Calculate approximate confidence intervals for the estimators in Topic 1(h), using asymptotic properties of the maximum likelihood estimators.

A point estimate of $\hat\mu = 0.012$ is useless without a range. Asymptotic MLE theory hands you that range almost for free, once you compute one second derivative.

Under mild regularity, the MLE $\hat\theta$ is consistent and asymptotically normal. The standardizing variance is $I(\theta)^{-1}$, where $I(\theta)$ is the Fisher information:

$$I(\theta) = -E\!\left[\frac{d^2 \ell(\theta)}{d\theta^2}\right]$$

In practice you rarely take the expectation. You evaluate the observed information $I(\hat\theta) = -\ell''(\hat\theta)$ at the MLE and invert it. That estimate is consistent and avoids integration over the data-generating distribution.

KEY: "Asymptotic" means the result holds as sample size grows. Treat any sample with fewer than 30 events warily; the Wald CI can undercover at small samples and may push limits past natural bounds.

A two-sided $100(1-\alpha)\%$ interval is:

$$\hat\theta \pm z_{\alpha/2} \sqrt{I(\hat\theta)^{-1}}$$

Critical values you must know cold: $z_{0.025}=1.96$ (95%), $z_{0.05}=1.645$ (90%), $z_{0.005}=2.576$ (99%).

For the constant-force model with $D$ deaths and central exposure $E^c$, the MLE is $\hat\mu = D/E^c$ and $I(\hat\mu) = E^c/\hat\mu$. The variance estimate is $\hat\mu/E^c = \hat\mu^2/D$.

Common mistakes

• Confusing $I$ and $I^{-1}$. Variance is the inverse. $I(\hat\mu)=1{,}450/0.02=72{,}500$, so $Var = 1/72{,}500 = 1.38\times10^{-5}$, not 72,500.
• Squaring $z$ instead of multiplying. The half-width is $z \cdot SE$, not $z^2 \cdot SE$. Using 1.96² gives 3.84, the chi-square critical value.
• Sign error in $g'$ for delta method. The sign vanishes when squared. Don't panic if $g'$ is negative; only $[g']^2$ enters the variance.

Bottom line

• Asymptotic normality: $\hat\theta \approx N(\theta, I(\theta)^{-1})$ for large samples under standard regularity conditions.
• Observed information: $I(\hat\theta) = -\ell''(\hat\theta)$; invert it for the variance estimate (use observed, not expected, from your data).
• Wald CI: $\hat\theta \pm z_{\alpha/2}\sqrt{\widehat{Var}(\hat\theta)}$; $z_{0.025}=1.96$.
• Delta method: $Var(g(\hat\theta)) \approx [g'(\hat\theta)]^2 \cdot Var(\hat\theta)$; only the squared derivative enters, so sign is irrelevant.

Exam shortcut

For constant force, write SE directly as $\hat\mu/\sqrt{D}$. With $D=29$ and $\hat\mu=0.02$, $SE = 0.02/\sqrt{29} = 0.003714$. Matches the long-form answer instantly. DECISION: CI for a parameter. Wald on $\hat\theta$. CI for a survival probability or annuity. Delta method on the function, optionally with a log transform. Memorize $z$-values 1.645, 1.96, 2.576 for 90%, 95%, 99%. Half the exam-day arithmetic in this LO collapses to one multiplication once you have the SE.

The full lesson (about 1,341 words, 9 min read) adds 2 worked examples, all 6 common mistakes, a self-check, free in the app.

Learning objectives

• 1i

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