Approximate Confidence Interval

Study guides: Exam ALTAM, Exam ASTAM

Approximate Confidence Interval is an interval for a parameter built from the large-sample behaviour of its maximum likelihood estimator rather than from an exact sampling distribution. The usual one is the normal interval, the estimate plus or minus a normal percentile times a standard error taken from the Fisher information. The non-normal intervals, log-transformed or based on the likelihood ratio, keep a positive parameter positive and become asymmetric when the likelihood is.

$\hat\theta \pm z_{\alpha/2}\sqrt{\widehat{\text{Var}}(\hat\theta)}$

$\widehat{\text{Var}}(\hat\theta) = \big[-\ell''(\hat\theta)\big]^{-1}$

$\Big(\hat\theta\,e^{-z_{\alpha/2}\text{SE}/\hat\theta},\ \hat\theta\,e^{z_{\alpha/2}\text{SE}/\hat\theta}\Big)$

$\Big\{\theta : \ell(\theta) \ge \ell(\hat\theta) - \tfrac{1}{2}\chi^2_{1,\,1-\alpha}\Big\}$

$\hat\mu = \frac{d}{v}$

$\widehat{\text{Var}}(\hat\mu) = \frac{d}{v^2}$

Symbols: $\ell$ is the log-likelihood, $z_{\alpha/2}$ is the upper $\alpha/2$ point of the standard normal (1.96 for 95%), and $\text{SE} = \sqrt{\widehat{\text{Var}}(\hat\theta)}$ . The variance can come from the observed information $-\ell''(\hat\theta)$ , as written, or from the expected information evaluated at $\hat\theta$ . $\chi^2_{1,\,1-\alpha}$ is the $100(1-\alpha)$ th percentile of the chi-squared distribution with one degree of freedom, which is 3.841 at 95%. For a function $g(\theta)$ , the delta method gives $\widehat{\text{Var}}[g(\hat\theta)] \approx g'(\hat\theta)^2\,\widehat{\text{Var}}(\hat\theta)$ .
Non-normal intervals: The log-transformed interval is the normal interval for $\ln\theta$ mapped back, since the delta method gives $\text{SE}(\ln\hat\theta) \approx \text{SE}/\hat\theta$ . It stays positive and reaches further above $\hat\theta$ than below. The likelihood-ratio interval keeps every $\theta$ that a likelihood ratio test would not reject. It follows the shape of the likelihood, needs no variance, and transforms with the parameter, and it is the non-normal interval of Klugman, Panjer & Willmot (Loss Models, chapter 11).
Transition intensities: In a multiple state model with a constant transition intensity $\mu$ out of a state, $v$ is the total time that lives were observed in the state and $d$ is the number of transitions observed. The likelihood is $\mu^d e^{-\mu v}$ , so $\hat\mu = d/v$ , and $-\ell''(\mu) = d/\mu^2$ gives the variance $\hat\mu^2/d = d/v^2$ (Dickson, Hardy & Waters, chapter 8). Intensities out of the same state share $v$ , but the cross derivatives of the log-likelihood are 0, so their estimators are asymptotically independent.
Accuracy: All of these intervals are large-sample approximations. With few observations the normal interval can include impossible values, such as a negative intensity, and the non-normal forms are the remedy.

Three Intervals for a Pareto Tail Index Ten losses above $500,000 are modelled as single-parameter Pareto with $\theta = 500{,}000$ known and $\alpha$ unknown. The data give $\sum \ln(x_i/500{,}000) = 4.0$ . Find the maximum likelihood estimate of $\alpha$ and 95% normal, log-transformed and likelihood-ratio intervals for it.
Answer Apart from a constant, the log-likelihood is $\ell(\alpha) = 10\ln\alpha - 4\alpha$ , so $\hat\alpha = 10/4 = 2.5$ and $-\ell''(\alpha) = 10/\alpha^2$ .
 $\begin{align*} \widehat{\text{Var}}(\hat\alpha) &= \frac{2.5^2}{10} \\ &= 0.625 \\ \text{Normal} &= 2.5 \pm 1.96(0.7906) \\ &= (0.950,\ 4.050) \\ e^{1.96(0.7906)/2.5} &= e^{0.6198} \\ &= 1.8586 \\ \text{Log} &= (2.5/1.8586,\ 2.5 \times 1.8586) \\ &= (1.345,\ 4.646) \end{align*}$ 
For the likelihood-ratio interval, $\ell(2.5) = 10\ln 2.5 - 10 = -0.8371$ , and the cutoff is $-0.8371 - 3.841/2 = -2.7576$ . Solving $10\ln\alpha - 4\alpha = -2.7576$ numerically gives the interval $(1.253,\ 4.385)$ . Both non-normal intervals sit further right than the normal one, because the log-likelihood falls away more steeply below $\hat\alpha$ than above it, and both keep well clear of $\alpha = 1$ , below which the Pareto mean is infinite.

