 Solution

For $S={\sum }_i{t}_i$, the likelihood is $L(\lambda )={\lambda }^n{e}^{-\lambda S}$, and its log derivative is $n/\lambda -S$. Thus, almost surely,The large-sample distribution is that of the estimator, rather than of the fixed parameter: $\sqrt{n}(\hat{\lambda }-\lambda )\stackrel{d}{\to }N(0,{\lambda }^2)$. A Wald confidence interval is consequentlyintersected with the positive parameter space if necessary. The unadorned parameter in the source's distribution request is interpreted as this estimator.

A single aphid survives to $T$ with probability $r=e^{-\lambda T}$, independently of the others. Hence $N_T\sim \mathrm{Bin}(n,r)$, andfor $0<n_T<n$. If $n_T=n$, the optimum is at the boundary $\lambda =0$; if $n_T=0$, the likelihood increases to its supremum as $\lambda \to \infty$, so there is no finite maximum likelihood estimate. Fisher information transforms from the binomial probability parameter asThe given strict inequality showsThresholding discards information about how long each survival time is. The second experiment therefore has a larger information bound on variance and larger asymptotic uncertainty for an efficient estimator. This is an expected-information comparison, rather than a claim that every realised finite-sample confidence interval must be wider.