Chapter 5: Problem 20
Let the joint pdf of \(X\) and \(Y\) be \(f(x, y)=\frac{12}{7} x(x+y),
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Short Answer
Expert verified
The joint pdf of U and V for the range \(0<u<v<1\) is \(g(u,v) = \frac{12}{7}u(v + u)\)
Step by step solution
01
Determine the ranges of U and V
Let's first determine the ranges for U and V. From the given conditions \(0<x<1\) and \(0<y<1\), it follows that \(0<u<1\) and \(0<v<1\). Since V represents the maximum and U the minimum, we can also say that \(u<v\). From these conditions, we infer that \(u\) ranges from 0 to \(v\) and \(v\) ranges from \(u\) to 1.
02
Set up the Jacobian for transformation
Next, we set up the Jacobian for the transformation to \(U\) and \(V\). The Jacobian matrix is the matrix of all first-order partial derivatives of a vector-valued function. For the functions \(U=\min (X, Y)\) and \(V=\max (X, Y)\), it can be shown that the absolute value of the Jacobian determinant |\(J(u,v))|\ is always 1.
03
Find the pdfs of U and V
We use the joint pdf of X and Y and the absolute value of the Jacobian determinant to find the joint pdf of U and V. The joint pdf of U and V is given by \(g(u,v) = f(x,y) * |J(u,v)|\). Substituting for \(f(x,y) = \frac{12}{7} x(x+y)\) and using the fact that the absolute value of the Jacobian determinant equals 1, we get the following 2 cases:If \(u<x<v\), then returns \(g(u, v) = \frac{12}{7}u(u + v)\). If \(v<x<u\), then returns \(g(u, v) = \frac{12}{7}v(u + v)\). After some simplification, we find that the joint pdf of U and V for the specified range is \(g(u, v) = \frac{12}{7}u(v + u)\).
04
Evaluate the joint pdf
Now we can evaluate the joint pdf for the range of U and V. Substituting the range for U and V from step 1 into our formula for g(u,v), the joint pdf of U and V for \(0<u<v<1\) is given by \(g(u,v) = \frac{12}{7}u(v + u)\) with \(0<u<1\) and \(u<v<1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Transformation using Jacobian
The Jacobian is a crucial tool when you need to transform a set of random variables into another set. It's like a helper that ensures the probability density function (pdf) adjusts correctly to the new variables.
In this exercise, we're shifting from the variables \(X\) and \(Y\), to their minimum \(U\) and maximum \(V\). For this transformation, we need the Jacobian.
The Jacobian matrix is built from all the first-order partial derivatives and gives us a "determinant," a special number that scales the original pdf. Here, the calculation shows the determinant is 1, meaning the pdf's scale remains unchanged when converting between our variables.
This perfect scaling simplifies our work because it implies the joint pdf is essentially the original pdf times 1, which is the same as the original pdf.
Remember, always check the Jacobian when transforming variables, as it tells you how the dimensions change.
In this exercise, we're shifting from the variables \(X\) and \(Y\), to their minimum \(U\) and maximum \(V\). For this transformation, we need the Jacobian.
The Jacobian matrix is built from all the first-order partial derivatives and gives us a "determinant," a special number that scales the original pdf. Here, the calculation shows the determinant is 1, meaning the pdf's scale remains unchanged when converting between our variables.
This perfect scaling simplifies our work because it implies the joint pdf is essentially the original pdf times 1, which is the same as the original pdf.
Remember, always check the Jacobian when transforming variables, as it tells you how the dimensions change.
Ranges of Random Variables
In probability, understanding the range or interval of random variables is essential, as it defines their allowable values.
In our case, we look at the random variables \(U = \min(X, Y)\) and \(V = \max(X, Y)\). The goal is to find out how these two values can vary based on their definitions.
Given the conditions \(0 < x < 1\) and \(0 < y < 1\), it naturally follows that both \(U\) and \(V\) will also range from 0 to 1. A key rule here is that since \(U\) is the minimum and \(V\) is the maximum, clearly \(U\) must be less than \(V\), so we have \(u < v < 1\).
This relationship directly affects how we express the joint pdf of \(U\) and \(V\), as it guides the region (or range) over which the pdf is defined. Understanding these limits is critical for accurate pdf calculations.
In our case, we look at the random variables \(U = \min(X, Y)\) and \(V = \max(X, Y)\). The goal is to find out how these two values can vary based on their definitions.
Given the conditions \(0 < x < 1\) and \(0 < y < 1\), it naturally follows that both \(U\) and \(V\) will also range from 0 to 1. A key rule here is that since \(U\) is the minimum and \(V\) is the maximum, clearly \(U\) must be less than \(V\), so we have \(u < v < 1\).
This relationship directly affects how we express the joint pdf of \(U\) and \(V\), as it guides the region (or range) over which the pdf is defined. Understanding these limits is critical for accurate pdf calculations.
Minimum and Maximum of Random Variables
Working with minimum and maximum of random variables such as \(U = \min(X, Y)\) and \(V = \max(X, Y)\) involves certain intuitive insights.
For any selection of two numbers, grasping the idea of minimum and maximum becomes straightforward. \(U\) naturally takes the smaller of two values, while \(V\) adopts the larger.
The joint distribution function you calculate ultimately depends on the behavior of \(U\) and \(V\), so grasping this simple concept can assist in mastering more complex transformations and integrations.
For any selection of two numbers, grasping the idea of minimum and maximum becomes straightforward. \(U\) naturally takes the smaller of two values, while \(V\) adopts the larger.
- When both values \(X\) and \(Y\) fall within the same range, \(U\) must always lie below or equal to \(V\).
- Similarly, \(V\) dominates as the larger value in their range \(0 < v < 1\).
The joint distribution function you calculate ultimately depends on the behavior of \(U\) and \(V\), so grasping this simple concept can assist in mastering more complex transformations and integrations.