Question

Suppose we want to simulate a point located at random in a circle of radius \(r\) centered at the origin. That is, we want to simulate \(X, Y\) having joint density $$ f(x, y)=\frac{1}{\pi r^{2}}, \quad x^{2}+y^{2} \leqslant r^{2} $$ (a) Let \(R=\sqrt{X^{2}+Y^{2}}, \theta=\tan ^{-1} Y / X\) denote the polar coordinates. Compute the joint density of \(R, \theta\) and use this to give a simulation method. Another method for simulating \(X, Y\) is as follows: Step 1: Generate independent random numbers \(U_{1}, U_{2}\) and set \(Z_{1}=\) \(2 r U_{1}-r, Z_{2}=2 r U_{2}-r\). Then \(Z_{1}, Z_{2}\) is uniform in the square whose sides are of length \(2 r\) and which encloses, the circle of radius \(r\) (see Figure 11.6). Step 2: If \(\left(Z_{1}, Z_{2}\right)\) lies in the circle of radius \(r\) -that is, if \(Z_{1}^{2}+Z_{2}^{2} \leqslant r^{2}-\) set \((X, Y)=\left(Z_{1}, Z_{2}\right) .\) Otherwise return to step \(1 .\) (b) Prove that this method works, and compute the distribution of the number of random numbers it requires.

Short answer

In part (a), we found the joint density of \(R, \theta\) using the given joint density of \(X, Y\), and derived a simulation method using the inverse transform method. The Cartesian coordinates \(X, Y\) can be obtained as follows: \[X = -\ln(1−U_1)\cos(2\pi U_2)\] and \[Y = -\ln(1−U_1)\sin(2\pi U_2),\] where \(U_1, U_2\) are independent uniformly distributed random numbers on \((0,1)\). In part (b), we proved that the outlined method works by showing that it generates a point \((Z_1, Z_2)\) uniformly distributed within the circle of radius \(r\). We also computed the distribution of the required random numbers, finding that it follows a geometric distribution with success probability \(p=\frac{\pi}{4}\) and probability mass function: \[P(N = n) = \left(1-\frac{\pi}{4}\right)^{n-1}\frac{\pi}{4}.\]

Step-by-step solution

Find Joint Density of R, θ

Begin by finding the joint density of \(R, \theta\) using the given joint density of \(X, Y\), i.e. \(f(x, y) = \frac{1}{\pi r^2}\) for \(x^2 + y^2 \leq r^2\). To do this, we'll use the transformation technique, with the Jacobian matrix for the transformations: \[J = \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{pmatrix} = \begin{pmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{pmatrix}.\] Then, compute the determinant of the Jacobian matrix to find the mapping between the transformed variables: \[\det(J) = r(\cos^2 \theta + \sin^2 \theta) = r.\]

Compute the Transformed Joint Density

The transformed joint density is given by: \[f_R(r, \theta) = f_{X,Y}(x(r, \theta),y(r, \theta))\cdot \| \det(J)\|.\] Substituting the relationship found in Step 1, we have: \[f_R(r, \theta) = \frac{1}{\pi r^2} \cdot r = \frac{r}{\pi r^2}.\] Note that this density is valid for \(0\leq r \leq r\) and \(0\leq\theta\leq2\pi\).

Use Inverse Transform Method

To simulate \(R\) and \(\theta\), we can use the inverse transform method. The marginal density for \(R\) and \(\theta\) are given by: \[f_R(r) = \int_0^{2\pi} \frac{r}{\pi r^{2}}d\theta= \frac{\cancel{2\pi r}}{\cancel{2\pi r}} =1,\] and \[f_{\theta}(θ)=\int_0^r \frac{r}{\pi r^{2}}dr=\frac{1}{\cancel{\pi}}\int_0^r \frac{r}{r^{\cancel{2}}}\cancel{dr} = \frac{1}{\pi}\cdot r\Big| _{0}^{r}=\frac{1}{\pi}.\] Then, the inverse transform method yields \(R = −\ln(1−U_1)\) and \(\theta = 2\pi U_2\), where \(U_1, U_2\) are independent uniformly distributed random numbers on \((0,1)\).

