\).

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