FRM Part I · FRM Exam Part I · Random Variables
A discrete random variable X takes the values 0, 1, 2 and 3 with probabilities 0.10, 0.30, k and 0.20 respectively. What is the expected value of X?
The expected value is 1.70. The missing probability must be 0.40 so the mass function sums to one. Weighting each outcome by its probability gives 0.30 + 0.80 + 0.60 = 1.70.
- A1.50
- B1.70Correct
- C1.90
- D1.60
Explanation
Probabilities must sum to 1, so k = 1 - 0.10 - 0.30 - 0.20 = 0.40. E[X] = 0(0.10) + 1(0.30) + 2(0.40) + 3(0.20) = 0.30 + 0.80 + 0.60 = 1.70. Using k = 0.30 by mistake would give 1.50, and the other values do not follow from the data.
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