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Test whether         can be the probability density function of a continuous random variable?

To determine whether the given function can be a probability density function (PDF) of a continuous random variable, we need to verify two conditions:

1. **Non-negativity**: \( f(x) \geq 0 \) for all \( x \) in the domain.
2. **Normalization**: The total area under the curve of \( f(x) \) over the entire real line must equal 1. Mathematically, this is expressed as:
   \[
   \int_{-\infty}^{\infty} f(x) \, dx = 1
   \]

Given the function \( f(x) \) in the image:

\[
f(x) = \begin{cases} 
x, & -1 \leq x \leq 1 \\
0, & \text{otherwise}
\end{cases}
\]

### Step 1: Check Non-Negativity
The function \( f(x) \) is defined as \( x \) within the interval \([-1, 1]\). In this interval:
- \( f(x) \) is non-negative because \( -1 \leq x \leq 1 \), and the value of \( x \) is non-negative for \( 0 \leq x \leq 1 \).

Outside this interval, \( f(x) = 0 \), which is also non-negative. Thus, the non-negativity condition is satisfied.

### Step 2: Check Normalization
We need to calculate the integral of \( f(x) \) over the interval \([-1, 1]\):

\[
\int_{-\infty}^{\infty} f(x) \, dx = \int_{-1}^{1} x \, dx
\]

The integral of \( x \) with respect to \( x \) is:

\[
\int_{-1}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{-1}^{1} = \frac{(1)^2}{2} - \frac{(-1)^2}{2} = \frac{1}{2} - \frac{1}{2} = 0
\]

Since the integral equals 0, this function does **not** satisfy the normalization condition required for a PDF (which must integrate to 1 over its domain).

### Conclusion
The given function **cannot** be the probability density function of a continuous random variable because it does not satisfy the normalization condition.

Do you want more details, or have any questions?

Here are some related questions:

1. How do you modify a function to make it a valid probability density function?
2. What are some examples of valid probability density functions?
3. How is the concept of a cumulative distribution function (CDF) related to the PDF?
4. What does it mean when the integral of a function is zero?
5. Can a function with a negative integral be a probability density function?

**Tip:** For any function to qualify as a probability density function, always check that its integral over the entire space equals 1, indicating that it covers the whole probability space.

Explore whether a given function can serve as a probability density function (PDF) of a continuous random variable. Understand the conditions required, such as non-negativity and normalization, through a detailed mathematical analysis. Learn how to verify these conditions and why they are crucial in probability theory.

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