Problem 21 - Calculus and its applications

Question

Find \(k\) such that each function is a probability density function over the given interval. Then write the probability density function. $$ f(x)=\frac{k}{x}, \quad[1,2] $$

Step-by-Step Solution

Answer

The PDF is \(f(x) = \frac{1}{x \cdot \ln(2)}\) for \([1, 2]\).

Step 1: Understand the Problem

To make the function a probability density function (PDF) over the interval \([1,2]\), the integral of the function over this interval must equal 1. We have \(f(x) = \frac{k}{x}\), so we need to find \(k\) such that \(\int_{1}^{2} \frac{k}{x} \, dx = 1\).

Step 2: Solve the Integral

Integrate \(\frac{k}{x}\) over the interval \([1, 2]\). This gives \(\int_{1}^{2} \frac{k}{x} \, dx = k \cdot \left[ \ln|x| \right]_{1}^{2}\). Evaluate this as \( k \cdot (\ln(2) - \ln(1)) \). Since \(\ln(1) = 0\), this simplifies to \(k \cdot \ln(2)\).

Step 3: Set the Integral Equal to 1

To satisfy the PDF condition, set the result of the integral equal to 1: \(k \cdot \ln(2) = 1\).

Step 4: Solve for k

Solve the equation \(k \cdot \ln(2) = 1\) for \(k\). Divide both sides by \(\ln(2)\) to obtain \(k = \frac{1}{\ln(2)}\).

Step 5: Write the Probability Density Function

Substitute the found value of \(k\) back into the function to define the PDF as \(f(x) = \frac{1}{x \cdot \ln(2)}\) over the interval \([1, 2]\).

Key Concepts

Understanding Integration in Probability

When dealing with probability density functions (PDFs), integration is key. A PDF, like the one we're discussing, describes a continuous probability distribution. To determine if a function is a PDF over a specific interval, we must ensure that the total area under the curve equals 1 within that interval. This is where integration comes into play, as it allows us to calculate the area under the curve. For the function given,

• We need to integrate \( \frac{k}{x} \) over the interval \( [1, 2] \)
• This integral calculates the weighted area between the curve and the x-axis over the specified interval.

Evaluating the integral ensures that the function correctly models a probability distribution over the given domain. The integral of \( \frac{k}{x} \) results in \( k \cdot \left[ \ln(x) \right] \), showcasing the beauty of linking integration with logarithmic functions.

Exploring the Natural Logarithm

The natural logarithm, represented as \( \ln \), is a mathematical function that provides insights into exponential growth and functions. In the context of our problem, the natural logarithm arises from the integral of \( \frac{1}{x} \). Why is this important?

• The integral of \( \frac{1}{x} \) is \( \ln|x| \), indicating that the natural logarithm plays a critical role in solving such integrals.
• Natural logs help convert multiplicative relationships into additive ones, simplifying calculations.

In our example, by evaluating \( \ln(x) \) from 1 to 2, we find \( \ln(2) - \ln(1) \). Knowing that \( \ln(1) = 0 \), this simplifies to \( \ln(2) \), which is crucial for defining the PDF correctly. This illustrates the usefulness of natural logarithms in mathematical problem-solving.

Introducing Continuous Probability

Continuous probability deals with outcomes that can take any value within a specified range. Unlike discrete probability, where outcomes are distinct and countable, continuous probability involves distributions over intervals. A probability density function (PDF) models this continuous nature when finding probabilities over a range. In our scenario,

• The PDF, \( f(x) = \frac{k}{x} \), represents a continuous distribution over the interval \( [1, 2] \).
• To qualify as a PDF, the integral over this interval must equate to 1.

This condition ensures that the total probability covered by the interval accounts for all possible outcomes in that range. Understanding this is essential for correctly interpreting continuous variables and their resulting probabilities. When we solve for \( k = \frac{1}{\ln(2)} \), we ensure that the function satisfies all requirements of a PDF, effectively capturing continuous probabilities over our selected interval.