Problem 13 - A First Course in Probability

Question

Let \((X, Y)\) be uniformly distributed in the circle of radius 1 centered at the origin. Its joint density is thus $$ f(x, y)=\frac{1}{\pi} \quad 0 \leq x^{2}+y^{2} \leq 1 $$ Let \(R=\left(X^{2}+Y^{2}\right)^{1 / 2}\) and \(\theta=\tan ^{-1}(Y / X)\) denote its polar coordinates. Show that \(R\) and \(\theta\) are independent with \(R^{2}\) being uniform on \((0,1)\) and \(\theta\) being uniform on \((0,2 \pi)\).

Step-by-Step Solution

Answer

We have shown that \(R\) and \(\theta\) are independent, with \(R^2\) being uniform on (0, 1) and \(\theta\) being uniform on (0, 2\(\pi\)). We found the joint density function of \(R\) and \(\theta\) using the transformation formula and calculated the Jacobian of the transformation. Then, we computed the marginal density functions and verified the product of marginal density functions equals the joint density function, proving their independence.

Step 1: Find the Jacobian of the transformation

We are given the Cartesian coordinates \((X, Y)\) of a point in the circle of radius 1 and we need to convert them to polar coordinates \((R, \theta)\). The transformation is defined by: \[X = R \cos{\theta}\] \[Y = R \sin{\theta}\] We will find the Jacobian of this transformation: \[J = \det\left(\begin{array}{cc} \frac{\partial X}{\partial R} & \frac{\partial X}{\partial \theta} \\ \frac{\partial Y}{\partial R} & \frac{\partial Y}{\partial \theta} \end{array}\right)\]

Step 2: Calculate the Jacobian

Let's compute the partial derivatives: \[\frac{\partial X}{\partial R} = \cos{\theta}\] \[\frac{\partial X}{\partial \theta} = -R \sin{\theta}\] \[\frac{\partial Y}{\partial R} = \sin{\theta}\] \[\frac{\partial Y}{\partial \theta} = R \cos{\theta}\] Now substitute these partial derivatives to find the Jacobian: \[J = \det\left(\begin{array}{cc} \cos{\theta} & -R \sin{\theta} \\ \sin{\theta} & R \cos{\theta} \end{array}\right) = R(\cos^2{\theta} + \sin^2{\theta})\] We know that \(\cos^2{\theta} + \sin^2{\theta} = 1\), so the Jacobian is \(J = R\).

Step 3: Find the joint density function of R and θ

Now, we will find the joint density function of R and θ using the transformation formula: \[f_{R,\theta}(r, \theta) = f_{X,Y}(x(r, \theta), y(r, \theta)) \cdot |J|\] Substitute the values for the transformation and the Jacobian: \[f_{R,\theta}(r, \theta) = \frac{1}{\pi} \cdot r\] This is the joint density function of R and θ.

Step 4: Compute the marginal density functions

To prove the independence of R and θ, we need to find their marginal density functions and verify that their product equals the joint density function. First, find the marginal density functions: \[f_R(r) = \int_0^{2\pi} f_{R,\theta}(r, \theta) d\theta = \int_0^{2\pi} \frac{r}{\pi} d\theta = r\int_0^{2\pi} \frac{1}{\pi} d\theta = r\] \[f_{\theta}(\theta) = \int_0^1 f_{R,\theta}(r, \theta) dr = \int_0^1 \frac{r}{\pi} dr = \frac{1}{\pi}\int_0^1 r dr = \frac{1}{\pi}\]

Step 5: Verify independence

Now we will check if the product of the marginal density functions equals the joint density function: \[f_R(r) \cdot f_{\theta}(\theta) = r \cdot \frac{1}{\pi} = \frac{r}{\pi}\] Since the product of the marginal density functions equals the joint density function, \( f_{R,\theta}(r, \theta) = f_R(r) \cdot f_{\theta}(\theta) \), R and θ are independent.

Step 6: Conclusion

We have demonstrated that R and θ are independent with \(R^2\) being uniform on (0, 1) and θ being uniform on (0, 2π) by finding the joint and marginal density functions and showing that the product of the marginal density functions is equal to the joint density function.

Key Concepts

Uniform Distribution

A uniform distribution is a type of probability distribution in which every outcome within a certain range is equally likely. In a uniform distribution over a circular region, like a disk, each point inside the circle has the same probability density. For this exercise, the joint probability density function of \(X\) and \(Y\)\ in the circle is defined as \(f(x, y) = \frac{1}{\pi}\) when \(0 \leq x^2 + y^2 \leq 1\). This implies that for any point within the circle, the probability of selecting that point is uniformly distributed.

• In simpler terms, if you were to randomly throw a dart at this circle, the chance of it landing at any two different locations within the circle is the same.
• This type of uniform distribution simplifies calculations because it assumes constant probability across the entire area.

Understanding this concept helps in converting from Cartesian to polar coordinates effectively when dealing with uniformly distributed random points in circular regions.
Hence, when we transform these coordinates and consider \(R\) and \(\theta\), uniform distribution characteristics affect their respective probability density functions.

Independence of Random Variables

Two random variables are said to be independent if the occurrence of one does not affect the probability of occurrence of the other. To check for independence in our context, we use the joint and marginal distribution functions.

• The joint density function \(f_{R,\theta}(r, \theta)\) is derived from the transformation, which involves the Jacobian in converting from Cartesian to polar coordinates.
• In this problem, we computed it as \(f_{R,\theta}(r, \theta) = \frac{r}{\pi}\).

The marginal density functions \(f_R(r)\) and \(f_{\theta}(\theta)\) were then calculated individually. These were \(f_R(r) = r\) and \(f_{\theta}(\theta) = \frac{1}{\pi}\). By multiplying these marginal densities, we obtained the same function as our joint density, \(\frac{r}{\pi}\). This conclusion showed that \(R\) and \(\theta\) are independent of each other, illustrating an important principle: independence can often be verified by demonstrating that the joint distribution equals the product of the marginal distributions.

Polar Coordinates Transformation

Transforming from Cartesian coordinates \((X, Y)\) to polar coordinates \((R, \theta)\) involves a mathematical process that uses the following formulas:

• \(X = R \cos{\theta}\)
• \(Y = R \sin{\theta}\)

This transformation is key in analyzing random variables defined in circular or radial regions because it simplifies expressing probabilities and relationships in terms of radius \(R\) and angle \(\theta\).
The exercise required us to find the joint density function through this transformation process using the Jacobian determinant method. The Jacobian expresses how volume elements in Cartesian coordinates change when transformed to polar coordinates. The transformation's Jacobian determinant was computed to be \(R\), which adjusted the uniform probability density as \(f_{R,\theta}(r, \theta) = \frac{r}{\pi}\). Understanding this transformation helps demonstrate characteristics like independence, as all variables are neatly expressed in these terms, expressing the entire parameter space of the circle efficiently.