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The problem from the image can be interpreted as follows:

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**Problem:**
Define a random variable \( X \) on a probability space. Let \( X \) be a random variable with the distribution function \( F \) defined as:

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

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

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### Solution Outline:

#### 1. **Type of the Random Variable \( X \):**
   - **Discrete**: A discrete random variable takes specific distinct values.
   - **Continuous**: A continuous random variable takes values in a continuous range, and its cumulative distribution function (CDF) \( F(y) \) is continuous.
   
   From the given distribution function \( F(y) \):
   - \( F(y) \) is a piecewise function. Specifically:
     - It is constant (0) for \( y < -1 \),
     - It increases linearly in the range \( -1 \leq y < 1 \) as \( F(y) = \frac{y + 1}{2} \),
     - It becomes constant (1) again for \( y \geq 1 \).
     
   Since \( F(y) \) is continuous and non-decreasing, and there are no jumps, this suggests that \( X \) is a **continuous random variable**.

#### 2. **Computing \( P(X = 0) \):**
   - To find \( P(X = 0) \), we use the fact that for continuous random variables, the probability at a single point is typically 0. However, let's verify it formally using the CDF.
   - The probability \( P(X = 0) \) is computed as:
     \[
     P(X = 0) = F(0) - F(0^{-}),
     \]
     where \( F(0^{-}) \) refers to the value of the CDF just before 0.
   - From the given function, for \( -1 \leq y < 1 \), we have \( F(y) = \frac{y + 1}{2} \), so:
     \[
     F(0) = \frac{0 + 1}{2} = \frac{1}{2}.
     \]
     Since the function is continuous at 0 (there is no jump), we have:
     \[
     F(0^{-}) = F(0) = \frac{1}{2}.
     \]
     Therefore:
     \[
     P(X = 0) = F(0) - F(0^{-}) = \frac{1}{2} - \frac{1}{2} = 0.
     \]
   
   Thus, \( P(X = 0) = 0 \).

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Would you like further details or have any questions? Here are some related questions you could explore:

1. What is the probability density function (PDF) corresponding to this CDF?
2. How can the CDF \( F(y) \) be used to compute \( P(X \leq a) \) for some value \( a \)?
3. What is the mean of this random variable \( X \)?
4. How would you verify if \( X \) has any discontinuities?
5. How does this distribution compare to a uniform distribution on the interval \([-1, 1]\)?

**Tip:** Remember, for continuous random variables, the probability of taking any exact value is typically 0, since probabilities are defined over intervals.

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 for -1 <= y < 1, and F(y) = 1 for y > 1. Verify whether X is discrete or continuous or neither. Compute the probability P(X = 0).

This problem defines a random variable with a piecewise cumulative distribution function and asks to determine whether the variable is discrete, continuous, or neither. It also computes the probability P(X = 0). Suitable for undergraduate students studying probability.

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