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Let's break down the problem step by step.

### Problem Overview:

You are given a random variable \( X \) with a distribution function \( F(y) \) defined as follows:

\[
F(y) =
\begin{cases}
0, & y < -1, \\
\frac{y+1}{2}, & -1 \leq y \leq 1, \\
1, & y > 1.
\end{cases}
\]

You are asked to:
1. Verify whether \( X \) is discrete, continuous, or neither.
2. Compute the probability \( P(X = 0) \).

### Step 1: Verifying the Type of Random Variable

To determine whether the random variable \( X \) is discrete, continuous, or neither, we can examine the properties of its cumulative distribution function (CDF), \( F(y) \).

- **Continuous Random Variable**: The CDF should be continuous for all values of \( y \), and its derivative (which would be the probability density function, PDF) should exist almost everywhere.
- **Discrete Random Variable**: The CDF would have jump discontinuities at specific points, corresponding to the probability mass at those points.

Looking at the CDF \( F(y) \):

- For \( y < -1 \), \( F(y) = 0 \), which indicates no probability mass below -1.
- For \( -1 \leq y \leq 1 \), the CDF is a linear function: \( F(y) = \frac{y+1}{2} \), which is continuous and smooth.
- For \( y > 1 \), \( F(y) = 1 \), which implies all the probability is concentrated between -1 and 1.

The CDF is continuous everywhere, and there are no jumps or discontinuities. This implies that \( X \) is a **continuous random variable**.

### Step 2: Computing \( P(X = 0) \)

To find \( P(X = 0) \), we need to check the probability density function (PDF), which is the derivative of the CDF \( F(y) \) with respect to \( y \), for \( -1 \leq y \leq 1 \).

The PDF \( f(y) \) is:

\[
f(y) = \frac{d}{dy} \left( \frac{y+1}{2} \right) = \frac{1}{2}, \quad \text{for } -1 \leq y \leq 1.
\]

This indicates that the random variable has a uniform distribution over the interval \([-1, 1]\).

Since \( X \) is continuous, the probability that \( X \) takes any exact value (such as 0) is 0. Therefore:

\[
P(X = 0) = 0.
\]

---

Would you like more details, or do you have any questions?

Here are 5 follow-up questions to expand on this topic:
1. How can we compute the probability \( P(a \leq X \leq b) \) for different values of \( a \) and \( b \)?
2. Can you explain how the concept of the CDF and PDF relate to each other?
3. How would this problem change if the CDF had discontinuities at certain points?
4. What other types of distributions might this CDF resemble, and why?
5. How can we use integration to compute probabilities in continuous distributions?

**Tip:** For continuous random variables, the probability of taking an exact value is always 0. Probabilities are computed over intervals instead of exact points.

Define a random variable on a probability space. Let X be a random variable with distribution function F defined as F(y) = 0, y < -1, F(y) = (y+1)/2, -1 ≤ y ≤ 1, and F(y) = 1, y > 1. Verify whether X is discrete or continuous or neither discrete nor continuous? Compute the probability P(X = 0).

Solve a probability problem involving a random variable with a defined CDF. Learn how to determine whether the variable is continuous or discrete, and compute the probability P(X = 0). Suitable for college-level probability theory students.

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