Estimation, Evaluation, and Selection of Actuarial Models
Estimation, Evaluation, and Selection of Actuarial Models
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