91Ó°ÊÓ

For a function to qualify as a probability density function (PDF), it must satisfy the condition of non-negativity. In simple terms, the function cannot take any negative values within its domain.

Let's consider the given function: \[\begin{equation} f(x) = 2(x-1). \rightarrow \text{Domain: } 1 \leq x \leq 2 \end{equation}\]
To check non-negativity:
- For \(x = 1\) : \(f(1) = 2(1-1) = 0\) - For \(x = 2\) : \(f(2) = 2(2-1) = 2\) For any other value within the range 1 < x < 2, f(x) remains positive. Thus, \( \forall x \in [1,2] \), \(2(x-1) \geq 0 \).
This means, \( f(x)\geq 0\) within its domain, satisfying the non-negativity criterion for a PDF.

Another fundamental property that determines if a function is a PDF is that its definite integral over its entire domain must be equal to 1. This is because a probability density function represents the probability distribution of a continuous random variable, and the total probability must always equal 1.

To find the definite integral: \[\begin{equation} \int_{1}^{2} 2(x-1) \, dx. \rightarrow \text{We need to integrate the function } f(x) = 2(x-1) \text{ over the interval } [1, 2] \end{equation}\].

Integration is the process of finding the integral of a function. In our context, we are finding the antiderivative of \( 2(x-1) \), which involves finding a function that, when differentiated, returns \( 2(x-1) \).

The steps are as follows: 1. Find the antiderivative: \[\begin{equation} \int 2(x-1) \, dx \rightarrow 2 \int (x-1) \, dx \rightarrow 2 \left( \frac{x^2}{2} - x \right) \rightarrow x^2 - 2x + C. \end{equation}\]
2. Evaluate the definite integral from 1 to 2:
\[\begin{equation} \left[ x^2 - 2x \right]_{1}^{2} \rightarrow ( (2^2 - 2 \cdot 2) ) - ( (1^2 - 2 \cdot 1) ) \rightarrow (4 - 4) - (1 - 2) \rightarrow 0 - (-1) \rightarrow 1. \text{ This integral equals 1, verifying the criteria. }\end{equation}\].

A function must meet two main criteria to be called a probability density function. Let's summarize these here:

1. **Non-Negativity**: The function must be non-negative within its domain. It means the function cannot return negative values. We verified this criterion for \( f(x) = 2(x-1) \) within the domain [1, 2].

2. **Normalization**: The integral of the function over its entire domain must be equal to 1. This ensures that the total probability represented by the function is 1. Through our integration steps for \( 2(x-1) \), we confirmed that its definite integral equals 1 from 1 to 2.

Only when both criteria are satisfied can we conclude that the function is a valid PDF.