FRM Part I · FRM Exam Part I · Random Variables
A discrete random variable X takes the values 2, 4 and 8 with probabilities 0.50, 0.25 and 0.25 respectively. What is the variance of X?
Placeholder
- A4.50Correct
- B5.00
- C3.75
- D25.00
Explanation
E[X] = 1 + 1 + 2 = 4. E[X^2] = 0.5*4 + 0.25*16 + 0.25*64 = 2 + 4 + 16 = 22. Variance = 22 - 16 = 6. Recheck: deviations -2, 0, 4 give squares 4, 0, 16; weighted: 0.5*4 + 0 + 0.25*16 = 2 + 4 = 6. So the correct value is 6, which is not listed.
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