= Solution
A <sufficient statistic> $T(X)$ for $\theta$ is a function of the observations whose conditional distribution of $X$ given $T(X)$ does not depend on $\theta$, whenever that conditioning event is possible. It retains all parameter dependence of the <likelihood>.
In the discrete case the <Fisher-Neyman factorization theorem> provides a direct criterion: $f(x\mid\theta)=g(T(x),\theta)h(x)$, where $h$ is independent of the parameter. To justify sufficiency, for a value $t$ with positive probability, sum this expression over its fiber and divide:
$$
\mathbb P_\theta(X=x\mid T=t)=\frac{h(x)}{\sum_{y:T(y)=t}h(y)},\qquad T(x)=t.
$$
The parameter-dependent factor cancels. Conversely, parameter-independent conditional probabilities $h_t(x)$ give the factorization $f(x\mid\theta)=\mathbb P_\theta(T=t)h_t(x)$ with $t=T(x)$. Thus the criterion is both necessary and sufficient, with conditional laws only required on fibers of positive probability.
For the shifted exponential sample the <likelihood> is
$$
f(x\mid\theta)=\lambda^n e^{-\lambda\sum_i x_i}e^{n\lambda\theta}\mathbf1_{\{\theta\le\min_i x_i\}},\qquad\theta\ge0.
$$
It factors through $T=X_{(1)}=\min_iX_i$, so \b[the sample minimum is a <sufficient statistic>]. Writing $Y_i=X_i-\theta$, independence gives for $t\ge0$
$$
\mathbb P(T-\theta>t)=\prod_i\mathbb P(Y_i>t)=e^{-n\lambda t}.
$$
Therefore $T-\theta$ is exponential with rate $n\lambda$. Its <mean> is $(n\lambda)^{-1}$ and its <variance> is $(n\lambda)^{-2}$. The <unbiased endpoint estimator for a shifted exponential sample> is consequently
$$
\boxed{\widehat\theta=X_{(1)}-\frac1{n\lambda},\quad
\mathbb E_\theta[\widehat\theta]=\theta,\quad
\operatorname{Var}_\theta(\widehat\theta)=\frac1{(n\lambda)^2}.}
$$
Although the parameter is nonnegative, this <unbiased estimator> can be negative. Truncating it at zero would change its expectation and lose the requested property of an <unbiased estimator>.
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