FRM Part I · FRM Exam Part I · Random Variables
A discrete random variable has the cumulative distribution function F(0) = 0.15, F(1) = 0.40, F(2) = 0.75, F(3) = 1.00, with support only at 0, 1, 2 and 3. What is P(1 ≤ X ≤ 2)?
P(1 ≤ X ≤ 2) is 0.60. The probabilities at 1 and 2 are 0.25 and 0.35 from differences in the cumulative distribution function, and together they equal 0.75 minus 0.15.
- A0.35
- B0.60Correct
- C0.50
- D0.25
Explanation
The mass at 1 is F(1) - F(0) = 0.25 and at 2 is F(2) - F(1) = 0.35. Their sum is 0.60, which equals F(2) - F(0) = 0.75 - 0.15. Using F(2) - F(1) = 0.35 alone omits the mass at 1.
Did you get it right without looking?
One question tells you little. A timed set on Random Variables shows your real accuracy, how long you take and where you lose marks.
More Random Variables questions
- A continuous random variable X has CDF F(x) = 1 − e^(−x/10) for x ≥ 0. What is the 90th percentile of X, closest to?
- A risk manager models a stock's price as lognormal, where the continuously compounded annual return is normal with mean 6% and standard devi…
- A portfolio's annual return R has mean 8% and standard deviation 12%. A risk manager defines a transformed variable Y = 3 - 2R, where R is i…
- A portfolio's returns have a variance of 9 (in squared percentage points) and a third central moment of -54. What is the skewness of the ret…
- For a random variable X, E[X] = 2, E[X^2] = 8 and E[X^3] = 44. What is the skewness of X?
- Daily P&L of a trading desk is normally distributed with mean USD 0 and standard deviation USD 2 million. Using z-values of 1.645 for the 95…