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IAI Actuarial Core Principles · Actuarial Statistics · Generating functions

Z ~ N(0,1) and Y = Z². The MGF of Y is found by integrating e^(tz²) times the standard normal density. For which values of t does it exist, and what is it?

The integral gives (1 - 2t)^(-1/2), valid only when t < 1/2, which is the chi-squared (1 d.f.) MGF. The expression exp(t²/2) belongs to Z, not Z².

  1. A(1 - 2t)^(-1/2) for t < 1/2Correct
  2. B(1 - t)^(-1/2) for t < 1
  3. C(1 - 2t)^(-1) for t < 1/2
  4. Dexp(t²/2) for all t
  5. (1 - 2t)^(1/2) for t < 1/2

Explanation

E(e^(tZ²)) = ∫ (2π)^(-1/2) exp(-z²(1-2t)/2) dz = (1-2t)^(-1/2), requiring 1-2t > 0. This is chi-squared with 1 degree of freedom. The exp(t²/2) option is the MGF of Z itself, not Z².

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