Problem 11 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

In this exercise we compute the MAD of the \(\operatorname{Exp}(\lambda)\) distribution. a. Let \(X\) have an \(\operatorname{Exp}(\lambda)\) distribution, with median \(m=(\ln 2) / \lambda\). Show that \(Y=|X-m|\) has distribution function $$ G(y)=\frac{1}{2}\left(\mathrm{e}^{\lambda y}-\mathrm{e}^{-\lambda y}\right) . $$ b. Argue that the MAD of the \(\operatorname{Exp}(\lambda)\) distribution is a solution of the equation \(\mathrm{e}^{2 \lambda y}-\mathrm{e}^{\lambda y}-1=0\). c. Compute the MAD of the \(\operatorname{Exp}(\lambda)\) distribution. Hint: put \(x=\mathrm{e}^{\lambda y}\) and first solve for \(x\).

Step-by-Step Solution

Answer

MAD is \(\frac{1}{\lambda}\ln\left(\frac{1+\sqrt{5}}{2}\right)\).

Step 1: Define Y and Determine Its Distribution

Let \( Y = |X - m| \), where the median \( m = \frac{\ln 2}{\lambda} \). We want to find the distribution function of \( Y \). We know that the exponential distribution has the cumulative distribution function (CDF) \( F(x) = 1 - e^{-\lambda x} \) for \( x \geq 0 \). Hence, the CDF for \( Y \) is given by splitting it into two cases: \( X \geq m \) and \( X < m \).

Step 2: Compute the CDF of Y for X ≥ m

For \( X \geq m \), \( Y = X - m \). Therefore, \( P(Y \leq y | X \geq m) = P(|X - m| \leq y) = P(X \leq m + y) \). This is given by \( F(m + y) = 1 - e^{-\lambda(m + y)} = 1 - e^{-\lambda y}e^{-\lambda m} \). Since \( e^{-\lambda m} = \frac{1}{2} \), \( F(m + y) = 1 - \frac{1}{2}e^{-\lambda y} \).

Step 3: Compute the CDF of Y for X < m

For \( X < m \), \( Y = m - X \). Hence, \( P(Y \leq y | X < m) = P(m - X \leq y) = P(X \geq m - y) \). This probability is \( 1 - F(m - y) = e^{-\lambda(m-y)} = e^{-\lambda m}e^{\lambda y} \). Since \( e^{-\lambda m} = \frac{1}{2} \), this becomes \( \frac{1}{2}e^{\lambda y} \).

Step 4: Combine CDF Components to Find G(y)

Combine the results from Steps 2 and 3, noting that \( G(y) = \frac{1}{2}(e^{\lambda y} - e^{-\lambda y}) \). Both components are given equal weight because the median cuts the distribution into two equal halves. This gives us the CDF of \( Y \).

Step 5: Solve for MAD Using y Equation

To find the median absolute deviation (MAD), we solve \( e^{2 \lambda y} - e^{\lambda y} - 1 = 0 \) by substituting \( x = e^{\lambda y} \). This gives the quadratic equation in terms of \( x \): \( x^2 - x - 1 = 0 \).

Step 6: Solve Quadratic Equation

Solve \( x^2 - x - 1 = 0 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = -1 \). This simplifies to: \( x = \frac{1 \pm \sqrt{5}}{2} \). Choose the positive solution, \( x = \frac{1 + \sqrt{5}}{2} \), since \( x = e^{\lambda y} \) must be positive.

Step 7: Calculate MAD from x Solution

Set \( e^{\lambda y} = \frac{1 + \sqrt{5}}{2} \). Take the natural logarithm to find \( \lambda y = \ln\left(\frac{1 + \sqrt{5}}{2}\right) \), so \( y = \frac{1}{\lambda} \ln\left(\frac{1 + \sqrt{5}}{2}\right) \). This \( y \) is the MAD.

Key Concepts

Exponential Distribution

An exponential distribution is a probability distribution that describes the time between events in a Poisson process, where events occur continuously and independently. It is often used to model scenarios such as the time until a radioactive particle decays, or the time between arrivals at a service point, such as a bank teller. In a Poisson process, the time between each pair of successive events follows an exponential distribution. The distribution is characterized by the parameter \( \lambda \), which is often called the rate parameter.

• Rate Parameter (\( \lambda \)): Determines how quickly events happen. A higher \( \lambda \) signifies more frequent events.
• Probability Density Function (PDF): The PDF of an exponential distribution is given by \( f(x) = \lambda e^{-\lambda x} \), where \( x \geq 0 \).
• Mean and Variance: The mean is \( 1/\lambda \) and the variance is \( 1/\lambda^2 \).

Understanding the characteristics of exponential distribution helps in modeling and solving problems involving time or space between independent events.

Distribution Function

A distribution function in statistics is a function that gives the probabilities of a given range of outcomes. It describes how the probability is spread over a set of possible values, whether for a discrete set or a continuous range. For continuous random variables, this is often represented by the cumulative distribution function (CDF).

• Purpose: Helps in determining the probability that a random variable will take a value less than or equal to a certain amount.
• PDF and CDF Connection: For continuous variables, the PDF provides the area under the curve at a specific point, and the CDF provides the area up to a certain point.

When working with specific distributions such as Exponential, different functions like the PDF and CDF need to be derived to fully understand the behavior of the distribution. This understanding allows us to derive important conclusions such as the distribution function of \( Y \) in exponential settings.

Cumulative Distribution Function

A cumulative distribution function (CDF) tells you the probability that a random variable takes on a value less than or equal to a particular number. The CDF ranges from 0 to 1 and is a non-decreasing function.

• Mathematical Representation: For a random variable \( X \), the CDF is denoted as \( F(x) = P(X \leq x) \).
• Exponential Cumulative Distribution Function: For exponential distribution, \( F(x) = 1 - e^{-\lambda x} \) when \( x \geq 0 \). This function is crucial in problems where time until an event is important.

In the exercise, we develop the CDF for the transformed variable \( Y = |X - m| \), leading to the combined result \( G(y) = \frac{1}{2}(e^{\lambda y} - e^{-\lambda y}) \) by combining specific conditions under the exponential distribution.

Quadratic Equation

A quadratic equation is a second-order polynomial equation of the form \( ax^2 + bx + c = 0 \). These equations arise frequently in various mathematical contexts and have solutions that can be found using the quadratic formula.

• Quadratic Formula: To solve \( ax^2 + bx + c = 0 \), use \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Discriminant \( (b^2 - 4ac) \): Determines the nature of roots. If the discriminant is positive, there are two real distinct solutions. If zero, there is one real repeated solution. If negative, no real solutions exist, but complex solutions may be present.

In the exercise, we encounter the equation \( e^{2 \lambda y} - e^{\lambda y} - 1 = 0 \), which translates into the quadratic equation \( x^2 - x - 1 = 0 \) when letting \( x = e^{\lambda y} \). Solving this provides critical information about the median absolute deviation for the exponential distribution.