Problem 24
Question
Suppose that the random variables \(X\) and \(Y\) have joint PDF $$ f(x, y)=\left\\{\begin{array}{ll} \frac{1}{4}, & \text { if } 0 \leq x \leq 2,0 \leq y \leq 2 \\ 0, & \text { otherwise } \end{array}\right. $$ that is, \(X\) and \(Y\) are uniformly distributed over the square \(0 \leq x \leq 2,0 \leq y \leq 2\). Find (a) the joint PDF of \(U=X+Y\) and \(V=X-Y\), and (b) the marginal PDF of \(U\). 25\. Suppose \(X\) and \(Y\) have joint \(\mathrm{PDF}\) $$ f(x, y)=\left\\{\begin{array}{ll} e^{-x-y}, & \text { if } x \geq 0, y \geq 0 \\ 0, & \text { otherwise } \end{array}\right. $$ Find (a) the joint PDF of \(U=X+Y\) and \(V=X\) (b) the marginal PDF of \(U\).
Step-by-Step Solution
Verified Answer
The joint PDF of \((U, V)\) is \(\frac{1}{8}\), and the marginal PDF of \(U\) is \(\frac{1}{2}\).
1Step 1: Identify the transformation
We are transforming the random variables from \((X, Y)\) to \((U, V)\), where \(U = X + Y\) and \(V = X - Y\). We aim to find the joint PDF of \(U\) and \(V\).
2Step 2: Determine the Jacobian of the transformation
For transformations involving random variables, we use the Jacobian matrix to find the joint PDF. The transformation is given by \(U = X + Y\) and \(V = X - Y\). The Jacobian matrix for this transformation is:\[J = \begin{bmatrix}\frac{\partial X}{\partial U} & \frac{\partial Y}{\partial U} \\frac{\partial X}{\partial V} & \frac{\partial Y}{\partial V}\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & \frac{1}{2} \\frac{1}{2} & -\frac{1}{2}\end{bmatrix}\]The determinant of this matrix (Jacobian determinant) is \(\frac{1}{2}(-\frac{1}{2}) - \frac{1}{2}(\frac{1}{2}) = -\frac{1}{2}\).
3Step 3: Set up joint PDF for \(U\) and \(V\)
The joint PDF of \((U, V)\) is given by:\[f_{UV}(u, v) = f_{XY}(x, y) \cdot |J|\]where \(f_{XY}(x,y) = \frac{1}{4}\) for the given region and \(|J|\) is the absolute value of the Jacobian determinant.Thus,\[f_{UV}(u, v) = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}\] inside the transformed region.
4Step 4: Identify the limits of integration for \((U, V)\)
The limits for \((U, V)\) are determined from the region \(0 \leq x \leq 2\) and \(0 \leq y \leq 2\). For these ranges, \(0 \leq U \leq 4\) and \(-2 \leq V \leq 2\) as the transformed region.
5Step 5: Find the marginal PDF of \(U\)
The marginal PDF of \(U\), \(f_U(u)\), is found by integrating out \(V\) from the joint PDF:\[f_U(u) = \int_{-2}^{2} f_{UV}(u, v) \, dv\]Since \(f_{UV}(u, v) = \frac{1}{8}\) in the defined region,\[f_U(u) = \int_{-2}^{2} \frac{1}{8} \, dv = \frac{1}{8} \times (2 - (-2)) = \frac{1}{8} \times 4 = \frac{1}{2}\] for \(0 \leq u \leq 4\).
Key Concepts
Marginal PDFJacobian MatrixRandom VariablesProbability Density Function
Marginal PDF
When we talk about the Marginal Probability Density Function (PDF), we're focusing on simplifying the scenario outlined in joint PDFs. Essentially, marginal PDFs allow us to consider the behavior of a single random variable from a pair without its counterpart. If you have two random variables, say \(X\) and \(Y\), the marginal PDF of a variable, say \(X\), is obtained by integrating out the other variable \(Y\) from their joint PDF.
In our exercise, after transitioning from \((X, Y)\) to \((U, V)\), we find the marginal PDF of \(U\) by integrating over the variable \(V\). This simplifies the scenario to understand how \(U\) behaves on its own, capturing all possible values of \(V\) at each point of \(U\).
