Someone had asked this earlier, but since it was good practice for my qualifying exam coming up, I figured I would ask and share my work on the problem. The problem is:
Suppose $X$ is Unif$(0,\theta)$. Let $T = X_{(1)}$ (minimum order statistic). Is $T$ complete?
My initial thought was no since usually lectures show $X_{(n)}$ is, but here was my attempt:
I assume that $E[g(T)]=0$ for some function $g$ and want to find if $P(g =0)=1$.
$$P(T \leq t) = 1 - P(T > t) = 1-\left(1-\frac{t}{n} \right)^n$$
So taking derivatives to get the pdf:
$$f_T(t) = \left(1-\frac{t}{n}\right)^{n-1}$$
So now working with expectations:
$$0=E[g(T)] = \int_0^\theta g(t)\left(1-\frac{t}{n}\right)^{n-1} dt$$
This integral is differentiable, and since the integral is $0$, then the derivative is $0$, so taking the derivative with respect to $\theta$:
$$0= g(\theta)\left(1-\frac{\theta}{n}\right)^{n-1}$$
Hence $g(\theta) =0$, so $T$ is a complete statistics.
Does this argument work? I basically modeled it after the proof of showing the max order statistic is complete. If $X_{(1)}$ is not complete, where does the proof go wrong?