I'm struggling with proving or disproving the following: Is the sufficient statistic $X_{(1)}$ also complete for the shift parameter in exponential distribution with pdf $$f(x,\delta) = \lambda e^{-\lambda(x - \delta)},$$ where $\lambda > 0$ is known.
I'm actually looking for MVUE (minimum variance unbiased estimator) of $\delta$.
Any hints would be greatly appreciated, thanks
EDIT:
Ok, from the definition I want that the implication $E(w(S)) = 0\;\Rightarrow\;w(S)=0\; a.e.$ holds true $ \forall w\;$ measurable, integrable $\forall \delta,\; $ $S$ being statistic.
If I have only one observation, then this means $$ \int_\delta^\infty w(x) \lambda e^{-\lambda(x-\delta)} \ d{x} = 0 $$ now i can differentiate with respect to $\delta$ to get $$ -w(\delta)\lambda e^{-\lambda}=0 \; a.e. $$ from that simply follows that $w(\delta) =0\; a.e.$
And since the distribution of $X_{(1)}$, where $X_i$ are i.i.d. exponentially distributed, is also exponentially distributed with parameter $n \lambda$. This procedure also works for the minimum.
Am I correct? Thanks