Problem 19 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

From the "north pole" \(N\) of a circle with diameter 1 , a point \(Q\) on the circle is mapped to a point \(t\) on the line by its projection from \(N\), as illustrated in Figure \(8.2\). Suppose that the point \(Q\) is uniformly chosen on the circle. This is the same as saying that the angle \(\varphi\) is uniformly chosen from the interval \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) (can you see this?). Let \(X\) be this angle, so that \(X\) is uniformly distributed over the interval \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). This means that \(\mathrm{P}(X \leq \varphi)=1 / 2+\varphi / \pi\) (cf. Quick exercise \(5.3)\). What will be the distribution of the projection of \(Q\) on the line? Let us call this random variable \(Z\). Then it is clear that the event \(\\{Z \leq t\\}\) is equal to the event \(\\{X \leq \varphi\\}\), where \(t\) and \(\varphi\) correspond to each other under the projection. This means that \(\tan (\varphi)=t\), which is the same as saying that \(\arctan (t)=\varphi .\) a. What part of the circle is mapped to the interval \([1, \infty)\) ? b. Compute the distribution function of \(Z\) using the correspondence between \(t\) and \(\varphi\). c. Compute the probability density function of \(Z\). The distribution of \(Z\) is called the Cauchy distribution (which will be discussed in Chapter 11).

Step-by-Step Solution

Answer

(a) The circle part from \( \varphi = \frac{\pi}{4} \) to \( \frac{\pi}{2} \). (b) \( F_Z(t) = \frac{1}{2} + \frac{\arctan(t)}{\pi} \). (c) \( f_Z(t) = \frac{1}{\pi(1+t^2)} \).

Step 1: Understanding the Problem Context

In the problem, we project a point on a circle with diameter 1 from its north pole onto a line. The angle \( \varphi \) between the negative x-axis and the point \( Q \) is uniformly distributed between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \). We need to find how this angle translates to the line as a random variable \( Z \).

Step 2: Set Up the Correspondence

To translate the angle \( \varphi \) into \( t \), the projection on the line, we use the relation \( t = \tan(\varphi) \). Hence, the angle \( \varphi \) is equivalent to \( \arctan(t) \).

Step 3: Interval Mapping for Part a

The interval \([1, \infty)\) on the line corresponds to angles \( \varphi \) for which \( \tan(\varphi) \geq 1\). Solving \( \tan(\varphi) = 1 \) gives \( \varphi = \frac{\pi}{4} \). Thus, the part of the circle from \( \varphi = \frac{\pi}{4} \) to \( \frac{\pi}{2} \) maps to \([1, \infty)\).

Step 4: Compute Distribution Function for Part b

The distribution function for \( Z \) is derived from the uniform distribution of \( X \). Since \( \mathrm{P}(X \leq \varphi) = \frac{1}{2} + \frac{\varphi}{\pi} \), we can substitute \( \varphi = \arctan(t) \) to get \( \mathrm{P}(Z \leq t) = \frac{1}{2} + \frac{\arctan(t)}{\pi} \).

Step 5: Compute Probability Density Function for Part c

To find the density function \( f_Z(t) \), we differentiate the distribution function. Therefore, \( f_Z(t) = \frac{d}{dt} \left(\frac{1}{2} + \frac{\arctan(t)}{\pi}\right) = \frac{1}{\pi(1+t^2)} \). This is the probability density function of the Cauchy distribution.

Key Concepts

Uniform Distribution

In this exercise, we start with a point randomly selected on a circle. The angle \( \varphi \) associated with this point is uniformly chosen within the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\). This means every angle in this range has an equal chance of being selected, characteristic of a uniform distribution. This symmetrical range around zero ensures that the distribution covers the entire possible range of angles on the semicircle.

• Equal Probability: A uniform distribution signifies that the variable has constant probability across its range. Here, any angle within \([-\frac{\pi}{2}, \frac{\pi}{2}]\) is equally likely.
• Application: The probability \(\mathrm{P}(X \leq \varphi)=1 / 2+\varphi / \pi\) reflects the uniform spread, changing linearly with \( \varphi \).

Understanding uniform distribution is key, as it provides the foundation for mapping angles to line projections in this problem.

Projection Mapping

Projection mapping transforms a point on the circle onto a line using the relationship \( t = \tan(\varphi) \). Here, \( t \) is the projected value on the line, and this mapping plays a central role in translating circular geometry to linear values.

• From Circle to Line: The function \( t = \tan(\varphi) \) allows the angle to translate directly into a point \( t \) on the line, expanding the angle's impact across a linear dimension.
• Reversibility: To reverse this, \( \varphi = \arctan(t) \) gives back the original angle from the projection, which is essential for finding relationships between distributions.

Projection mapping simplifies complex angular movements to straightforward linear ones, facilitating easier manipulation of data across different dimensions.

Angle to Line Transformation

This concept involves transforming angles from the circle to specific intervals on the line. For example, angles \( \varphi \) that meet \( \tan(\varphi) = 1 \) (i.e., \( \varphi = \frac{\pi}{4} \)) demarcate a critical change, mapping to \([1, \infty)\) on the line. This transformation helps understand how sections of the circle translate into linear intervals.

• Translation to Intervals: As \( \varphi \) approaches \( \frac{\pi}{4} \) and beyond, the point \( t \) on the line surpasses 1, moving into the interval \([1, \infty)\).
• Interval Mapping: The mapping requires evaluating the tangent function to link specific angles to line segments, unveiling the impact of angle shifts in linear terms.

This transformation is crucial in understanding how evenly distributed angles can produce varying densities along a line.

Probability Density Function

In this step, we compute the probability density function (PDF) of the distribution represented by \( Z \). Derived from the relation \( \tan(\varphi) = t \), the function describes the likelihood of various projections occurring on the line.

• Derivation: By differentiating the cumulative distribution function (CDF) \( \mathrm{P}(Z \leq t) = \frac{1}{2} + \frac{\arctan(t)}{\pi} \), we obtain the PDF: \( f_Z(t) = \frac{1}{\pi(1+t^2)} \).
• Understanding Cauchy Distribution: The resulting PDF is characteristic of a Cauchy distribution, a distribution known for heavy tails and undefined mean. No matter how far out you go, probabilities never entirely diminish.

The probability density function is essential as it provides a complete account of randomness within the projection mapping framework.