Problem 23 - Calculus Concepts

Question

Indicate whether the function could be a probability density function. Explain. \(p(t)=\left\\{\begin{array}{ll}\frac{\ln 3}{t} & \text { when } 1 \leq t \leq 3 \\ 0 & \text { elsewhere }\end{array}\right.\)

Step-by-Step Solution

Answer

The function cannot be a PDF because its integral over the defined range does not equal 1.

Step 1: Understanding the Conditions

To determine if the given function is a probability density function (PDF), it must satisfy two conditions: (1) the function should be non-negative for all values of the variable, and (2) the integral over its range should equal 1.

Step 2: Check Non-negativity

The given function, \( p(t) = \frac{\ln 3}{t} \), is defined for \( 1 \leq t \leq 3 \). Since \( t \) is positive within this range, and since \( \ln 3 \) is a constant positive value, \( \frac{\ln 3}{t} \) remains positive for all \( t \) between 1 and 3. Thus, the function is non-negative over the interval.

Step 3: Compute the Integral

We need to verify that the integral of the function from 1 to 3 equals 1:\[\int_{1}^{3} \frac{\ln 3}{t} \, dt.\]This integral can be computed as:\[\ln 3 \cdot \left[ \ln |t| \right]_{1}^{3} = \ln 3 \cdot (\ln 3 - \ln 1).\]Since \( \ln 1 = 0 \), this simplifies to \( \ln 3 \cdot \ln 3 = (\ln 3)^2 \).

Step 4: Check Normalization Condition

The result \((\ln 3)^2\) must equal 1 in order for the function to be a PDF. However, since \( (\ln 3)^2 eq 1 \) (approximate value \( (\ln 3)^2 \approx 1.216 \)), the integral of the function over its range does not equal 1. Therefore, the function is not normalized.

Key Concepts

Non-Negativity of Probability Density Functions

A probability density function (PDF) must meet a fundamental requirement: being non-negative over its entire domain. A non-negative function ensures that probabilities, which are essentially areas under the curve of the function, are not negative. This is because a negative probability does not make sense in real-life scenarios.
For the considered function \( p(t) = \frac{\ln 3}{t} \), its domain is defined from \( t = 1 \) to \( t = 3 \). Within this defined domain, both the natural logarithm of 3 (\( \ln 3 \)) and the variable \( t \) remain positive. As a result, the fraction \( \frac{\ln 3}{t} \) is always positive for all values of \( t \) between 1 and 3. Thus, the function meets the non-negativity condition within its domain.

• If \( t \) were negative or zero, the function could potentially fall short of this requirement, but within the interval \( [1, 3] \), \( t \) remains positive.
• This effectively means that the probability aspect is maintained, as no piece of the function curve dips below the horizontal axis in this interval.

Recognizing non-negativity is a simple yet crucial step in the validation of any probability-based function.

Normalization Condition of Probability Density Functions

A critical aspect of any probability density function is the normalization condition, which states that the total area under the curve of the function must equal 1 over its given range. This signifies that the total probability across all possible outcomes sums up to 1, indicating certainty that one of these outcomes will occur.
For our function \( p(t) = \frac{\ln 3}{t} \), the range is from \( t = 1 \) to \( t = 3 \). To check if it satisfies the normalization condition, we need to calculate the integral of this function over the specified interval.

• Imagine this integral as the total area trapped between the function curve and the \( t \)-axis from 1 to 3.
• If the computed area equates to precisely 1, the function is perfectly normalized, implying that the function accurately represents a complete probability distribution.

For this function, however, the integral evaluation revealed that \((\ln 3)^2 \), which approximates to 1.216, does not equal 1. Hence, despite any non-negativity, the function fails the normalization test. The integral exceeding 1 indicates that the function mis-distributes the probability it seeks to represent.

Integral Calculation for Probability Density Functions

Integral calculation is pivotal in probability theory, particularly when verifying whether a function is eligible to be a PDF. We need to compute the integral of the function over its applicable domain to check for proper normalization, meaning the integral should result in a value of 1.
For the function \( p(t) = \frac{\ln 3}{t} \), we evaluated:\[\int_{1}^{3} \frac{\ln 3}{t} \, dt.\]Performing the integral computation involves understanding how the function behaves -- essentially, looking at the area under the curve from \( t = 1 \) to \( t = 3 \). The steps include:

• The ln constant factor allows us to take it out of the integral. Hence, we compute \( \ln 3 \cdot \int_{1}^{3} \frac{1}{t} \, dt \).
• The integral of \( \frac{1}{t} \) is \( \ln |t| \), which once evaluated from 1 to 3 results in \( \ln 3 \).
• Thus, the calculation simplifies to \( \ln 3 \cdot (\ln 3 - \ln 1) = (\ln 3)^2 \).

Unfortunately, since \((\ln 3)^2\) does not equal 1, the integral fails to confirm the normalization condition, indicating a misalignment with probability standards.Exploring integral calculations like this helps in ascertaining the reliability and suitability of potential PDFs.