 Solution

The joint likelihood is ${\lambda }^n{e}^{-\lambda T_n}{\prod }_i{1}_{x_i>0}$, so the factorization criterion for sufficiency proves sufficiency. For any two positive sample vectors its likelihood ratio is $e^{-\lambda (T_n(x)-T_n(y))}$, independent of $\lambda$ exactly when their sums agree. The likelihood-ratio criterion for minimal sufficiency therefore makes $T_n$ minimal sufficient.

Its gamma distribution has density ${\lambda }^n{t}^{n-1}{e}^{-\lambda t}/\Gamma (n)$ on $t>0$. If $g(T_n)$ is integrable for every $\lambda$ and has expectation zero for every $\lambda$, thenBy uniqueness of the Laplace transform, $g(t)t^{n-1}=0$ almost everywhere. One may first multiply by $e^{-{\lambda }_{0}t}$ to obtain an integrable function and use transform parameters $\lambda >{\lambda }_0$; this handles the transform's integrability hypothesis. Since the gamma densities are positive on $(0,\infty )$, $g(T_n)=0$ almost surely for every $\lambda$. $T_n$ is complete as well as minimal sufficient.