Y is distributed as an exponential family member with b(θ) = −ln(−θ) for θ 0. Using E(Y) = b'(θ) and Var(Y) = φ b''(θ), which statement is correct?

E(Y) = −1/θ and Var(Y) = (1/ν)/θ², so V(μ) = μ²

E(Y) = 1/θ and Var(Y) = (1/ν)/θ², so V(μ) = μ

E(Y) = −1/θ and Var(Y) = −(1/ν)/θ², so V(μ) = μ²

E(Y) = −θ and Var(Y) = 1/ν, so V(μ) = 1

E(Y) = −1/θ and Var(Y) = (1/ν)/θ³, so V(μ) = μ³

IAI Actuarial Core Principles · Actuarial Statistics · Generalised linear models

Y is distributed as an exponential family member with b(θ) = −ln(−θ) for θ < 0 and φ = 1/ν for a constant ν > 0. Using E(Y) = b'(θ) and Var(Y) = φ b''(θ), which statement is correct?

E(Y) = −1/θ and Var(Y) = φ/θ² = (1/ν)/θ², so V(μ) = μ². Differentiating b(θ) = −ln(−θ) gives b' = −1/θ and b'' = 1/θ² = μ². This identifies the gamma family with a quadratic variance function.

AE(Y) = −1/θ and Var(Y) = (1/ν)/θ², so V(μ) = μ²Correct
BE(Y) = 1/θ and Var(Y) = (1/ν)/θ², so V(μ) = μ
CE(Y) = −1/θ and Var(Y) = −(1/ν)/θ², so V(μ) = μ²
DE(Y) = −θ and Var(Y) = 1/ν, so V(μ) = 1
E(Y) = −1/θ and Var(Y) = (1/ν)/θ³, so V(μ) = μ³

Explanation

b(θ) = −ln(−θ), so b'(θ) = −(1/(−θ))(−1)... directly: d/dθ[−ln(−θ)] = −1/θ, which is positive for θ < 0. Then b''(θ) = 1/θ². Since μ² = 1/θ², V(μ) = μ². Var(Y) = φ/θ². This is the gamma family.

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Y is distributed as an exponential family member with b(θ) = −ln(−θ) for θ < 0 and φ = 1/ν for a constant ν > 0. Using E(Y) = b'(θ) and Var(Y) = φ b''(θ), which statement is correct?

Explanation

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