A Sickness Intensity and a Probability of Staying Healthy In a healthy–sick–dead model with constant intensities over ages 60 to 65, a study observes 1,250 person-years in the healthy state, with 30 transitions to sick and 5 deaths. Give 95% intervals for $\mu^{01}$ and for the probability ${}_1p_{60}^{\overline{00}} = e^{-(\mu^{01} + \mu^{02})}$ that a healthy life stays healthy for a year.
Answer
 $\begin{align*} \hat\mu^{01} &= \frac{30}{1{,}250} \\ &= 0.024 \\ \text{SE}(\hat\mu^{01}) &= \frac{\sqrt{30}}{1{,}250} \\ &= 0.004382 \\ \text{CI} &= 0.024 \pm 1.96(0.004382) \\ &= (0.0154,\ 0.0326) \end{align*}$ 
The total intensity is $\hat\mu^{01} + \hat\mu^{02} = 35/1{,}250 = 0.028$ , and because the two estimators are asymptotically independent its variance is $35/1{,}250^2$ , with standard error 0.004733. The derivative of $e^{-s}$ is $-e^{-s}$ , so the delta method multiplies that standard error by $\hat p = e^{-0.028} = 0.97239$ .
 $\begin{align*} \text{SE}(\hat p) &= 0.97239(0.004733) \\ &= 0.004602 \\ \text{CI} &= 0.97239 \pm 1.96(0.004602) \\ &= (0.9634,\ 0.9814) \end{align*}$ 
Thirty transitions estimate the sickness intensity only to within about 36% either way, while the one-year probability is pinned down far more tightly in absolute terms, because it is close to 1.

How it appears on the exam

ASTAM asks candidates to estimate the variance of estimators and construct normal and non-normal confidence intervals. A question fits a distribution by maximum likelihood, asks for the variance from the information, and then asks for an interval for the parameter or, through the delta method, for a probability, a percentile or a limited expected value.

ALTAM asks for approximate confidence intervals for the maximum likelihood estimates of transition intensities and probabilities in multiple state and multiple decrement models. A question gives waiting times and transition counts for each state and asks for $\hat\mu = d/v$ with its interval, or for an interval for a probability built from several intensities.

Likelihood-ratio interval: When it is asked for, the interval is found by solving $\ell(\theta) = \ell(\hat\theta) - \tfrac12\chi^2$ for its two endpoints, numerically or by checking given candidate values.

Exam traps

Information or its inverse: The variance is the inverse of the information. Using $-\ell''(\hat\theta)$ itself as the variance gives a nonsense interval, which is easy to miss only when the information happens to be close to 1.

Half the chi-squared value: The likelihood-ratio cutoff is $\ell(\hat\theta) - \tfrac12\chi^2_{1,\,1-\alpha}$ . Subtracting the full 3.841 gives an interval that is much too wide.

Exposure in the right state: In $\hat\mu = d/v$ , $v$ is the time spent in the state the transition leaves, not the total time in the study. Using all person-years understates an intensity out of the healthy state.

Every way out of the state: A probability of staying in a state depends on every intensity out of it. In the second example the variance of the total intensity is $(30 + 5)/1{,}250^2$ , and using only the 30 sicknesses leaves out the deaths, which also end a stay in the healthy state.