Chapter 5: Problem 19
Let \(X\) and \(Y\) denote independent random variables with respective
probability density functions \(f(x)=2 x, 0
Short Answer
Expert verified
The joint probability density function of U and V is \(h(u, v) = 6u*v^2 + 6u^2*v\), for \(0<u<v<1\), zero elsewhere.
Step by step solution
01
Set Up The Equations
Define the following system of equations based on the definition of U and V. \n\(u = \min(x,y)\) and \(v = \max(x,y)\). The inverse transformations can be given as, Case 1: If \(x=y\), then \(u = y\) and \(v = x\)
02
Calculate Jacobian
Find the Jacobian of transformation by taking the derivative of u and v w.r.t x and y. For case 1, Jacobian \(J_1 = |(du/dx)(dv/dy) - (du/dy)(dv/dx)| = |1|\) . For case 2, Jacobian \(J_2 = |(du/dx)(dv/dy) - (du/dy)(dv/dx)| = |1| \). For both cases the Jacobian equals to 1.
03
Evaluate Joint PDF
Then, determine the joint pdf of U and V using the formula: \(h(u, v) = f(x(u, v), y(u, v))*|J|\). For case 1: \(h_1(u,v) = f(x)*g(y)*|J_1| = 2u*3v^2*1 = 6u*v^2\) which is valid for \(0<u<v<1\). And for case 2: \(h_2(u,v) = f(x)*g(y)*|J_2| = 2v*3u^2*1 = 6u^2*v\) which is valid for \(0<u<v<1\).
04
Combine the Two Cases
Because both of the cases are valid for \(0<u<v<1\), the joint pdf is the sum of these two cases. Therefore, \(h(u, v) = h_1(u, v) + h_2(u, v) = 6u*v^2 + 6u^2*v\), for \(0<u<v<1\), zero elsewhere.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Independent Random Variables
Random variables are said to be independent if the occurrence of one does not affect the probability of occurrence of another. When two variables, say \(X\) and \(Y\), are independent, their joint probability density function is simply the product of their individual probability density functions. This means:
- \( f(x, y) = f(x)g(y) \)
Transformation of Variables
Transformation of variables is a technique used to change variables from one set to another, which can often make solving a problem more convenient. In this problem, \(U = \min(X, Y)\) and \(V = \max(X, Y)\) are new variables that stem from \(X\) and \(Y\).
An important aspect of transformation is using inverse transformations to switch back to original variables. We can express the original variables in terms of the transformed ones. Here, the transformations lead to two scenarios:
An important aspect of transformation is using inverse transformations to switch back to original variables. We can express the original variables in terms of the transformed ones. Here, the transformations lead to two scenarios:
- Case 1: If \(x < y\), then \(u = x\) and \(v = y\).
- Case 2: If \(x \geq y\), then \(u = y\) and \(v = x\).
Jacobian Determinant
The Jacobian determinant is a mathematical tool used when transforming variables. It accounts for how the volume or "space" is distorted during the transformation of variables.
In our case, the transformation involves finding \(U\) and \(V\) based on \(X\) and \(Y\). The Jacobian determinant here is crucial because it quantifies this transformation's effect.
To compute the Jacobian, take the partial derivatives of the new variables (\(u, v\)) with respect to the old ones (\(x, y\)). For both cases in this transformation, the determinant simplifies to:
In our case, the transformation involves finding \(U\) and \(V\) based on \(X\) and \(Y\). The Jacobian determinant here is crucial because it quantifies this transformation's effect.
To compute the Jacobian, take the partial derivatives of the new variables (\(u, v\)) with respect to the old ones (\(x, y\)). For both cases in this transformation, the determinant simplifies to:
- Case 1 and 2: \(|(du/dx)(dv/dy) - (du/dy)(dv/dx)| = |1|\)
Probability Density Function
The probability density function (PDF) provides a density function for continuous random variables. It describes the likelihood of a random variable to take on a particular value.
In the exercise, after establishing transformations \(U = \min(X, Y)\) and \(V = \max(X, Y)\), we need their joint PDF. The construction of the joint PDF follows considering both cases identified with transformation:
In the exercise, after establishing transformations \(U = \min(X, Y)\) and \(V = \max(X, Y)\), we need their joint PDF. The construction of the joint PDF follows considering both cases identified with transformation:
- For Case 1 (\(x < y\)): The PDF is \(6u\,v^2\) within the bounds \(0 < u < v < 1\).
- For Case 2 (\(x \geq y\)): The PDF is \(6u^2\,v\) for the same bounds.