Question

Let \( \Phi(.) \) denote the cumulative distribution function of a standard normal random variable. If the random variable \( X \) has the cumulative distribution function \[ F(x) = \begin{cases} \Phi(x) & \text{if } x < -1, \\ \Phi(x + 1) & \text{if } x \geq -1, \end{cases} \] then which one of the following statements is true?

• A. \( P(X \leq -1) = \frac{1}{2} \)
• B. \( P(X = -1) = \frac{1}{2} \)
• C. \( P(X<-1) = \frac{1}{2} \)
• D. \( P(X \leq 0) = \frac{1}{2} \)

Hint:

- The CDF \( F(x) \) is defined in pieces, and at \( x = -1 \), it is computed using the function \( \Phi(x+1) \).
- For a continuous random variable, \( P(X = x) = 0 \) for any point \( x \).
- The value of \( \Phi(0) \) is always \( 0.5 \), which helps in evaluating probabilities for standard normal variables.

Answer 1

The Correct Option is A

1) Understanding the CDF:
The function \( F(x) \) is a piecewise function defined as \( \Phi(x) \) for \( x < -1 \) and \( \Phi(x+1) \) for \( x \geq -1 \). The value of the CDF at \( x = -1 \) is computed from the second piece of the function, i.e., \( F(-1) = \Phi(-1 + 1) = \Phi(0) \). Since \( \Phi(0) = 0.5 \), we can immediately say that \( P(X \leq -1) = 0.5 \).
2) Analysis of the options:
(A) \( P(X \leq -1) = \frac{1}{2} \): This is true because, as mentioned earlier, \( F(-1) = \Phi(0) = 0.5 \). Hence, \( P(X \leq -1) = 0.5 \).
(B) \( P(X = -1) = \frac{1}{2} \): This is incorrect because for continuous random variables, \( P(X = x) = 0 \) for any point \( x \). The probability at a single point is always zero.
(C) \( P(X < -1) = \frac{1}{2} \): This is incorrect because \( P(X < -1) = F(-1^-) = \Phi(-1) \), which is not equal to \( 0.5 \). \( \Phi(-1) \) is approximately \( 0.1587 \).
(D) \( P(X \leq 0) = \frac{1}{2} \): This is incorrect because \( P(X \leq 0) = F(0) = \Phi(1) \), which is approximately \( 0.8413 \), not \( 0.5 \).
The correct answer is (A) \( P(X \leq -1) = \frac{1}{2} \).