IAI Actuarial Core Principles · Actuarial Statistics · Generating functions
X ~ N(2, 9). Using the MGF M(t) = exp(2t + 4.5t^2), what is E[e^(X)] , i.e. M(1)?
For a normal with mean 2 and variance 9, M(t) = exp(2t + 4.5t^2). At t = 1 this is exp(6.5), so E[e^X] = e^6.5. Ignoring either the mean or the variance term gives wrong options.
- Ae^2
- Be^4.5
- Ce^6.5Correct
- De^11
- e^8.5
Explanation
M(1) = exp(2 + 4.5) = e^6.5. It is not e^2 (ignores variance) and e^4.5 ignores the mean. Using sigma instead of sigma^2 would give 2 + 1.5 = 3.5, also not offered.
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