Existence of complete sufficient statistic I'm trying to solve a problem in problem list of qualifying exam. 
Here is my problem : Let $X_1, \ldots, X_n$ be a random sample from a p.d.f $f(x,\theta) = \theta f_1(x) 1_{(-\infty,0)} + (1-\theta) f_2(x) 1_{(0,\infty)}$
where $f_1 \geq 0 , f_2 \geq 0$ and $\int_{-\infty}^0 f_1=0$,  $\int_0^\infty f_2=0.$
Prove or disprove that there exists a complete sufficient statistic for $\theta$.
I'm trying to show that there is only one sufficient statistic $T(X_1, \ldots, X_n) =(X_1, \ldots, X_n)$ for $\theta$ and that such $T$ is not complete. But I cannot even show the uniqueness of such sufficient statistic. Anyone can help me?
 A: I'm going to guess that where you wrote $\int_{-\infty}^0 f_1=0,$  $\int_0^\infty f_2=0,$ you meant $\int_{-\infty}^0 f_2=0,$  $\int_0^\infty f_1=0.$
Thus you have
\begin{align}
\int_{-\infty}^\infty f_1 = \int_{-\infty}^0 f_1 = 1, \\[10pt]
\int_{-\infty}^\infty f_2 = \int_0^\infty f_2 = 1.
\end{align}
Thus for $i=1,\ldots,n$ you have $\Pr(X_i<0)=\theta$ and $\Pr(X_i>0) = 1-\theta.$
Let $Y$ be the number of negative observations in $\{X_1,\ldots,X_n\}.$ Then
$$
Y\sim\operatorname{Binomial}(n,\theta).
$$
The conditional density of $X_i$ given that $X_i<0$ is $f_1$, and given that $X_i>0$ is $f_2$.
See if you can show that the conditional distribution of $X_1,\ldots,X_n$ given that $Y=y$ is the same as the distribution of $X_1,\ldots,X_n$ when $\theta= y/n.$ Thus this conditional distribution does not depend on $\theta.$
Next there's the question of completeness. You have
$$
\operatorname{E}(g(Y)) = \sum_{y=0}^n g(y) \binom n y \theta^y (1-\theta)^{n-y}.
$$
If a polynomial in $\theta$ is equal to $0$ regardless of the value of $\theta$ then all of the coefficients are $0$. See if you can use that to show that $g(y)=0$ for every $y\in\{0,1,2,\ldots,n\}.$