In our exercise, after transitioning from \((X, Y)\) to \((U, V)\), we find the marginal PDF of \(U\) by integrating over the variable \(V\). This simplifies the scenario to understand how \(U\) behaves on its own, capturing all possible values of \(V\) at each point of \(U\).
- Steps involve integrating across the limits of the transformed range of \(V\).
- This integration leaves us with a function dependent solely on \(U\).
- For many applications, the marginal PDF provides insights into the probability distribution of each transformed variable independently.
Jacobian Matrix
The Jacobian Matrix is a key concept in transforming variables, especially when dealing with changes from one set of random variables to another. If you transform a pair of random variables \(X\) and \(Y\) into new variables \(U\) and \(V\), the Jacobian Matrix helps correct for changes in the scale and orientation during this transformation.
With any transformation of variables, like moving from \((X, Y)\) to \((U, V)\), it's crucial to understand how these coordinates relate to one another.
With any transformation of variables, like moving from \((X, Y)\) to \((U, V)\), it's crucial to understand how these coordinates relate to one another.
- The Jacobian Matrix is formulated using partial derivatives of the new variables with respect to the original ones.
- In our exercise, the matrix involves the partial derivatives of \(U = X + Y\) and \(V = X - Y\) with respect to \(X\) and \(Y\).
- The determinant of this matrix, called the Jacobian determinant, provides a correction factor for the area measure in the variable transformation.
Random Variables
Random Variables are a cornerstone concept in probability and statistics. A random variable is a numerical outcome of a random phenomenon and can be either discrete or continuous. In our exercise, we deal with random variables \(X\) and \(Y\) which are jointly continuous.
These random variables are often used to model real-world phenomena where the outcomes aren't deterministic. Each set of outcomes is represented by a random variable, allowing us to apply probability theory to assess various questions regarding these phenomena.
These random variables are often used to model real-world phenomena where the outcomes aren't deterministic. Each set of outcomes is represented by a random variable, allowing us to apply probability theory to assess various questions regarding these phenomena.
- They are described using probability distributions that show likelihoods of various outcomes.
- They help in transforming real problems into mathematical frameworks that can be analyzed and solved with precision.
- In transformations, new random variables like \(U = X + Y\) and \(V = X - Y\) are introduced to provide alternative insights or solve specific problems, like calculating total values or differences.
Probability Density Function
The Probability Density Function (PDF) describes how the probability density of a continuous random variable is distributed over a range of values. Unlike discrete probabilities, PDFs deal with a continuum of possible outcomes.
In the given exercise, both \(X\) and \(Y\) are associated with a joint PDF because they are defined over a continuous space. This joint PDF informs us about the likelihood of outliers within a defined region of \((X, Y)\) pairs.
In the given exercise, both \(X\) and \(Y\) are associated with a joint PDF because they are defined over a continuous space. This joint PDF informs us about the likelihood of outliers within a defined region of \((X, Y)\) pairs.
- A PDF isn’t a probability, but a density; it needs to be integrated over a range to obtain a probability.
- For transformation purposes, a new joint PDF like \(f_{UV}(u, v)\) is established, capturing probabilities in the new variable space.
- For both marginal and joint distributions, PDFs provide a full description of the distribution of continuous variables, assisting in calculations and interpretations of probabilistic models.
Other exercises in this chapter
Problem 23
Use the comparison property of double integrals to show that if \(f(x, y) \geq 0\) on \(R\) then \(\iint f(x, y) d A \geq 0\)
View solution Problem 24
Use spherical coordinates to find the indicated quantity. Find the volume of the solid inside both of the spheres \(\rho=2 \sqrt{2} \cos \phi\) and \(\rho=2\).
View solution Problem 24
Evaluate by using polar coordinates. Sketch the region of integration first. $$ \int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} \sin \left(x^{2}+y^{2}\right) d x d y $$
View solution Problem 24
Sketch the indicated solid. Then find its volume by an iterated integration. Solid in the first octant bounded by the coordinate planes and the planes \(2 x+y-4
View solution