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The invariance property of maximum likelihood estimation givesSince $\hat{\theta }-\theta \sim N(0,{\sigma }_{\theta }^2)$, the ratio $\hat{h}/h$ has a log-normal distribution. Its exact variance isand the lowest-order delta method approximation isFor fixed ${\sigma }_{\mathrm{tot}}$, case (i) has $A={\sigma }_{\mu }^2+{\sigma }_{\mathrm{tot}}^2$, whereas case (ii) has $A={\sigma }_{\mu }^2+{\sigma }_{\mathrm{tot}}^2/K$. Because $K\ge 2$ and $B$ is the same in both cases, independent supernova-level variation in case (ii) gives the smaller fractional variance: averaging reduces it, while a shared galaxy fluctuation does not average away.

Continuous differentiability gives the first-order expansionThe assumed convergence in distribution implies $\sqrt{n}({\hat{\theta }}_n-{\theta }_0)=O_p(1)$, so after multiplication by $\sqrt{n}$ the remainder is $o_p(1)$. The Slutsky theorem gives the multivariate delta method

For observations $X_1,\dots ,X_n$, the log-likelihood is, up to an additive constant,Its unique maximum isNow $\mathbb{E}[X_i^2]={\sigma }^2$ and, using $\mathbb{E}[Z^4]=3$, $\mathrm{Var}(X_i^2)=2{\sigma }^4$. The central limit theorem followed by the delta method for $g(x)=\sqrt{x}$ gives

For $Y_{\mu }\sim \mathrm{Poisson}(\mu )$, the delta method with $g(y)=2\sqrt{y}$ givesThus the transformation stabilizes variance at large $\mu$.

A Gaussian linear model for $\sqrt{Y}$ treats the transformed observations as having additive constant-variance errors and models $\mathbb{E}(\sqrt{Y}\mid X)$ as linear. A Poisson regression with square-root link retains the Poisson likelihood and modelsThe two procedures therefore have different likelihoods, fitted means, and constraints; transforming the response is not the same as transforming its conditional mean.

The central limit theorem givesApplying the delta method to $g(x)=x^2$ yieldsFor $\mu =0$ this denotes the degenerate distribution at zero, which is the correct limit at the $\sqrt{n}$ scale.

LetIts gradient, in the coordinate order $(a,c,b)$, isApply the Delta method to the limit in part (c). Multiplying the displayed covariance matrix there on both sides by this gradient givesConsequently the asymptotic distribution of the Gaussian sample correlation is

The transformation $g(t)=e^{t/2}$ satisfies $g(\theta )=\sigma$ and $g^{\prime }(\theta )=\sigma /2$. Applying the delta method to part c givesThus exponentiating half the robust location estimate produces an asymptotically normal scale estimator.

The delta method givesIf ${\phi }^{\prime }(\psi )=0$, this denotes the degenerate distribution at zero.

Since $\mathbb{E}{X}_i=1/\theta$ and $\mathrm{Var}(X_i)=1/{\theta }^2$, the central limit theorem givesApplying the delta method to $g(x)=1/x$ yieldsThus, with $z_{1-\alpha /2}={\Phi }^{-1}(1-\alpha /2)$, an asymptotic interval centered at the MLE isoptionally intersected with $(0,\infty )$. Slutsky's theorem gives coverage tending to $1-\alpha$.

Put $r=e^{\theta }$. The geometric moments areFor $g(y)=\log ((1+y)/y)$, one has $g(m)=\theta$ andThe central limit theorem for $\overline{Y}$ followed by the delta method therefore gives

For transition-intensity vector $q$ and generator $Q(q)$, the expected infectious occupancy over three months from $S$ is the Markov reward model quantityInsert the fitted intensities to obtain $m(\hat{q})$, evaluating the matrix exponential and integral numerically if necessary.

If $\eta =(\log q_{SE},\log q_{EI},\log q_{IR})$ has estimated covariance $V$, calculate the numerical gradient $g={\nabla }_{\eta }m(e^{\eta })$ at $\hat{\eta }$. The delta method gives estimated variance $g^{T}Vg$ and the approximate confidence intervalSimulation from $N(\hat{\eta },V)$ followed by transformation through $m$ gives a useful asymmetric alternative.

Since $X_j\sim \mathrm{Poisson}(n_j{\lambda }_j)$, the rate MLE has asymptotic variance ${\lambda }_j/n_j$. The delta method for the logarithm givesIndependence of the arms therefore yieldsUnder $H_0:{\lambda }_T={\lambda }_C=\lambda$, the asymptotic variance of the log rate ratio is