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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

Assume independent survival times and positive rate $\lambda$. With $S={\sum }_i{t}_i$, the likelihood is $L(\lambda )={\lambda }^n{e}^{-\lambda S}$ on the observed nonnegative data. The score and negative expected second derivative are $n/\lambda -S$ and $n/{\lambda }^2$. For $S>0$, which holds almost surely under the model,The usual independent-sample likelihood asymptotics give $\sqrt{n}(\hat{\lambda }-\lambda )\Rightarrow N(0,{\lambda }^2)$. Thus an approximate Wald interval isFor a positive interval with the same first-order justification, apply the delta method to $\log \hat{\lambda }$, obtaining endpoints $\hat{\lambda }\exp (\pm z_{1-\alpha /2}/\sqrt{n})$.

A single individual survives until $T$ with probability $s=e^{-\lambda T}$. Thus $N_T\sim \mathrm{Bin}(n,s)$ and the new likelihood is $\left(\genfrac{}{}{0ex}{}{n}{n_T}\right)e^{-\lambda T{n}_T}(1-e^{-\lambda T}{)}^{n-n_T}$. For $0<n_T<n$, its maximum occurs at $s=n_T/n$, givingIf $n_T=0$, the supremum occurs as $\lambda \to \infty$ and there is no finite maximum. If $n_T=n$, it occurs at the boundary $\lambda =0$ of the extended parameter space, or is only a supremum when the rate is required to be strictly positive.

A binomial experiment has information $n/[s(1-s)]$ for $s$. Since $ds/d\lambda =-Ts$, parameter transformation givesAs quantified by Fisher information in an exponential survival snapshot, the ratio of this information to the complete-data information is $z^2/(e^z-1)$ with $z=\lambda T>0$, strictly less than one by the supplied inequality. Hence every fixed survival snapshot loses information. Under regular large-sample conditions its efficient rate estimator has larger asymptotic variance, and correspondingly wider uncertainty intervals, than the estimator based on all individual survival times. This comparison concerns information and asymptotic uncertainty, not a claim that every realized interval in one experiment is wider.

First establish the limiting law needed by the delta method. For fixed $t$, put $u_n=1/2+t/\sqrt{n}$ and let $N_n$ count how many of the $U_i$ are at most $u_n$. Then $N_n$ has a binomial distribution with expectation $n/2+t\sqrt{n}$ and variance $n/4-t^2$. The event $M_U\le u_n$ is exactly $N_n\ge (n+1)/2$. Its standardized threshold tends to $-2t$, so the central limit theorem for these binomial counts givesHere $\Phi$ is the cumulative distribution function of the standard normal distribution. Equivalently,Use the usual local regularity for the delta method: $f$ is continuous at the population median $\theta$ and $f(\theta )>0$. The inverse function theorem, applied at $F(\theta )=1/2$, givesThereforeThe asymptotic normality gives the requested leading variance approximationHere the variance parameter for the rescaled limiting law is $1/(4f(\theta {)}^2)$, while the source states the corresponding $1/n$ approximation for the unscaled estimator. Strict increase of $F$ alone does not force $f(\theta )>0$; that local condition is needed for this form of the asymptotic distribution of a sample median.

For a Bernoulli distribution with fixed $0<p<1$, the mean and variance are $p$ and $p(1-p)$. The odds transformation $g(x)=x/(1-x)$ has derivative $g^{\prime }(x)=(1-x{)}^{-2}$. Applying the delta method from part (ii),More precisely, $\sqrt{n}\{\overline{X}/(1-\overline{X})-p/(1-p)\}$ converges to a normal distribution of mean zero and variance $p/(1-p{)}^3$, after defining the statistic arbitrarily on $\overline{X}=1$. That exceptional event has probability $p^n\to 0$, so this extension does not change the limit. The unextended ratio is undefined there and has no finite exact variance if interpreted as infinite. At $p=0$ it is zero almost surely; at $p=1$ the denominator is always zero. This is the delta method for Bernoulli odds.

