 Solution

For the fixed observation interval, $N_3$ has Poisson distribution with mean and variance $3\lambda$. Hence $N_3/3$ is unbiased and has variance $\lambda /3$. Its score function is $N_3/\lambda -3$, and its Fisher information is $3/\lambda$. The Cramér-Rao bound for unbiased estimation of $\lambda$ therefore gives variance at least $\lambda /3$, attained by $N_3/3$.

For the stopping-time observation, $T_{10}$ has Gamma distribution with shape $10$ and rate $\lambda$. Direct integration of its density givesThus $k=9$, andThe score function is $10/\lambda -T_{10}$, so the Fisher information is $10/{\lambda }^2$. Its Cramér-Rao lower bound is ${\lambda }^2/10$, strictly below this estimator's variance.

Nevertheless, $9/T_{10}$ is uniformly minimum-variance unbiased. The observed $T_{10}$ is a sufficient statistic, and its family is complete: if ${\mathbb{E}}_{\lambda }h(T_{10})=0$ for all $\lambda >0$, then ${\int }_0^{\infty }h(t)t^9{e}^{-\lambda t}dt=0$ for all $\lambda >0$. Uniqueness of the Laplace transform, applied after an exponential tilt to make an Lebesgue integrable function, implies $h=0$ almost everywhere. The Lehmann–Scheffé theorem now proves the minimum-variance assertion. Failure to attain the Cramér–Rao bound does not disprove minimum variance within the class of unbiased estimators.

For the actual observations, the two likelihood functions areThey differ by a factor independent of $\lambda$. The likelihood principle consequently requires the same inferential conclusions, since the observed data carry proportional likelihoods. But the two minimum-variance unbiased estimates are $\boxed{10/3\text{ and }3}$ respectively. Minimum-variance unbiased estimation therefore violates the likelihood principle: its unbiasedness and variance refer to the sampling plan, including the stopping rule, rather than solely to the observed Likelihood function.