Parameter Estimation

Study guides: Exam MAS-I, Exam 7

Parameter Estimation uses sample data to put values on the unknown parameters of a probability model, such as a claim-size distribution, a regression, or the development pattern behind an unpaid claim distribution. A point estimator \hat{\theta} is the statistic used to do it, and its value on a particular sample is the estimate.

\hat{\theta} = T(X_1, \ldots, X_n)

\mathrm{MSE}(\hat{\theta}) = \mathrm{Var}(\hat{\theta}) + \left[\mathrm{Bias}(\hat{\theta})\right]^2

• Criteria (MAS-I). An estimator is judged on several properties:

  • Bias. \mathrm{Bias}(\hat{\theta}) = E[\hat{\theta}] - \theta, and the estimator is unbiased when this is zero.
  • Consistency: \hat{\theta} converges to \theta as n grows. MSE tending to zero is enough.
  • Efficiency: an unbiased estimator whose variance reaches the Cramér-Rao Lower Bound 1/[nI(\theta)] is efficient. The unbiased estimator with the smallest variance of all is the minimum variance unbiased estimator (MVUE).
  • Sufficiency: the estimator uses a Sufficient Statistic that carries all the sample's information about \theta.
  • Mean Square Error combines bias and variance, so a slightly biased estimator can beat an unbiased one.

• Methods. The Method of Moments equates sample moments to theoretical ones; it is quick but not always efficient. Maximum Likelihood Estimation maximizes \ell(\theta) and is invariant under transformation. Under regularity conditions — every f(x \mid \theta) has the same support, the true \theta is an interior point, the log-likelihood is differentiable in \theta, and \hat\theta is the unique root of \ell'(\theta) = 0 — it is also consistent, asymptotically normal and asymptotically efficient. The uniform example below breaks the first condition: the support (0, \theta) moves with \theta, and the MLE X_{(n)} has MSE 2\theta^2/[(n+1)(n+2)], shrinking like 1/n^2 rather than the 1/n of the regular case. MLE handles incomplete data directly: a censored value contributes S(u), and a loss truncated at a deductible d contributes f(x)/S(d).

• The normal sample. For a normal sample, \bar{X} is the MLE of \mu. It is unbiased and sufficient, and it attains the Cramér-Rao bound. The MLE of \sigma^2 divides by n and is biased low, while the sample variance S^2 divides by n-1 and is unbiased.

• Unpaid claim distributions (Exam 7). Reserve variability runs on the same ideas:

  • Mack estimates volume-weighted factors \hat{f}_k = \sum_i C_{i,k+1} / \sum_i C_{i,k}, and variance parameters \hat{\alpha}_k^2 from the spread of individual link ratios around \hat{f}_k.
  • Clark fits a growth curve and an ELR (or one ultimate per year) by MLE, assuming over-dispersed Poisson increments.
  • The ODP bootstrap takes its scale parameter from Pearson residuals.

  Error in these estimates is parameter risk, one part of prediction error.

Two Estimators of a Loss Ceiling Five claims (in $000s) of 2, 7, 4, 9, 3 are assumed uniform on (0, \theta). Estimate \theta by the method of moments and by MLE, and compare the two estimators' MSE.

Answer Method of moments: E[X] = \theta/2, so \tilde{\theta} = 2\bar{x} = 2(5) = 10. It is unbiased, so its MSE is its variance:

\begin{align*}
\mathrm{MSE}(\tilde{\theta}) &= \frac{4}{n} \cdot \frac{\theta^2}{12} \\
&= \frac{\theta^2}{15}
\end{align*}

MLE: L(\theta) = \theta^{-5} for \theta \geq \max x_i, which is largest at \hat{\theta} = X_{(5)} = 9. Since E[X_{(n)}] = n\theta/(n+1), the bias is -\theta/6:

\begin{align*}
\mathrm{MSE}(\hat{\theta}) &= \frac{n\theta^2}{(n+1)^2(n+2)} + \frac{\theta^2}{36} \\
&= \frac{5\theta^2}{252} + \frac{7\theta^2}{252} \\
&= \frac{\theta^2}{21}
\end{align*}

The biased MLE has the lower MSE (\theta^2/21 < \theta^2/15), and the method of moments can even return a ceiling below the largest claim. Rescaling to \tfrac{6}{5}X_{(5)} = 10.8, a function of the sufficient statistic X_{(n)}, removes the bias and cuts the MSE to \theta^2/35.

MLE with a Deductible and a Policy Limit Ground-up losses are exponential with mean \theta. Only losses above a \$500 deductible are reported, and losses are capped at \$10{,}000. The five reported losses are \$1{,}500, \$2{,}500, \$4{,}000, and two recorded at the \$10{,}000 cap. Find \hat{\theta}.

Answer Truncation at d = 500 turns each term into an excess over d. Exponential losses are memoryless, so f(x)/S(d) = \theta^{-1}e^{-(x-500)/\theta}. Each capped loss contributes S(10{,}000)/S(500) = e^{-9{,}500/\theta}:

\begin{align*}
\ell(\theta) &= -3\ln\theta - \frac{1{,}000 + 2{,}000 + 3{,}500 + 2(9{,}500)}{\theta} \\
&= -3\ln\theta - \frac{25{,}500}{\theta}
\end{align*}

Setting \ell'(\theta) = -3/\theta + 25{,}500/\theta^2 = 0 gives \hat{\theta} = 25{,}500/3 = 8{,}500. Only the uncensored losses count in the divisor; treating the capped losses as exact would give 5{,}100, understating the mean.

Clark's Cape Cod ELR by Maximum Likelihood Premiums for accident years 1 to 3 are 1{,}000, 1{,}100 and 1{,}200. A fitted growth curve gives G(12) = 0.40, G(24) = 0.70 and G(36) = 0.85. Incremental paid losses are 260, 190, 90 (AY1), 300, 200 (AY2) and 325 (AY3). Holding the curve fixed, find the MLE of the ELR, the scale parameter \sigma^2 (with p = 3 parameters: ELR, \omega, \theta), and the reserve.

Answer Under the over-dispersed Poisson likelihood, the ELR estimate sets total fitted losses equal to total actual losses:

\begin{align*}
\widehat{\text{ELR}} &= \frac{540 + 500 + 325}{1{,}000(0.85) + 1{,}100(0.70) + 1{,}200(0.40)} \\
&= \frac{1{,}365}{2{,}100} \\
&= 0.65
\end{align*}

Each fitted increment is \mu = P \cdot \text{ELR} \cdot \Delta G. That gives 260, 195, 97.5; 286, 214.5; and 312. The terms (c-\mu)^2/\mu are 0, 0.128, 0.577, 0.685, 0.980 and 0.542, which sum to 2.912. So

\begin{align*}
\sigma^2 &= \frac{2.912}{6 - 3} \\
&= 0.971
\end{align*}

The reserve is \sum P \cdot \text{ELR}\,(1 - G) = 97.5 + 214.5 + 468 = 780. Its process standard deviation is \sqrt{0.971 \times 780} = 27.5, or 3.5\% of the reserve. Parameter variance from the information matrix comes on top of that.