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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.

Given the complete stopped sample $(x_1,\dots ,x_n)$, the likelihood function isThe stopping factor is positive and independent of $\theta$. The Fisher-Neyman factorization theorem makes $(N,S)$ sufficient. For two possible stopped samples of lengths $n,m$ and sums $S,S^{\prime }$, their likelihood ratio is a parameter-free factor times ${\theta }^{n-m}{e}^{-\theta (S-S^{\prime })}$. This is independent of $\theta$ exactly when $n=m$ and $S=S^{\prime }$: differentiating its logarithm gives $(n-m)/\theta -(S-S^{\prime })=0$ for all $\theta >0$. By the likelihood-ratio criterion,The likelihood principle says proportional likelihoods give the same evidence about the parameter. Thus with the full sample observed, this parameter-free stopping rule adds no likelihood evidence beyond the realized values and length; it gives the same likelihood-based inference as a fixed-length sample with those values. This statement does not assert that frequentist sampling distributions are unchanged.

Let $q(\theta )={\int }_0^{\infty }\alpha (x)\theta e^{-\theta x}dx\in (0,1)$ be the chance of stopping at a given stage. Independence of successive stages givesTherefore $a(y)=\log \alpha (y)$ and $k(\theta )=\log {\int }_0^{\infty }{e}^{a(y)-\theta y}dy$ give the requested exponential family form.

Writing $p=p_Y$, differentiation in $y$ gives $p^{\prime }=(a^{\prime }-\theta )p$. Integrating and using $p(0)=p(\infty )=0$ shows ${\mathbb{E}}_{\theta }{a}^{\prime }(Y)=\theta$. LikewiseThe stipulated vanishing of $p^{\prime }$ at both boundaries givesThese integrations require the displayed moments and integrals to exist, the usual regularity interpretation of the question; when establishing a finite variance this assumption should be explicit.

Differentiating the normalization in $\theta$ gives $k^{\prime }(\theta )=-{\mathbb{E}}_{\theta }Y$ and $k^{\prime \prime }(\theta )={\mathrm{Var}}_{\theta }Y$. The score function is $-k^{\prime }(\theta )-Y$, so the Fisher information is $I(\theta )=k^{\prime \prime }(\theta )$. The Cramér-Rao bound states that a regular unbiased estimator of $\theta$ has variance at least $1/I(\theta )$. Apply it to $a^{\prime }(Y)$:Equivalently differentiate ${\mathbb{E}}_{\theta }{a}^{\prime }(Y)=\theta$ and use the Cauchy-Schwarz inequality to get $1\le \sqrt{\mathrm{Var}(a^{\prime }(Y))\mathrm{Var}(Y)}$, which gives the same conclusion.