Problem 52 - A First Course in Probability

Question

Let \(X\) and \(Y\) denote the coordinates of a point uniformly chosen in the circle of radius 1 centered at the origin. That is, their joint density is $$f(x, y)=\frac{1}{\pi} \quad x^{2}+y^{2} \leq 1$$ Find the joint density function of the polar coordinates \(R=\left(X^{2}+Y^{2}\right)^{1 / 2}\) and \(\Theta=\tan ^{-1} Y / X\)

Step-by-Step Solution

Answer

The joint density function of the polar coordinates \(R\) and \(\Theta\) is given by: \[g(R, \Theta) = \frac{R}{\pi}\] for \(0 \leq R \leq 1\) and \(0 \leq \Theta \leq 2\pi\).

Step 1: Review the relationship between Cartesian and polar coordinates

Recall that the Cartesian coordinates \(X\) and \(Y\) can be represented in polar coordinates \(R\) and \(\Theta\) as follows: \[X = R \cos \Theta\] \[Y = R \sin \Theta\] Since we are given the joint density function of \(X\) and \(Y\) and we want to find the joint density function of \(R\) and \(\Theta\), we will need to make use of the Jacobian transformation.

Step 2: Use the Jacobian transformation technique

The Jacobian transformation technique is used to find the joint density function of a new set of variables formed from an existing set of variables. The Jacobian transformation formula is given by: \[g(u, v) = f(x, y) |J(u, v)|\] where \(J(u, v)\) is the determinant of the Jacobian matrix of the transformation. First, we need to find the Jacobian matrix of the transformation from Cartesian to polar coordinates. Let \[u = R\] \[v = \Theta\] The inverse transformation from polar to Cartesian coordinates is given by: \[x = u \cos v\] \[y = u \sin v\] Now, we can find the partial derivatives resulting in the Jacobian matrix: \[\begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix}\] Computing the partial derivatives, we get: \[\begin{bmatrix} \cos v & -u \sin v \\ \sin v & u \cos v \end{bmatrix}\] Now, compute the determinant of the Jacobian matrix (\(|J(u, v)|\)): \[ |J(u,v)| = \left(\cos{v}\right)(u \cos{v}) - \left(-u \sin{v}\right)(\sin{v}) = u \] The joint density function of \(X\) and \(Y\) is \(f(x, y) = \frac{1}{\pi}\) when \(x^2 + y^2 \leq 1\), and 0 otherwise. We can now compute the joint density function of \(R\) and \(\Theta\) using the given transformation technique formula. The transformed joint density function will be: \[g(u, v) = f(u \cos v, u \sin v) |J(u, v)|\] For \(x^2 + y^2 \leq 1\), substitute \(x = u\cos{v}\) and \(y = u\sin{v}\): \[u^2 \leq 1\] As \(u = R\), this implies that \(0 \leq R \leq 1\). Now, using the relation \[g(u, v) = f(u \cos v, u \sin v) |J(u, v)|\] We get: \[g(u, v) = \frac{1}{\pi} \cdot u\] So, the joint density function of \(R\) and \(\Theta\) is given by: \[g(R, \Theta) = \frac{R}{\pi}\] for \(0 \leq R \leq 1\) and \(0 \leq \Theta \leq 2\pi\).

Key Concepts

Joint Density Function

In probability theory, the joint density function describes how two random variables are distributed together. When dealing with continuous random variables, a joint density function is a mathematical function that gives the probability that two random variables fall within a certain range, particularly over a given area in the plane. This is critical for understanding how one variable affects the other.

• The joint density function of two random variables, say, \(X\) and \(Y\), is denoted as \(f(x, y)\).
• This function describes how the probability that \(X\) takes a particular value while simultaneously \(Y\) takes another value is distributed over the plane.

In this exercise, we begin with a joint density function for \(X\) and \(Y\), which are the Cartesian coordinates of a point chosen uniformly from within a unit circle. This function is a simple constant \(\frac{1}{\pi}\) over its valid region because the point is uniformly distributed across the circle. Finding the joint density function for different sets of random variables involves transformations, which we'll explore next.

Polar Coordinates

Polar coordinates are a system in mathematics for representing points in a plane using a distance from a reference point and an angle from a reference direction. Unlike the Cartesian coordinate system, which uses two perpendicular axes, polar coordinates use a central point, called the pole (analogous to the origin in Cartesian coordinates), and an angle \(\Theta\) from a reference direction.

• The radius \(R\) represents the distance from the origin to a point.
• The angle \(\Theta\) is measured from the positive x-axis, with angles increasing counterclockwise.

For the given problem, the transformation from Cartesian coordinates \((X, Y)\) to polar coordinates \((R, \Theta)\) is helpful because it can simplify problems related to circular regions or angle-dependent situations. Here, \(R\) is calculated by the equation \(R = \sqrt{X^2 + Y^2}\), and \(\Theta\) is given by \(\Theta = \tan^{-1} (Y / X)\). Polar coordinates make it easier to describe circular or radially symmetrical regions, assisting in effectively solving the initial problem of determining the joint density function for \(R\) and \(\Theta\).

Jacobian Transformation

The Jacobian transformation is a mathematical method used to transform coordinates for multivariable functions. It's particularly helpful when transitioning from one coordinate system to another, such as from Cartesian to polar coordinates in this exercise. The Jacobian is used to adjust the scale of the transformation, allowing us to map one region in one coordinate system to another in a new system while maintaining the probability density.Here are the steps simplified:

• Identify the transformation equations. In our case, the transformation from Cartesian \((x, y)\) to polar \((R, \Theta)\) was used.
• Calculate the Jacobian matrix, which contains partial derivatives of the transformation functions with respect to the original variables.
• Compute the determinant of the Jacobian matrix. This number scales the density function accordingly during the transformation process.

For our transformation, the Jacobian determinant was \(|J(u, v)| = u\) for the change from \( (X, Y) \) to \( (R, \Theta) \). This determinant appropriately scales the joint probability density from Cartesian coordinates to polar coordinates, facilitating the conversion to find the new joint density \(g(R, \Theta)\).

Uniform Distribution

A uniform distribution is a type of probability distribution in which every outcome is equally likely. It can be continuous, where the probability is spread over an interval or a shape, such as a rectangle or circle.In this exercise, the coordinates \(X, Y\) are uniformly distributed over the unit circle. This means that no specific region within the circle is more likely to be chosen than any other. Every point within the circle has an equal chance of being selected.

• The joint density function for a uniform distribution over a continuous area, like a circle, is constant within its bounds and zero outside this region.
• For the unit circle, the density was \( \frac{1}{\pi} \), reflecting uniform probability spread across the circle.

Transforming this uniform distribution from Cartesian coordinates to polar coordinates using the Jacobian transformation ensures the distribution remains uniform but now accommodates the radial logic of polar systems. This preserves the equal-likelihood characteristic across the circle when expressed as \( (R, \Theta) \), providing a holistic understanding of probability within this circular domain, resulting in \( g(R, \Theta) = \frac{R}{\pi} \). This continues to ensure the point retained has an even likelihood over its bounded area in the polar form.