Question

Let \((X, Y)\) be uniformly distributed in a circle of radius \(r\) about the origin. That is, their joint density is given by $$ f(x, y)=\frac{1}{\pi r^{2}}, \quad 0 \leqslant x^{2}+y^{2} \leqslant r^{2} $$ Let \(R=\sqrt{X^{2}+Y^{2}}\) and \(\theta=\arctan Y / X\) denote their polar coordinates. Show that \(R\) and \(\theta\) are independent with \(\theta\) being uniform on \((0,2 \pi)\) and \(P\\{R

Short answer

In conclusion, the polar coordinates \(R\) and \(\theta\) are not independent for the given uniformly distributed point \((X,Y)\) in a circle of radius \(r\) about the origin. The distribution of \(\theta\) is uniform on the interval \((0, 2\pi)\), and the probability that \(R < a\) for \(0 < a < r\) is given by \(P\{R<a\} = \frac{a^2}{r^2}\).

Step-by-step solution

Obtain the Jacobian of the transformation from Cartesian coordinates to polar coordinates

To find the marginal densities of \(R\) and \(\theta\), we need to find the Jacobian of the transformation from the Cartesian coordinates \((x,y)\) to the polar coordinates \((R,\theta)\), which can be obtained by calculating the determinants of the partial derivatives of the functions \(x(R,\theta) = R\cos \theta\) and \(y(R,\theta) = R\sin \theta\). Calculate the Jacobian as follows: \[ J(R, \theta) = \begin{vmatrix} \frac{\partial x}{\partial R} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial R} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos \theta & -R\sin \theta \\ \sin \theta & R\cos \theta \end{vmatrix} = R(\cos^2 \theta + \sin^2 \theta) = R \]

Calculate the joint density of \(R\) and \(\theta\)

Now we can obtain the joint density of \(R\) and \(\theta\) by using the Jacobian transformation formula for the probability density function: \[ g(R, \theta) = f(x(R, \theta), y(R, \theta))|J(R, \theta)| = \frac{1}{\pi r^2} R \]

Obtain the marginal densities of \(R\) and \(\theta\)

With the joint density of \(R\) and \(\theta\) available, we can now calculate the marginal densities. For \(R\), integrate the joint density function as follows: \[ g_R(R) = \int_0^{2\pi} g(R, \theta) d\theta = \frac{R}{\pi r^2} \int_0^{2\pi} d\theta = \frac{2\pi R}{\pi r^2} = \frac{2 R}{r^2} \quad\text{for}\, 0 \leq R \leq r \] For \(\theta\), integrate the joint density function: \[ g_\theta(\theta) = \int_0^r g(R, \theta) dR = \frac{1}{\pi r^2} \int_0^r R^2 dR = \frac{1}{3\pi}r^2 \quad\text{for}\, 0 \leq \theta \leq 2\pi \]

Show the independence of \(R\) and \(\theta\)

Now that we have the marginal densities of \(R\) and \(\theta\), we can show their independence by checking if the joint density of \(R\) and \(\theta\) is equal to the product of their marginal densities. \[ g_R(R)g_\theta(\theta) = \frac{2 R}{r^2} \cdot \frac{1}{3\pi}r^2 = \frac{2 R}{6\pi} = \frac{R}{3\pi} \] Comparing this result with the joint density we calculated in Step 2, we notice that \(g_R(R)g_\theta(\theta) \neq g(R, \theta)\). Hence, \(R\) and \(\theta\) are not independent.

Calculate \(P\{R < a\}\)

Finally, we calculate the probability that \(R < a\) for \(0 < a < r\). To do this, integrate the function \(g_R(R)\) (density of \(R\)) over the interval \((0, a)\) to find the cumulative distribution function of \(R\). \[ P\{R<a\} = \int_0^a g_R(R) dR = \int_0^a \frac{2 R}{r^2} dR = \frac{2}{r^2} \int_0^a R dR = \frac{a^2}{r^2} \] Thus, the probability that \(R < a\) is \(\frac{a^2}{r^2}\) for \(0 < a < r\).