Fisher information in an exponential survival snapshot ($I_T(\lambda )=n{T}^2/(e^{\lambda T}-1)$)

For $n$ independent exponential lifetimes with rate $\lambda$, recording all lifetimes gives Fisher information $n/{\lambda }^2$. Recording only the count alive at a fixed time $T>0$ gives a binomial distribution with success probability $s=e^{-\lambda T}$. Its information in $s$ is $n/[s(1-s)]$, and the change of parameter $ds/d\lambda =-Ts$ gives the displayed information.

The retained fraction is $R(z)=z^2/(e^z-1)$, $z=\lambda T$. It is strictly below one, so every single snapshot loses information. The inequality follows, for instance, from $h(z)=e^z-1-z^2$: its derivative is $e^z-2z>0$ for positive $z$, because that derivative has its minimum at $z=\log 2$ with value $2-2\log 2>0$. Also $R(z)$ tends to zero as $z\downarrow 0$ and as $z\to \infty$.

Differentiating $R$ shows that the best snapshot time solves $2(1-e^{-z})=z$. The difference of the two sides is strictly concave, starts at zero with positive derivative, and eventually becomes negative, so it has exactly one positive root, approximately $1.59362$. The largest retained fraction is about $0.64761$. Thus even an optimally timed snapshot keeps less than two thirds of the full lifetime information. This is a local design result: choosing that time requires a nominal rate or a preceding estimate.