Question

Let \( X \) be a positive valued continuous random variable with finite mean. If \( Y = \lfloor X \rfloor \), the largest integer less than or equal to \( X \), then which of the following statements is/are true?

• A. \( P(Y \leq u) \leq P(X \leq u) \) for all \( u \geq 0 \)
• B. \( P(Y \geq u) \leq P(X \geq u) \) for all \( u \geq 0 \)
• C. \( E(Y)<E(X) \)
• D. \( E(X)>E(Y) \)

Hint:

- For continuous random variables, truncating the variable (e.g., using the floor function) generally results in a lower expected value for the truncated variable.
- The cumulative distribution function for the truncated variable is also generally smaller than for the original continuous variable.

Answer 1

The Correct Option is B

For the problem, we know that \( Y = \lfloor X \rfloor \), the greatest integer less than or equal to \( X \). This means that \( Y \) is always less than or equal to \( X \), which affects the cumulative distribution function and the expected values.
(A) \( P(Y \leq u) \leq P(X \leq u) \): This is not true because \( Y \) is the greatest integer less than or equal to \( X \), so the probability \( P(Y \leq u) \) is generally less than or equal to \( P(X \leq u) \).
(B) \( P(Y \geq u) \leq P(X \geq u) \): This is true because the integer part of \( X \) is less than or equal to \( X \), so the probability that \( Y \geq u \) is always less than or equal to the probability that \( X \geq u \).
(C) \( E(Y)<E(X) \): This is true because the expected value of \( Y \), being the greatest integer less than or equal to \( X \), is less than the expected value of \( X \).
(D) \( E(X)>E(Y) \): This is true because \( E(X) \) is always greater than \( E(Y) \) due to the truncation of the continuous variable \( X \).
Thus, the correct answer is (B) and (D).