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The problem states that \( X \) and \( Y \) are independent random variables uniformly distributed on the interval \((-1, 1)\), and it asks to prove that \( Z = X + Y \) is an absolutely continuous random variable with the given probability density function (PDF):

\[
f_Z(z) = \frac{1}{4}(2 - |z|) \cdot \mathbf{1}_{[-2, 2]}(z)
\]

### Step-by-Step Approach:

1. **Sum of Independent Uniform Random Variables**:
   We are tasked with finding the PDF of the sum \( Z = X + Y \), where \( X \) and \( Y \) are independent and uniformly distributed over \((-1, 1)\). The PDF of each random variable, \( X \) and \( Y \), is:

   \[
   f_X(x) = f_Y(y) = \mathbf{1}_{[-1, 1]}(x)
   \]
   This indicates that the random variables \( X \) and \( Y \) are uniformly distributed, with their values ranging from \(-1\) to \( 1 \).

2. **Convolution of PDFs**:
   Since \( Z = X + Y \) is the sum of two independent random variables, the PDF of \( Z \) can be found using the convolution of the PDFs of \( X \) and \( Y \):

   \[
   f_Z(z) = (f_X * f_Y)(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z - x) dx
   \]

   Given that \( f_X(x) = \mathbf{1}_{[-1, 1]}(x) \) and \( f_Y(y) = \mathbf{1}_{[-1, 1]}(y) \), the convolution integral simplifies to:

   \[
   f_Z(z) = \int_{-1}^{1} \mathbf{1}_{[-1, 1]}(z - x) dx
   \]

   This integral calculates the area of overlap between the intervals where the indicator functions are non-zero.

3. **Support of \( Z \)**:
   The sum of two random variables \( X \) and \( Y \) uniformly distributed on \((-1, 1)\) will have a support \( Z \in [-2, 2] \). Therefore, the PDF of \( Z \) will be non-zero only on this interval.

4. **Computing the PDF**:
   The convolution results in the following piecewise function for \( f_Z(z) \):

   \[
   f_Z(z) = 
   \begin{cases} 
   \frac{1}{4}(2 - |z|) & \text{for } z \in [-2, 2] \\
   0 & \text{otherwise}
   \end{cases}
   \]

   This matches the given PDF \( f_Z(z) = \frac{1}{4}(2 - |z|) \cdot \mathbf{1}_{[-2, 2]}(z) \).

Thus, \( Z = X + Y \) is an absolutely continuous random variable with the specified PDF.

---

Would you like more details on any step, or have questions on the derivation?

Here are some related questions for further exploration:
1. How does the convolution of two independent random variables' PDFs work in general?
2. How would the solution change if \( X \) and \( Y \) had different distributions (not uniform)?
3. Can we compute the cumulative distribution function (CDF) for \( Z \)?
4. How would you simulate the distribution of \( Z \) using a programming language like Python?
5. What is the interpretation of the piecewise function in the context of probability distributions?

**Tip:** When performing convolution of PDFs, always check the overlap between the ranges of the two random variables. This will help in setting the limits of integration.

Siano X e Y variabili aleatorie indipendenti, uniformemente distribuite nell'intervallo (-1, 1). Si mostri che Z = X + Y è una variabile aleatoria assolutamente continua, con densità f_Z(z) = 1/4 (2 - |z|) 1_{[-2, 2]}(z).

Prove that Z = X + Y, where X and Y are independent random variables uniformly distributed on (-1, 1), is absolutely continuous with the given PDF. Explore convolution of PDFs, probability theory concepts, and uniform distributions.

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