Question

Let \(X\) and \(Y\) denote independent random variables with respective probability density functions \(f(x)=2 x, 0

Short answer

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

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 \(v = x\)

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.

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 \(0<u<v<1\).

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.

Key concepts

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:

• \( f(x, y) = f(x)g(y) \)

For this exercise, \(X\) and \(Y\) are independent random variables. Thus, the joint probability density function is straightforward to handle. The simplicity in dealing with independent variables lies in how we manage to separate their influences on outcomes, making it easier to compute functions like expected values or variances.

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:

• Case 1: If \(x < y\), then \(u = x\) and \(v = y\).
• Case 2: If \(x \geq y\), then \(u = y\) and \(v = x\).

Once the transformation equations are identified, you can apply the Jacobian determinant to correctly transform the probability densities.

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:

• Case 1 and 2: \(|(du/dx)(dv/dy) - (du/dy)(dv/dx)| = |1|\)

The result, \(1\), indicates that there is no change in volume. Thus, the joint probability density function just includes this Jacobian as a multiplicative factor of 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:

• 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.

Adding these, the complete joint PDF is: \[ h(u, v) = 6u v^2 + 6u^2 v, \quad \text{for} \; 0 < u < v < 1 \]Anywhere outside these bounds, the PDF is zero. The sum reflects all possibilities of \(U\) and \(V\) within specified constraints.