Question:

Let \( X \) be a random variable with cumulative distribution function \[ F(x) = \begin{cases} 0 & \text{if } x < -1, \\ \frac{1}{4}(x + 1) & \text{if } -1 \leq x < 0, \\ \frac{1}{4}(x + 3) & \text{if } 0 \leq x < 1, \\ 1 & \text{if } x \geq 1 \end{cases} \] Which one of the following statements is true?

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- For continuous distributions, the probability at a single point is always zero.
- The CDF can be used to calculate probabilities over intervals.
Updated On: Aug 30, 2025
  • \( \lim_{n \to \infty} P\left( -\frac{1}{2} + \frac{1}{n}<X<-1 - \frac{1}{n} \right) = \frac{5}{8} \)
  • \( \lim_{n \to \infty} P\left( -\frac{1}{2} - \frac{1}{n}<X<1 - \frac{1}{n} \right) = \frac{5}{8} \)
  • \( \lim_{n \to \infty} P\left( X = -\frac{1}{n} \right) = \frac{1}{2} \)
  • \( P(X = 0) = \frac{1}{3} \)
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The Correct Option is B

Solution and Explanation

1) Understanding the cumulative distribution function (CDF):
The given CDF describes a piecewise function for \( X \). We will use this function to calculate the probabilities for different intervals.
2) Analysis of the options:
(A) Incorrect: This option asks for the probability \( P\left( -\frac{1}{2} + \frac{1}{n}<X<-1 - \frac{1}{n} \right) \). As \( n \to \infty \), the interval becomes a small neighborhood around \( -1 \), but the probability is not equal to \( \frac{5}{8} \).
(B) Correct: This option asks for the probability \( P\left( -\frac{1}{2} - \frac{1}{n}<X<1 - \frac{1}{n} \right) \). As \( n \to \infty \), the interval approaches the entire range \( [-\frac{1}{2}, 1] \), for which the CDF gives a probability of \( \frac{5}{8} \).
(C) Incorrect: This option concerns \( P(X = -\frac{1}{n}) \), but \( P(X = x) \) for a continuous random variable is always zero.
(D) Incorrect: The probability \( P(X = 0) \) is not \( \frac{1}{3} \) because the CDF at a single point does not give a nonzero value for continuous random variables.
The correct answer is (B).
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