Actuarial Notes

Estimator

Study guides: Exam ALTAM, Exam ASTAM

An estimator \hat\theta is a rule — a function of the sample X_1, \ldots, X_n — that produces a value for an unknown parameter \theta; being a function of random data, it is itself a random variable, and its quality is judged by the distribution it has before the data are seen. The number it produces from a particular sample is the estimate.

\text{bias}_{\hat\theta}(\theta) = E[\hat\theta \mid \theta] - \theta

\text{MSE}_{\hat\theta}(\theta) = E\big[(\hat\theta - \theta)^2 \mid \theta\big]

\text{MSE}_{\hat\theta}(\theta) = \text{Var}(\hat\theta \mid \theta) + \big[\text{bias}_{\hat\theta}(\theta)\big]^2

  • Properties. An estimator is unbiased if its bias is zero for every \theta, asymptotically unbiased if the bias vanishes as n \to \infty, and consistent if \Pr(|\hat\theta_n - \theta| > \delta) \to 0 for every \delta > 0. Among unbiased estimators, the one with smallest variance is the most efficient; the Cramér-Rao Lower Bound is the floor.
  • MSE trades bias against variance, so a biased estimator can beat an unbiased one (Bias-Variance Tradeoff).
  • Variance of an MLE. Under regularity conditions the MLE is asymptotically normal with variance 1/I_n(\theta), estimated by the inverse of the observed information -\ell''(\hat\theta) (Fisher Information). That gives the normal confidence interval \hat\theta \pm z_{1-\alpha/2}\sqrt{\widehat{\text{Var}}(\hat\theta)}; for a function g(\hat\theta) use the delta method.
  • Non-normal interval. Klugman's likelihood-based alternative keeps every \theta with \ell(\theta) \ge \ell(\hat\theta) - \tfrac{1}{2}\chi^2_{1,\,1-\alpha}; it need not be symmetric about \hat\theta.

Three Estimators of a Maximum Claim Claim sizes are uniform on (0, \theta). From a sample of n = 4, compare by MSE: (a) 2\bar X, (b) the sample maximum X_{(4)} (the MLE), (c) \tfrac{5}{4}X_{(4)}. Use E[X_{(n)}] = \frac{n\theta}{n+1} and \text{Var}(X_{(n)}) = \frac{n\theta^2}{(n+1)^2(n+2)}.

Answer (a) E[2\bar X] = \theta, so it is unbiased and

\begin{align*}
\text{MSE} &= 4 \cdot \frac{\theta^2}{12 \cdot 4} \\
&= 0.0833\,\theta^2
\end{align*}

(b) The maximum is biased low:

\begin{align*}
\text{bias} &= \tfrac{4}{5}\theta - \theta \\
&= -0.2\,\theta \\
\text{Var} &= \frac{4\theta^2}{25 \cdot 6} \\
&= 0.02667\,\theta^2 \\
\text{MSE} &= 0.02667\,\theta^2 + 0.04\,\theta^2 \\
&= 0.0667\,\theta^2
\end{align*}

(c) Scaling removes the bias:

\begin{align*}
\text{MSE} &= \left(\tfrac{5}{4}\right)^2 (0.02667\,\theta^2) \\
&= 0.0417\,\theta^2
\end{align*}

The biased MLE beats the unbiased 2\bar X, and the bias-corrected maximum beats both: unbiasedness alone does not decide which estimator is better.

Confidence Interval for a Transition Intensity In a healthy–sick model with a constant intensity \mu from healthy to sick, a study records d = 45 transitions in v = 1{,}500 years spent healthy. The log-likelihood is \ell(\mu) = d \ln \mu - \mu v. Estimate \mu and give an approximate 95\% confidence interval.

Answer

\begin{align*}
\hat\mu &= \frac{d}{v} \\
&= \frac{45}{1{,}500} \\
&= 0.03
\end{align*}

The observed information is -\ell''(\hat\mu) = d/\hat\mu^2, so

\begin{align*}
\widehat{\text{Var}}(\hat\mu) &= \frac{\hat\mu^2}{d} \\
&= \frac{d}{v^2} \\
\text{SE} &= \frac{\sqrt{45}}{1{,}500} \\
&= 0.004472 \\[4pt]
\text{CI} &= 0.03 \pm 1.96(0.004472) \\
&= (0.0212,\ 0.0388)
\end{align*}

The data place the annual sickness intensity between about 2.1\% and 3.9\%; a tighter interval needs more transitions, not just more exposure time.

How it appears on the exam

  • ASTAM (estimate the variance of the estimators and construct normal and non-normal confidence intervals): fit a severity or frequency by maximum likelihood, invert the (observed or expected) information for the variance, then build \hat\theta \pm z\,\text{SE}; a variant asks for the likelihood-based interval instead.
  • ASTAM also asks for the bias, variance or MSE of a stated estimator — sample mean, sample maximum, an empirical survival probability — where the answer turns on \text{MSE} = \text{Var} + \text{bias}^2.
  • ALTAM (approximate confidence intervals for the estimators in Topic 1(h)): the MLE of a constant transition intensity d/v from multiple-state or multiple-decrement data, its variance d/v^2, and an interval for the intensity or, through the delta method, for a transition probability.

Exam traps

  • Squaring the bias into the wrong place. \text{MSE} = \text{Var} + \text{bias}^2; adding the bias unsquared, or subtracting it because it is negative, gives a wrong answer that is often an offered choice.
  • Confusing \text{Var}(X) with \text{Var}(\bar X). The variance of the sample mean divides by n; an interval built on \sigma instead of \sigma/\sqrt n is \sqrt n times too wide.
  • Dividing by exposure instead of transitions. For \hat\mu = d/v the variance is d/v^2 (equivalently \hat\mu^2/d), not \hat\mu/v^2 or \hat\mu^2/v.
  • Equating unbiased with best. An unbiased estimator with large variance can lose to a biased one on MSE, as the sample maximum shows.