Obtain X and Y

Finally, obtain the Cartesian coordinates \(X\) and \(Y\) by converting from the polar coordinates found in Step 3: \[X = R\cos{\theta} = -\ln(1−U_1)\cos(2\pi U_2)\] and \[Y = R\sin{\theta} = -\ln(1−U_1)\sin(2\pi U_2).\] Part (b):

Prove the Method Works

To prove that the outlined method works, we must show that it generates a point \((Z_1, Z_2)\) uniformly distributed within the circle of radius \(r\). We know that \(Z_1\) and \(Z_2\) are uniform in the square that encloses the circle. By only accepting points \((Z_1, Z_2)\) such that \(Z_1^2 + Z_2^2 \leq r^2\), we restrict their joint distribution to be within the circle with radius \(r\). This retains the joint uniform distribution within the circle, thus proving the method works.

Compute the Distribution of Required Random Numbers

To compute the distribution of the required random numbers, we must find the probability of needing \(n\) random numbers to generate a valid point within the circle. Note that the ratio of the circle's area to the square's area that encloses the circle is given by: \[\frac{\pi r^2}{(2r)^2} = \frac{\pi r^2}{4r^2} = \frac{\pi}{4}.\] Since generating a valid point is a success and generating an invalid point is a failure, we can model the distribution of required random numbers as a geometric distribution with success probability \(p=\frac{\pi}{4}\). The probability mass function of the geometric distribution is given by: \[P(N = n) = (1-p)^{n-1} p = \left(1-\frac{\pi}{4}\right)^{n-1}\frac{\pi}{4},\] where \(N\) is the number of random numbers needed for a valid point.

Key concepts

Probability Density Function

Understanding the probability density function (PDF) is crucial when simulating random points, as it is a tool used to specify the probability of a random variable falling within a particular range. In the case of our circle with radius r, the joint PDF for the Cartesian coordinates X and Y is uniform within the circle and defined as:

\[ f(x, y)=\frac{1}{\pi r^{2}} \text{for } x^{2}+y^{2} \leq r^{2} \]
Here, the function f(x, y) represents the density of points over the area of the circle, indicating that the likelihood of a point occurring at any location is constant across the circle's area. For a PDF to be valid, it must meet the following criteria: The probability is non-negative everywhere, and the integral of the PDF over all possible values equals one. In this case, integrating f(x, y) over the circle's area indeed yields one, validating our PDF.

Polar Coordinate Transformation

When working within a circular region, polar coordinates are often more convenient than Cartesian coordinates. The polar coordinate transformation converts the Cartesian coordinates, X and Y, into radial R and angular θ components.

\[ R=\sqrt{X^{2}+Y^{2}}, \quad \theta=\tan^{-1}\frac{Y}{X} \]
This transformation is particularly useful for our simulation since the original problem's geometry is circular. The joint density in polar coordinates must also consider the Jacobian of the transformation, which accounts for the two-dimensional 'stretching' that occurs when mapping between coordinate systems:

\[ \det(J) = r \]
where J is the determinant of the Jacobian matrix for the transformation. This gives us the transformed joint density function f_R(r, θ), necessary for simulating points in polar coordinates.

Inverse Transform Sampling

Inverse transform sampling is a method to generate observations from a specified probability distribution. To apply this method to our simulation of random points within a circle, we first need the cumulative distribution function (CDF), which we obtain by integrating the density function. From the given uniform distribution, we can directly deduce the CDFs for R and θ:

\[ F_R(r) = r, \text{ and } F_{\theta}(\theta) = \frac{\theta}{2\pi} \text{ for } 0 \leq r \leq r, 0 \leq \theta \leq 2\pi \]
Then, by generating random numbers U_1 and U_2 uniformly distributed between 0 and 1, and applying the inverse transform method, we can simulate random radii and angles:

\[ R = -\ln(1−U_1) \text{ and } \theta = 2\pi U_2 \]
This technique aids in producing points that follow the original probability distributions of R and θ in the simulated circle.

Geometric Distribution

Lastly, in order to understand the efficiency of our simulation method, we analyze the geometric distribution. This distribution models the number of trials needed to get the first success in a series of independent and identically distributed Bernoulli trials. In the context of simulating our random points, each trial is an attempt to generate a valid point within the circle, and a success occurs when a point falls inside the circle's area.

The probability of success on a single trial is the ratio of the area of the circle to the area of the enclosing square, p = π/4. The probability mass function for the geometric distribution is:

\[ P(N = n) = (1-p)^{n-1} p \]
where N represents the number of trials needed to obtain the first success. In this way, we can calculate the expected number of attempts we must make before generating a point (X, Y) within our circle, illustrating the simulation's efficiency.