FRM Part I · FRM Exam Part I · Random Variables
A discrete random variable X takes values 1, 2, 3 and 4 with probabilities 0.10, 0.30, k and 0.20 respectively. What is the value of the probability mass function at X = 3, and what is E[X]?
The PMF must sum to one, so f(3) = 0.40. The expected value is the probability-weighted sum of outcomes: 0.10 + 0.60 + 1.20 + 0.80 = 2.70. A simple average of the values would wrongly ignore the probabilities.
- Af(3) = 0.40; E[X] = 2.70Correct
- Bf(3) = 0.40; E[X] = 2.50
- Cf(3) = 0.60; E[X] = 2.70
- Df(3) = 0.30; E[X] = 2.60
Explanation
Probabilities must sum to 1, so k = 1 - 0.10 - 0.30 - 0.20 = 0.40. E[X] = 1(0.10) + 2(0.30) + 3(0.40) + 4(0.20) = 0.10 + 0.60 + 1.20 + 0.80 = 2.70. Using 2.50 would come from the simple average of the values, ignoring the probabilities.
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