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An ellipsoidal confidence region defined by a positive-definite matrix $A$ and a calibrated threshold $c$. For a known-covariance multivariate normal distribution sample, $A=n{\Sigma }^{-1}$ and the $1-\alpha$ quantile of the chi-squared distribution with $d$ degrees of freedom give exact coverage for the $d$-dimensional mean.

In a regular model let ${\ell }_N(\theta )$ be the log-likelihood, $U_N(\theta )=\nabla {\ell }_N(\theta )$ its score, $I_N(\theta )$ its total Fisher information, and $\hat{\theta }$ the unrestricted maximum-likelihood estimator. For a point null $\theta ={\theta }_0$ in a p-dimensional parameter space, the standard statistics areThese define the Wald test, score test and likelihood-ratio test. With an interior true parameter, increasing information, suitable differentiability and asymptotic normality, all have asymptotic chi-squared distribution with p degrees of freedom under the null. Observed information can replace expected information when it is asymptotically equivalent.

For q smooth independent restrictions $g(\theta )=0$, replace the Wald statistic by $g(\hat{\theta }{)}^T[G{I}_N(\hat{\theta }{)}^{-1}{G}^T{]}^{-1}g(\hat{\theta })$, where G is the derivative of g at the estimate; compare with chi-squared q. The likelihood-ratio statistic uses the difference between unrestricted and null-restricted maxima. The score statistic is computed at the restricted maximum and uses efficient information for the tested coordinates: with nuisance coordinates lambda this is $I_{\psi \psi }-I_{\psi \lambda }{I}_{\lambda \lambda }^{-1}{I}_{\lambda \psi }$. Inverting the acceptance inequalities for point nulls produces approximate coverage-$1-\alpha$ confidence regions; for example the Wald region consists of theta satisfying $(\hat{\theta }-\theta {)}^T{I}_N(\hat{\theta })(\hat{\theta }-\theta )\le {\chi }_{p,1-\alpha }^2$. Analogous score and likelihood-ratio inequalities give their confidence regions.

For the Poisson family, put $Y_{-1}=1$ to include the initial observation, and defineMultiplying the initial and conditional Poisson probabilities gives, on the supported observation sequences,The parameter-free term is ${\sum }_{i=0}^n[Y_i\log Y_{i-1}-\log (Y_i!)]$, interpreting $0\log 0$ as zero. A positive count after a zero preceding count is outside the support.

For $S_n>0$, strict concavity gives unconstrained maximum $\tilde{\theta }=S_n/T_n$, and the upper endpoint restriction givesThere is an endpoint qualification in the printed question. With probability $e^{-\theta }$, $Y_0=0$ and all later counts are zero. Then $S_n=0$, $T_n=1$ and the likelihood is $e^{-\theta }$, which has no maximum on the stated space $0<\theta \le 1$. The formula produces zero only as an extended estimate on the closure $[0,1]$. This is a likelihood supremum at an excluded boundary, and zero cannot be called an actual maximum in the original space.

The observed negative second derivative is $S_n/{\theta }^2$. Conditional expectation gives $\mathbb{E}{Y}_0=\theta$ and $\mathbb{E}{Y}_i=\theta \mathbb{E}{Y}_{i-1}={\theta }^{i+1}$. Therefore for $0<\theta <1$,This is the bounded information in a subcritical Poisson branching process. Observing more generations of one family does not provide unbounded information. Indeed $\mathbb{P}(Y_n>0)\le \mathbb{E}{Y}_n={\theta }^{n+1}\to 0$, and zero is absorbing, so extinction occurs at a finite generation almost surely. The total observed progeny $S_{\infty }$ is finite; eventually $S_n=S_{\infty }$, $T_n=1+S_{\infty }$, and the extended estimate equals $S_{\infty }/(1+S_{\infty })$, a discrete nondegenerate random variable rather than a consistent increasingly precise estimate.

Here is a direct proof of the failure of a chi-squared Wald limit which does not rely solely on the information bound. Under any interior null ${\theta }_0$, the event $Y_0=1$, $Y_1=0$ has probability ${\theta }_0{e}^{-2{\theta }_0}>0$ and forces every later count to be zero. On it the genuine constrained estimate is $1/2$ for every $n\ge 1$, so the usual expected-information Wald statistic isThus it has a persistent atom tending to $4(1/2-{\theta }_0{)}^2$. The fact that atoms obstruct a continuous distributional limit rules out a chi-squared limit, including when ${\theta }_0$ equals one half and that atom is at zero. Any definition adopted for the separate all-zero sample event leaves this positive-count counterexample intact. Hence the Wald statistic does not have the usual asymptotic chi-squared null law for this increasing observation horizon.

One presentation is the cost-effectiveness plane, with $\Delta E$ horizontally and $\Delta C$ vertically. Plot the point estimate $(45,1800)$ with a joint confidence region or paired bootstrap cloud. Its position shows the tradeoff quadrant, while the cloud shows uncertainty, dependence and possible dominance. A willingness-to-pay threshold $\lambda$ is the line $\Delta C=\lambda \Delta E$ through the origin. Points below that line have positive incremental net monetary benefit, including points in quadrants where a ratio rule would fail. The point-estimated switch occurs at $\lambda =40$ pounds per day.

A second presentation is a cost-effectiveness acceptability curve, giving an uncertainty-based measure of cost-effectiveness for a range of thresholds. For paired bootstrap increments $(\Delta E^{\ast (b)},\Delta C^{\ast (b)})$, plotThis counts positive net benefits, not merely ratios below the threshold, so it handles all quadrants. Using Bayesian inference the curve is the posterior probability that incremental net monetary benefit is positive; a frequentist bootstrap curve is a resampling summary, not literally a probability that a fixed parameter is positive.

Here the estimated incremental net monetary benefit and its standard error areThus a normal approximation uncertainty curve isIt passes through one half at £40 per day. Alternatively display the incremental net monetary benefit curve itself with approximate pointwise $95\%$ bands $45\lambda -1800\pm 1.96s_{\mathrm{INMB}}(\lambda )$. The plane preserves the joint uncertainty; the acceptability curve shows how the threshold changes the cost-effectiveness assessment. Neither presentation chooses society's willingness-to-pay threshold.