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For a function to qualify as a probability density function (PDF), it must first adhere to the non-negativity criterion. This means that the function should never take on negative values throughout its entire domain. Essentially, no part of the graph of the function should dip below the x-axis within the specified range.

Taking the function from our exercise, which is defined as \[ f(x) = \frac{1}{18}x \text{ for } x \text{ over }[4,8]\],we need to check its behavior on this interval. The function's formula, \(\frac{1}{18}x\), is a simple linear equation with a positive slope. Since both \(\frac{1}{18}\) and \(x\) in the interval \([4, 8]\) are positive, this guarantees the entire function remains above the x-axis.

In summary:

• The non-negativity criterion ensures PDF values are always zero or greater.
• Our example function, \(f(x)\), remains non-negative throughout the given interval.

One of the crucial steps in determining if a function is a valid probability density function is calculating the integral of the function over its entire domain. This integral must equal 1 for the function to be considered a PDF. Essentially, this integral represents the total "probability mass" over the range, ensuring it fully covers the possibility space.

In our example, let's calculate the integral: \[\int_{4}^{8} \frac{1}{18}x \, dx\]The antiderivative or integral of \(\frac{1}{18}x\) is \(\frac{1}{36}x^2\). Evaluating this antiderivative from 4 to 8 gives:\[\left. \frac{1}{36}x^2 \right|_4^8 = \frac{1}{36}(8^2) - \frac{1}{36}(4^2) = \frac{1}{36} \times (64 - 16) = \frac{1}{36} \times 48 = \frac{4}{3}\]Unfortunately, \(\frac{4}{3}\) is not 1. This means the total probability "content" of \(f(x)\) on the interval \([4,8]\) exceeds the standard requirement, indicating a failure of this crucial PDF condition. This demonstrates that integrating a function can quickly verify if the total probability is valid.

To determine if a function can serve as a probability density function, two primary criteria must be satisfied: non-negativity and normalization. The function should be non-negative across its entire domain and integrate to 1 over that domain.

Let's consider our function from the exercise.
1. **Non-negativity Check**: This function, \(f(x) = \frac{1}{18}x\), maintains positivity over the interval \([4,8]\). This satisfies the first condition of the PDF criteria.

2. **Normalization Check**: This involves calculating the integral of the function over its specified domain. As we found, the integral equaled \(\frac{4}{3}\), not 1. This is a critical shortcoming, as PDFs must be normalized to represent total probability, which is exactly 1.
To summarize:

• Even though \(f(x)\) meets non-negativity, it fails the normalization condition.
• Both conditions are necessary for a valid probability density function.
• Therefore, \(f(x)\) cannot be classified as a PDF based on this assessment.