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Estimation, Evaluation, and Selection of Actuarial Models

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56 CHAPTER 3. SAMPLING PROPERTIES OF ESTIMATORS<br />

distribution is obtained from the inverse <strong>of</strong> the matrix with rsth element<br />

·<br />

∂ 2 ¸ ·<br />

∂ 2<br />

¸<br />

I(θ) rs = −E l(θ) = −nE ln f(X; θ)<br />

∂θ s ∂θ r ∂θ s ∂θ r<br />

· ∂<br />

= E l(θ) ∂ ¸ · ∂<br />

l(θ) = nE ln f(X; θ) ∂<br />

¸<br />

ln f(X; θ) .<br />

∂θ r ∂θ s ∂θ r ∂θ s<br />

The first expression on each line is always correct. The second expression assumes that the likelihood<br />

is the product <strong>of</strong> n identical densities. This matrix is <strong>of</strong>ten called the information matrix. The<br />

information matrix also forms the Cramér—Rao lower bound. That is, under the usual conditions,<br />

no unbiased estimator has a smaller variance than that given by the inverse <strong>of</strong> the information.<br />

Therefore, at least asymptotically, no unbiased estimator is more accurate than the mle.<br />

Example 3.30 Estimate the covariance matrix <strong>of</strong> the maximum likelihood estimator for the lognormal<br />

distribution. Then apply this result to Data Set B.<br />

The likelihood function <strong>and</strong> its logarithm are<br />

nY<br />

·<br />

1<br />

L(µ, σ) =<br />

x j σ √ 2π exp − (ln x j − µ) 2 ¸<br />

2σ 2<br />

l(µ, σ) =<br />

j=1<br />

j=1<br />

nX<br />

− ln x j − ln σ − 0.5ln(2π) − 1 µ ln xj − µ 2<br />

.<br />

2 σ<br />

j=1<br />

The first partial derivatives are<br />

∂l<br />

nX<br />

∂µ = ln x j − µ<br />

σ 2<br />

The second partial derivatives are<br />

∂ 2 l<br />

∂µ 2 = − n σ 2<br />

∂ 2 l<br />

nX<br />

∂σ∂µ = −2<br />

<strong>and</strong> ∂l<br />

∂σ = −n σ + nX<br />

j=1<br />

∂ 2 l<br />

∂σ 2 = n nX<br />

σ 2 − 3<br />

ln x j − µ<br />

σ 3<br />

j=1<br />

j=1<br />

(ln x j − µ) 2<br />

σ 4 .<br />

(ln x j − µ) 2<br />

σ 3 .<br />

The expected values are (ln X j has a normal distribution with mean µ <strong>and</strong> st<strong>and</strong>ard deviation σ)<br />

E[∂ 2 l/∂µ 2 ] = −n/σ 2<br />

E[∂ 2 l/∂σ∂µ] = 0<br />

E[∂ 2 l/∂σ 2 ] = −2n/σ 2 .<br />

Changing the signs <strong>and</strong> inverting produces an estimate <strong>of</strong> the covariance matrix (it is an estimate<br />

because Theorem 3.29 only provides the covariance matrix in the limit). It is<br />

· ¸<br />

σ2 /n 0<br />

0 σ 2 .<br />

/2n

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