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

Question

Compute the median of an \(\operatorname{Exp}(\lambda)\) distribution.

Step-by-Step Solution

Answer

The median of an \( \operatorname{Exp}(\lambda) \) distribution is \( \frac{\ln(2)}{\lambda} \).

Step 1: Understand the Exponential Distribution

An exponential distribution with rate parameter \( \lambda \) has a probability density function (pdf) given by \( f(x; \lambda) = \lambda e^{-\lambda x} \) for \( x \geq 0 \). We are interested in finding its median, which is the value \( m \) such that \( P(X \leq m) = 0.5 \).

Step 2: Set up the Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) of an exponential random variable is \( F(x) = 1 - e^{-\lambda x} \). We need to find \( m \) such that \( F(m) = 0.5 \).

Step 3: Solve for the Median

Set the CDF equal to 0.5: \( 1 - e^{-\lambda m} = 0.5 \). This leads to the equation \( e^{-\lambda m} = 0.5 \).

Step 4: Isolate the Exponent

Solve for \( m \) by isolating the exponent: take the natural logarithm of both sides to get \( -\lambda m = \ln(0.5) \).

Step 5: Solve for m

Divide both sides by \(-\lambda\) to solve for \( m \): \( m = -\frac{\ln(0.5)}{\lambda} \).

Step 6: Simplify the Expression

Recall that \( \ln(0.5) = -\ln(2) \). Therefore, \( m = \frac{\ln(2)}{\lambda} \).

Key Concepts

Probability Density Function

In the realm of probability, the Probability Density Function (PDF) is a fundamental concept, especially when dealing with continuous distributions. For an exponential distribution, which is commonly used to model the time until an event occurs, the PDF provides the likelihood of the time being a specific value. The formula for the exponential distribution's PDF is given by:\[ f(x; \lambda) = \lambda e^{-\lambda x} \]where

• \( x \geq 0 \) denotes non-negative values since time can't be negative,
• \( \lambda \) is the rate parameter,
• \( e \) is the base of natural logarithms, approximately equal to 2.718.

This formula helps us understand how the probability of the occurrence of an event varies over time. The function decreases exponentially, which means the likelihood of waiting a long time for an event to happen gets smaller as more time passes. Such a PDF is crucial for grasping how events like radioactive decay or the time between bus arrivals could be statistically analyzed.

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) is a useful tool to describe the probability that a random variable takes on a value less than or equal to a particular number. In the context of the exponential distribution, the CDF is expressed as:\[ F(x) = 1 - e^{-\lambda x} \]This formula is derived from integrating the PDF over the interval from zero to \( x \), integrating the area under the curve.
Here's what it tells us in detail:

• At \( x = 0 \), \( F(0) = 0 \), showing zero probability of a time less than zero.
• As \( x \to \infty \), \( F(x) \to 1 \), indicating near certainty that the event occurs at some point over time.
• Steepness and shape depend on \( \lambda \); larger \( \lambda \) means events happen more frequently."

By using a CDF, one can answer questions about the distribution of probabilities over a range. In practice, it helps to determine probabilities such as waiting time for the next phone call at a service desk or the duration before a machine part needs replacement.

Median Calculation

The median in probability theory is the value at which the cumulative probability reaches 50%. For our work with exponential distributions, finding the median involves solving for \( m \) in the CDF equation:\[ 1 - e^{-\lambda m} = 0.5 \]This equation denotes that the event has a 50% probability of occurring within time \( m \).
Steps to find the median:

• Set the CDF equal to 0.5 to reflect the definition of the median.
• Solve for \( m \) by taking the natural logarithm of both sides: \( -\lambda m = \ln(0.5) \).
• Rearrange to find \( m \): \( m = -\frac{\ln(0.5)}{\lambda} \).

Finally, the solution simplifies to \( m = \frac{\ln(2)}{\lambda} \). Having the median gives a central point of the distribution, indicating half of the events occur before and half occur after this time. It is particularly valuable when the mean is influenced by a skew in data.