For $0<{\sigma }^2<\infty$, the central limit theorem givesIf $g$ is differentiable at $\mu$, the delta method applies to the first-order expansionwhere the remainder notation means that its product with $\sqrt{n}$ tends to zero in probability. HenceThe useful nondegenerate normal approximation requires $g^{\prime }(\mu )\ne 0$. When $g^{\prime }(\mu )=0$, the first-order limit is zero and a higher-order expansion may be needed.

Write $S^2=Q/(n-1)$ and $S=\sqrt{S^2}$, with $n\ge 2$. Two exact facts about a normal sample areand independence of the normal sample mean and sample variance holds. A chi-squared distribution with $n-1$ statistical degrees of freedom is a sum of $n-1$ independent ${\chi }_1^2$ variables, each with mean one and variance two. The central limit theorem therefore givesFor $g(x)=\sqrt{x}$, the derivative at ${\mu }^2$ is $1/(2\mu )$ because $\mu >0$. The delta method gives the asymptotic normality of a normal sample standard deviation:The exact independence of $\overline{X}$ and $S$ makes their limiting normal distributions independent. Consequently,ThusThis has the same leading asymptotic variance as $\hat{\mu }$. The weights $1/3$ and $2/3$ are those in the optimal asymptotic combination of a normal sample mean and standard deviation. This does not require $S$ to be an exactly unbiased estimator of $\mu$.

Write $A=-{\sum }_i\log X_i>0$, with $\log$ the natural logarithm. The likelihood function and log-likelihood function areThe score function is ${\ell }^{\prime }(\theta )=n/\theta -A$ and the second derivative is ${\ell }^{\prime \prime }(\theta )=-n/{\theta }^2<0$. Thus the log-likelihood function is strictly concave, its stationary point is the unique global maximum, and the maximum-likelihood estimator isThe one-observation Fisher information is $I(\theta )=1/{\theta }^2$. The support of a probability distribution here is independent of $\theta >0$, so the regular asymptotic normality of a maximum likelihood estimator givesThere is also a direct check: $U_i=-\log X_i$ has an exponential distribution with rate $\theta$, since $\mathbb{P}(U_i>u)=\mathbb{P}(X_i<e^{-u})=e^{-\theta u}$ for $u>0$. Its mean and variance are $1/\theta$ and $1/{\theta }^2$. Apply the central limit theorem to $A/n$ and the delta method to $h(u)=1/u$ to obtain the same result.

Using the raw moments from part (i), the variance of one observation isThe central limit theorem for the independent statistical sample givesFor $g(x)=x/(1-x)$, its derivative at the mean is $g^{\prime }(\mu )=(1-\mu {)}^{-2}=(\theta +1{)}^2$. The delta method therefore givesEquivalently, the large-sample normal approximation isHere $v$ denotes the leading asymptotic variance, rather than an assertion about an exact finite-sample variance.

For independent samples of $n$ observations from Poisson distributions with positive means $a$ and $b$, the maximum-likelihood estimator of $\nu =\log (b/a)$ is the displayed difference when both totals are positive. Its asymptotic variance is $(a^{-1}+b^{-1})/n$, obtained by the delta method or by inverting the two-parameter Fisher information matrix. A zero group total makes its log-mean estimate a boundary value; finite normal intervals require nonzero totals.

If a response has mean $\mu$ and variance $V(\mu )$, the delta method gives $\mathrm{Var}(g(Y))\approx g^{\prime }(\mu {)}^{2}V(\mu )$. Choosing $g^{\prime }(\mu )$ proportional to $V(\mu {)}^{-1/2}$ makes this first-order variance independent of $\mu$. For $V(\mu )=k{\mu }^2$ on positive means, the resulting transformation is the natural logarithm. This is a local approximation and requires the transformation to be defined on the observations; it does not make arbitrary normal responses positive or preserve an exact normal distribution.