Exercise "Mathematical Statistics - Jun Shao" I'm trying to solve this problem:

Let $X_1, ...,X_n, (n \ge 2)$ be i.i.d. random variables having the
  normal distribution $N(\theta, 2)$ when $\theta=0$ and the normal
  distribution $N(\theta, 1)$ when $\theta \in {\rm I\!R}-\{0\}$. Show that the sample mean $\bar{X}$ is not a sufficient
  statistic for $\theta$.

So, first I found the sample distributions, 
$\bar{X} \sim N(0,2/n)$, when $\theta = 0$, and $\bar{X} \sim N(\theta,1/n)$, when $\theta \neq 0$.
Then I wrote the function as
\begin{align}
f_{ \theta  }(x)={ \left[ { (4\pi ) }^{ -1/2 }\exp\{ \frac { -{ x }^{ 2 } }{ 4 } \}  \right]  }^{ I_{ \{ \theta =0\}  } }{ \left[ { (2\pi ) }^{ -1/2 }\exp\{ \frac { -{ (x-\theta ) }^{ 2 } }{ 2 } \}  \right]  }^{ I_{ \{ \theta \neq 0\}  } }
\end{align}
But I'm not sure how to procedure with that. I thought of using the factorization theorem, but I'm stuck in this density. Any hint?
 A: Another way of expressing the density of a single observation is to write $$X \sim \operatorname{Normal}(\theta, 1 + \mathbb 1(\theta = 0));$$ that is to say, the variance is a function of $\theta$: $$\sigma_\theta^2 = 1 + \mathbb 1 (\theta = 0) = \begin{cases} 2, & \theta = 0 \\ 1, & \theta \ne 0. \end{cases}$$  Then the joint density of a sample $\boldsymbol x \in \mathbb R^n$ is given by $$\begin{align*} f(\boldsymbol x \mid \theta) &= (2\pi)^{-n/2}\sigma_\theta^{-n} \exp\left(-\sum_{i=1}^n \frac{(x_i - \theta)^2}{2\sigma_\theta^2} \right) \\ &= (2\pi)^{-n/2} (1 + \mathbb 1 (\theta = 0))^{-n} \exp \left(-\frac{\sum_{i=1}^n (x_i - \theta)^2}{2(1 + \mathbb 1 (\theta = 0))^2} \right). \end{align*}$$  I leave it as an exercise to show that $$\sum_{i=1}^n (x_i - \theta)^2 = \sum_{i=1}^n (x_i - \bar x)^2 + \sum_{i=1}^n (\bar x - \theta)^2,$$ so that $$f(\boldsymbol x \mid \theta) = (2\pi)^{-n/2} \sigma_\theta^{-n} \exp\left(-\frac{n \hat \sigma^2}{2\sigma_\theta^2}\right) \exp\left(-\frac{\sum_{i=1}^n (\bar x - \theta)^2}{2\sigma_\theta^2}\right),$$ where $\hat\sigma^2 = \frac{1}{n}\sum_{i=1}^n (x_i - \bar x)^2$ is the (biased) sample variance.  If $T(\boldsymbol x) = \bar x$ were sufficient for $\theta$, then we would not have this additional factor containing $\hat\sigma^2$.  Unfortunately, we cannot express $f$ in terms of $h(\boldsymbol x) g(T(\boldsymbol x) \mid \theta)$ for suitable $h$ and $g$ in such a case because the factor $\exp(-n\hat\sigma^2/(2\sigma_\theta^2))$ depends on $\theta$ through $\sigma_\theta$, and this is the crux of the problem, since if the variance did not depend on $\theta$, this factor does not depend on the sample through $\theta$ and would not need to be included in $g$; instead, it would be included in $h$.  However, the two-dimensional statistic $$\boldsymbol T(\boldsymbol x) = (\bar x, \hat \sigma^2)$$ is sufficient for $\theta$ because now we can choose, for example, $h(\boldsymbol x) = (2\pi)^{-n/2}$ and $$g(T_1, T_2 \mid \theta) = \sigma_\theta^{-n} \exp \left(-\frac{-n T_2}{2\sigma_\theta^2}\right) \exp \left(-\frac{\sum_{i=1}^n (T_1 - \theta)^2}{2\sigma_\theta^2}\right).$$
The above, of course, is not entirely rigorous because we have not formally proven that $\hat\sigma^2$ is not itself a function of $\bar x$.  To show this, it suffices to furnish an example of two samples $\boldsymbol x_1$, $\boldsymbol x_2 \in \mathbb R^n$, for which $\bar x_1 = \bar x_2$ but $\hat \sigma_1^2 \ne \hat \sigma_2^2$.  This is quite trivial; e.g., let $n = 2$ and $\boldsymbol x_1 = (-1,1)$ and $\boldsymbol x_2 = (-10, 10)$.  Then $\bar x_1 = \bar x_2 = 0$ but obviously $\hat\sigma_1^2 < \hat\sigma_2^2$.  Consequently, $\hat \sigma^2$ is not a uniquely determined function of $\bar x$ for any $n \ge 2$ (since in $n > 2$ we can choose all the other observations to be equal, reducing to the $2$-dimensional case).
