Estimation, Evaluation, and Selection of Actuarial Models

3.3. VARIANCE AND CONFIDENCE INTERVALS 59
is a function of the parameters. For example, we might be interested in the lognormal mean as
an estimate of the population mean. That is, we want to use exp(ˆµ +ˆσ 2 /2) as an estimate of
the population mean, where the maximum likelihood estimates of the parameters are used. It is
very difficult to evaluate the mean and variance of this random variable because it is a complex
function of two variables that already have complex distributions. The following theorem (from
Linear Statistical Inference and its Applications by C. R. Rao (Wiley, 1965)) can help. The method
is often called the delta method.
Theorem 3.33 Let X n =(X 1n ,...,X kn ) T be a multivariate random variable of dimension k based
on a sample of size n. Assume that X n is asymptotically normal with mean θ and covariance matrix
Σ/n, where neither θ nor Σ depend on n. Letg be a function of k variables that is totally differentiable.
Let G n = g(X 1n ,...,X kn ).ThenG n is asymptotically normal with mean g(θ) and variance
(∂g) T Σ(∂g)/n, where∂g is the vector of first derivatives, that is, ∂g =(∂g/∂θ 1 ,...,∂g/∂θ k ) T and
it is to be evaluated at θ, the true parameters of the original random variable. ¤
The statement of the Theorem is hard to decipher. The Xs are the estimators and g is the
function of the parameters that are being estimated. For a model with one parameter, the Theorem
reduces to the following statement: Let ˆθ be an estimator of θ that has an asymptotic normal
distribution with mean θ and variance σ 2 /n. Then g(ˆθ) has an asymptotic normal distribution
with mean g(θ) and asymptotic variance [g 0 (θ)](σ 2 /n)[g 0 (θ)] = g 0 (θ) 2 σ 2 /n.
Example 3.34 Use the delta method to approximate the variance of the maximum likelihood estimator
of the probability that an observation from an exponential distribution exceeds 200. Apply
this result to Data Set B.
From Example 2.32 we know that the maximum likelihood estimate of the exponential parameter
is the sample mean. We are asked to estimate p =Pr(X>200) = exp(−200/θ). The maximum
likelihood estimate is ˆp =exp(−200/ˆθ) =exp(−200/¯x). Determining the mean and variance of
this quantity is not easy. But we do know that that Var( ¯X) =Var(X)/n = θ 2 /n. Furthermore,
and therefore the delta method gives
For Data Set B,
g(θ) = e −200/θ
g 0 (θ) = 200θ −2 e −200/θ
Var(ˆp) . = [200θ −2 e −200/θ ] 2 θ 2 /n =40, 000θ −2 e −400/θ /n.
¯x = 1, 424.4
ˆp = exp(−200/1, 424.4) = 0.86900
dVar(ˆp) = 40, 000(1, 424.4) −2 exp(−400/1, 424.4)/20 = 0.0007444.
A 95% confidence interval for p is 0.869 ± 1.96 √ 0.0007444 or 0.869 ± 0.053. ¤
Example 3.35 Construct a 95% confidence interval for the mean of a lognormal population using
Data Set B. Compare this to the more traditional confidence interval based on the sample mean.