IAI Actuarial Core Principles · Actuarial Statistics · Generating functions
X has MGF M(t) = exp(2t + 3t²). Let Y = 3X − 1. What is P(Y > 5) approximately, given Φ(0.5) = 0.6915 and Φ(1.0) = 0.8413?
The MGF identifies X as normal with mean 2 and variance 6. Y = 3X − 1 is normal with mean 5, so P(Y > 5) equals 0.5 by symmetry about the mean, regardless of the variance.
- A0.3085
- B0.1587Correct
- C0.6915
- D0.8413
- 0.5000
Explanation
X ~ N(2, 6) since the exponent is μt + σ²t²/2, so σ² = 6. Y has mean 3×2−1 = 5 and variance 9×6 = 54, so P(Y>5) = 0.5. Check: the mean of Y is exactly 5, so the probability is 0.5; hence the correct option is 0.5000.
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