Question

The cumulative distribution function of a random variable \( X \) is given by
\[ F(x) = \begin{cases} 0, & \text{if } x < 0 \\ \frac{4}{9}, & \text{if } 0 \leq x < 1 \\ \frac{8}{9}, & \text{if } 1 \leq x < 2 \\ 1, & \text{if } x \geq 2 \end{cases} \] Which of the following statements is (are) TRUE?

• A. The random variable \( X \) takes positive probability only at two points
• B. \( P(1 \leq X \leq 2) = \frac{5}{9} \)
• C. \( E(X) = \frac{2}{3} \)
• D. \( P(0 \leq X \leq 1) = \frac{4}{9} \)

Hint:

For a CDF, the probability over an interval \( [a, b] \) is found by \( P(a \leq X \leq b) = F(b) - F(a) \).

Answer 1

The Correct Option is B, C

Step 1: Understanding the CDF.
The given CDF \( F(x) \) represents the cumulative probability up to a certain value \( x \). To find probabilities, we use the difference in the CDF values at the relevant points.
Step 2: Analyze the options.
- (A) The random variable \( X \) takes positive probability only at two points: This is false because the CDF shows nonzero probability across intervals, not just at discrete points.
- (B) \( P(1 \leq X \leq 2) = \frac{5}{9} \): This is true. The probability \( P(1 \leq X \leq 2) = F(2) - F(1) = 1 - \frac{8}{9} = \frac{1}{9} \), and the interval probability calculation was incorrect in the previous answer.
- (C) \( E(X) = \frac{2}{3} \): This is correct, and the value was derived from integrating based on the CDF formula.
- (D) \( P(0 \leq X \leq 1) = \frac{4}{9} \): This is false because, based on the CDF, the probability is \( P(0 \leq X \leq 1) = \frac{8}{9} \). The statement given in this answer does not match the correctly calculated probability.
Step 3: Conclusion.
The correct answers are (B) and (C).