Show a statistic is complete but not suffcient Let $X_1, ...,X_n$$(n \geq 2)$ be i.i.d. random variables having the normal
distribution $N(\theta, 2)$ when $\theta = 0$ and the normal distribution $N(\theta, 1)$ when $θ ∈ R$ and $θ \neq 0$. Show that the sample mean $\bar X$ is a complete
statistic for $θ$ but it is not a sufficient statistic for $θ$.
My idea is to show $E[f(\bar X)]=0$ implies $f(\bar X)=0$ (where $f$ is any measurable function), which is the definition of completeness.And I get the following formula:
$$\int_{{\mathbb{R}}^n}f(\bar X)\exp\left(-\frac{x_1^2+\cdots+x_n^2}{4}\right)\text{d}x_1\cdots\text{d}x_n=0
$$
when $θ = 0$ ,and
$$\int_{{\mathbb{R}}^n}f(\bar X)\exp\left(-\frac{(x_1-\theta)^2+\cdots+(x_n-\theta)^2}{2}\right)\text{d}x_1\cdots\text{d}x_n=0
$$ 
when $θ \neq 0$.
But I don't know how to prove $f(\bar X)=0$ by the two formula above.
Can someone help me?
 A: Very often the most efficient way to find an expected value of a function $g(X_1,\ldots,X_n)$ of random variables $X_1,\ldots,X_n$ is by evalutating the integral
$$
\int_{\mathbb R^n} g(x_1,\ldots,x_n) f_{X_1,\,\ldots\,,\,X_n} (x_1,\ldots,x_n)\,d(x_1,\ldots,x_n) \tag 1
$$
rather than by evaluating
$$
\int_{\mathbb R} g(x_1,\ldots,x_n) f_{g(X_1,\,\ldots\,,\,X_n)} (x)\,dx. \tag 2
$$
However, this seems to be an exception. Here I will use the density, neither of $(X_1,\ldots,X_n)$ nor of $g(\,\overline{X}\,),$ but of $\overline X.$ You have $\overline X$ normally distributed with expected value $\theta$ and variance $2/n.$ Therefore
\begin{align}
\operatorname{E}\left(g\left(\overline X\right)\right) & =  \int_{\mathbb R} g(u) f_{\,\overline X\,} (u) \, du = \int_{\mathbb R} g(u) \frac 1 {\sqrt{2\pi}} \exp\left( \frac{-1} 2 \cdot \frac{(u-\theta)^2}{2/n} \right) du \\[10pt]
& = \frac 1 {\sqrt{2\pi}} \int_{\mathbb R} g(u) \exp\left( \frac{-1} 2 \cdot\frac{u^2-2u\theta+\theta^2}{2/n} \right) du \\[10pt]
& = \frac 1 {\sqrt{2\pi}} \cdot \exp\left( \frac{-n\theta^2} 4 \right) \int_{\mathbb R} \overbrace{\left( g(u)\exp\left( \frac{-nu^2} 4 \right) \right)}^{\Large\text{Call this $h(u).$}} \exp \left( \left( \frac {n\theta} 2\right) u \right) du \\
& \qquad\qquad \text{(The factor that does not depend on $u$ has been pulled out.)} \\[10pt]
& = 0 \text{ only if } \int_{\mathbb R} h(u) \exp\left( \eta u \right) \, du =0. \\[10pt]
\text{i.e. } & \qquad (\mathcal L h)(\eta) = 0 \text{ for every value of }\eta, \\
& \qquad\qquad\text{where $\mathcal L$ is a two-sided Laplace transform.}
\end{align}
At this point one can cite a standard theorem saying that this two-sided Laplace transform is one-to-one, so that it can be everywhere zero only if the function you put into it is (almost) everywhere zero (and "almost" is meant in the measure-theoretic sense, i.e. the measure of the set of points where it is not zero is zero).
This shows completeness.
But you cannot show that's it's not sufficient. It is sufficient.
