Math Problem Statement

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).

Solution

The problem from the image can be interpreted as follows:

Problem: Define a random variable XX on a probability space. Let XX be a random variable with the distribution function FF defined as:

\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)$$. --- ### Solution Outline: #### 1. **Type of the Random Variable $$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$$. --- Would you like further details or have any questions? Here are some related questions you could explore: 1. What is the probability density function (PDF) corresponding to this CDF? 2. How can the CDF $$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.

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Math Problem Analysis

Mathematical Concepts

Random Variables
Probability Distribution Functions
Continuous and Discrete Variables

Formulas

F(y) = 0, y < -1
F(y) = (y + 1)/2 for -1 <= y < 1
F(y) = 1 for y > 1

Theorems

Definition of Cumulative Distribution Function (CDF)
Definition of Continuous Random Variables

Suitable Grade Level

Undergraduate level (Probability and Statistics)

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