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IAI Actuarial Core Principles · Actuarial Statistics · Generalised linear models

Y is gamma distributed with mean μ = 500 and coefficient of variation 0.4. In a GLM, Var(Y) = φ V(μ) with V(μ) = μ². What is the dispersion parameter φ, and the variance of Y?

The dispersion parameter is 0.16 and the variance is 40,000. For a gamma GLM, the squared coefficient of variation equals φ, so φ = 0.4² = 0.16. Then Var(Y) = 0.16 × 500² = 40,000, giving standard deviation 200.

  1. Aφ = 0.16, Var(Y) = 40,000Correct
  2. Bφ = 0.4, Var(Y) = 100,000
  3. Cφ = 0.16, Var(Y) = 80
  4. Dφ = 0.16, Var(Y) = 250,000
  5. φ = 0.4, Var(Y) = 200

Explanation

For gamma, the CV² equals φ, so φ = 0.4² = 0.16. Then Var(Y) = 0.16 × 500² = 0.16 × 250,000 = 40,000. Check: SD = 200 = 0.4 × 500. Using φ = 0.4 wrongly would give 100,000.

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