Estimation, Evaluation, and Selection of Actuarial Models

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56 CHAPTER 3. SAMPLING PROPERTIES OF ESTIMATORS distribution is obtained from the inverse of the matrix with rsth element · ∂ 2 ¸ · ∂ 2 ¸ I(θ) rs = −E l(θ) = −nE ln f(X; θ) ∂θ s ∂θ r ∂θ s ∂θ r · ∂ = E l(θ) ∂ ¸ · ∂ l(θ) = nE ln f(X; θ) ∂ ¸ ln f(X; θ) . ∂θ r ∂θ s ∂θ r ∂θ s The first expression on each line is always correct. The second expression assumes that the likelihood is the product of n identical densities. This matrix is often called the information matrix. The information matrix also forms the Cramér—Rao lower bound. That is, under the usual conditions, no unbiased estimator has a smaller variance than that given by the inverse of the information. Therefore, at least asymptotically, no unbiased estimator is more accurate than the mle. Example 3.30 Estimate the covariance matrix of the maximum likelihood estimator for the lognormal distribution. Then apply this result to Data Set B. The likelihood function and its logarithm are nY · 1 L(µ, σ) = x j σ √ 2π exp − (ln x j − µ) 2 ¸ 2σ 2 l(µ, σ) = j=1 j=1 nX − ln x j − ln σ − 0.5ln(2π) − 1 µ ln xj − µ 2 . 2 σ j=1 The first partial derivatives are ∂l nX ∂µ = ln x j − µ σ 2 The second partial derivatives are ∂ 2 l ∂µ 2 = − n σ 2 ∂ 2 l nX ∂σ∂µ = −2 and ∂l ∂σ = −n σ + nX j=1 ∂ 2 l ∂σ 2 = n nX σ 2 − 3 ln x j − µ σ 3 j=1 j=1 (ln x j − µ) 2 σ 4 . (ln x j − µ) 2 σ 3 . The expected values are (ln X j has a normal distribution with mean µ and standard deviation σ) E[∂ 2 l/∂µ 2 ] = −n/σ 2 E[∂ 2 l/∂σ∂µ] = 0 E[∂ 2 l/∂σ 2 ] = −2n/σ 2 . Changing the signs and inverting produces an estimate of the covariance matrix (it is an estimate because Theorem 3.29 only provides the covariance matrix in the limit). It is · ¸ σ2 /n 0 0 σ 2 . /2n
3.3. VARIANCE AND CONFIDENCE INTERVALS 57 For the lognormal distribution, the maximum likelihood estimates are the solutions to the two equations nX ln x j − µ σ 2 =0and − n nX σ + (ln x j − µ) 2 σ 3 =0. j=1 From the first equation, ˆµ = n 1 P n j=1 ln x j and from the second equation, ˆσ 2 = n 1 P n j=1 (ln x j − ˆµ) 2 . For Data Set B the values are ˆµ =6.1379 and ˆσ 2 =1.9305 or ˆσ =1.3894. With regard to the covariance matrix the true values are needed. The best we can do is substitute the estimated values to obtain dVar(ˆµ, ˆσ) = j=1 · 0.0965 0 0 0.0483 ¸ . (3.4) The multiple “hats” in the expression indicate that this is an estimate of the variance of the estimators. ¤ The zeroes off the diagonal indicate that the two parameter estimates are asymptotically uncorrelated. For the particular case of the lognormal distribution, that is also true for any sample size. One thing we could do with this information is construct approximate 95% confidence intervals for the true parameter values. These would be 1.96 standard deviations either side of the estimate: µ : 6.1379 ± 1.96(0.0965) 1/2 =6.1379 ± 0.6089 σ : 1.3894 ± 1.96(0.0483) 1/2 =1.3894 ± 0.4308. To obtain the information matrix, it is necessary to take both derivatives and expected values. This is not always easy to do. A way to avoid this problem is to simply not take the expected value. Rather than working with the number that results from the expectation, use the observed data points. The result is called the observed information. Example 3.31 Estimate the covariance in the previous example using the observed information. Substituting the observations into the second derivatives produces ∂ 2 l ∂µ 2 = − n σ 2 = −20 σ 2 ∂ 2 l nX ∂σ∂µ = −2 j=1 ∂ 2 l ∂σ 2 = n nX σ 2 − 3 ln x j − µ 122.7576 − 20µ σ 3 = −2 σ 3 j=1 (ln x j − µ) 2 σ 4 = 20 − 245.5152µ +20µ2 − 3792.0801 σ2 σ 4 . Inserting the parameter estimates produces the negatives of the entries of the observed information, ∂ 2 l ∂µ 2 = −10.3600 ∂ 2 l ∂σ∂µ = 0 ∂ 2 l ∂σ 2 = −20.7190.
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