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For independent observations from the exponential distribution with rate $\theta$, the log-likelihood is $n\log \theta -\theta {\sum }_i{X}_i$, soThe central limit theorem and delta method give

Let independent exposure sums $A,B$ have mean $N$ ADU at each detector pixel, with pure Poisson distribution noise and fixed gain $g$. The delta method gives $A/B\simeq 1+(A-N)/N-(B-N)/N$, so $\mathrm{Var}(A/B)\simeq 2/(gN)$ and $g\simeq 2/(N{E}^2)$. For nonuniform detector pixel means $N_i$, an unweighted spatial ratio variance is instead $(2/g)⟨1/N_i⟩$. This high-count approximation requires linear unsaturated data, bias subtraction and negligible read noise.

The log odds ratio is the natural logarithm of an odds ratio. For two independent binomial groups with positive event and non-event counts $(a,b)$ and $(c,d)$, its estimate is $\log (ad/(bc))$, with first-order variance $1/a+1/b+1/c+1/d$ by the delta method. Zero cells need a method that handles boundary data; the displayed normal approximation is not valid there.

For two independent binomial distributions with event and nonevent counts $a,b$ and $c,d$, the odds ratio is $ad/(bc)$. The delta method applied to $\log (p/(1-p))$, whose derivative is $1/[p(1-p)]$, gives a group variance $1/[Np(1-p)]$. Substitution of the fitted proportion gives $1/a+1/b$ in the first group and $1/c+1/d$ in the second. Their independence gives the displayed sum. A normal approximation then gives a confidence interval on the log odds ratio scale; exponentiation produces an odds-ratio interval. Zero cells require another approximation or model.

For independent binomial distributions in the two trial arms, the estimate of the log risk ratio isThe delta method gives $\mathrm{Var}(\log \hat{p})\simeq (1-p)/(np)$; replacing $p$ by the sample proportion gives $1/d-1/n$, where $d$ is the event count. Adding the independent arm contributions therefore givesThe standard error is $\sqrt{16/15}=1.0328$. Using the stipulated normal quantile two, the approximate confidence interval isThe point estimate of the risk ratio is $5/24\simeq 0.208$, corresponding to approximately 79% lower mortality risk in the transfusion arm. Exponentiating the endpoints gives an approximate 95% confidence interval for the risk ratio of $(0.0264,1.6437)$. It includes one, so the trial is compatible with no difference as well as substantial benefit or some harm. The point estimate favours transfusion, but the data are too imprecise to establish a difference at the 5% level. These are large-sample approximations, especially rough with only one event in an arm. A frequentist confidence interval describes repeated-sampling coverage, not a 95% posterior probability for this fixed parameter.

The event and nonevent cell counts are $(a,b)=(30,60)$ and $(c,d)=(20,40)$. Thus the estimated odds ratio and log odds ratio areFor independent binomial groups, the delta method givesUsing the permitted normal approximation quantile of 2, the approximate 95% interval isThe observed odds of death are equal between groups. The confidence interval includes zero and allows appreciable effects in either direction; it does not demonstrate equivalence. Exponentiating gives an odds ratio interval approximately $(0.493,2.028)$, so the data are compatible with roughly half to twice the death odds for the first treatment relative to the second. The tabulated standard error is rounded.

Let $k=m/2$. After offset subtraction, each summed image has mean $N$ analogue-to-digital units per detector pixel. For a constant detector conversion gain $g$ and independent Poisson distribution detected counts, $\mathrm{Var}(A)=\mathrm{Var}(B)=N/g$: summing $k$ independent frames adds both their means and variances. Here $N$ is the mean of the sum, not one frame.

Write $A=N+a$, $B=N+b$. At high counts, the delta method gives $A/B\simeq 1+a/N-b/N$. Independence therefore givesThus the formula is a high-signal gain estimation from flat-field ratios, not an exact identity for a ratio of random variables. A Poisson distribution denominator can even be zero at low counts. The exposures must be linear and unsaturated, offsets removed, and read noise negligible. Being close to saturation increases the signal but does not establish linearity; clipping or charge-induced correlations biases a variance estimate. A large ensemble of suitable detector pixels supplies the measured standard deviation. In normal photodetector terminology $g$ is collected Electrons per ADU, consistent with detected photons here only under the stated unit-yield convention; see the instrument gain definition.

The almost-sure bound gives ${\tilde{\theta }}_n-{\hat{\theta }}_n=o_{\mathbb{P}}(n^{-1/2})$, so Slutsky theorem transfers both consistency and the asymptotic normal law to ${\tilde{\theta }}_n$. The continuous mapping theorem and delta method then giveIf ${\psi }^{\prime }({\theta }_0)=0$, the latter is the Dirac measure at zero, conventionally a degenerate normal distribution; a nondegenerate second-order limit would need a different normalization and extra smoothness.

Let $x$ be the parameter-contrast vector for the specified game: it has entries $+1$ for ${\beta }_i,{\beta }_j,{\beta }_{\{i,j\}}$, entries $-1$ for ${\beta }_k,{\beta }_{\ell },{\beta }_{\{k,\ell \}}$, and zero elsewhere. PutUnder the stated regularity assumptions, asymptotic normality of a maximum likelihood estimator giveswhere $i_N(\beta )/N\to I(\beta )$. Since $g^{\prime }(\eta )=p(1-p)$, the delta method givesEquivalently, for large $N$,with $p=e^{x^T\beta }/(1+e^{x^T\beta })$.

Let $S_n={\sum }_{i=1}^n{X}_i$. The log-likelihood iswith score function $n/\theta -S_n$ and second derivative $-n/{\theta }^2<0$. The unique maximum-likelihood estimator is thereforeFor an exponential distribution of rate $\theta$,The central limit theorem givesApplying the delta method to $h(x)=1/x$, for which $h^{\prime }(1/\theta )=-{\theta }^2$, yields the exponential-rate maximum-likelihood estimator limit

PutThe central limit theorem gives $\sqrt{n}({\hat{\mu }}_n-\mu ({\theta }_0))\Rightarrow N(0,{\sigma }_h^2)$. Provided ${\mu }^{\prime }({\theta }_0)\ne 0$, the delta method applied to ${\mu }^{-1}$ yieldsStrict monotonicity alone does not exclude ${\mu }^{\prime }({\theta }_0)=0$; in that exceptional case this root-$n$ limit need not hold and higher-order analysis is required.

One does not necessarily have $\mathrm{Var}({\hat{\theta }}_n)\ge [nI({\theta }_0){]}^{-1}$ for every $n$, because ${\hat{\theta }}_n$ need not be unbiased. The Cramer-Rao bound for an estimator with bias $B_n(\theta )$ is insteadunder the corresponding regularity assumptions.

The log-likelihood is, up to constants, $n\alpha \log \beta -\beta {\sum }_i{X}_i$, so the maximum-likelihood estimator is $\hat{\beta }=\alpha /\overline{X}$. By the central limit theorem,Applying the delta method to $g(x)=\alpha /x$ gives