Show that $T$ is not a sufficient statistic Let $X_1\: , X_2$ a random sample for $N(\theta ,1).\:$ Show that $\:T=X_1 + 2X_2$ is not a sufficient statistic.
I've tried to prove it by contradiction:
I assumed that $T$ is sufficient. That means that two nonnegative functions $g,\:h$ can be found such that $$f_{\theta}(x)=g_{\theta}(T(x))*h(x)$$
So
$$f_{\theta}(x)=\frac{1}{2\pi}\:exp\{-\frac{1}{2}[(x_1-\theta)^2+(x_2-\theta)^2]\}$$
Then
$$f_{\theta}(x)=\frac{1}{2\pi}\:exp\{-\frac{(x_1^2+x_2^2)}{2}+\theta(x_1+x_2)-\theta^2\}$$
Then I multiply the $\:\theta(x_1+x_2)$ term by $\frac{2}{2}$ so:
$$f_{\theta}(x)=\frac{1}{2\pi}\:exp\{-\frac{(x_1^2+x_2^2)}{2}+\frac\theta 2(2x_1+2x_2)-\theta^2\}$$
Therefore
$$f_{\theta}(x)=\frac{1}{2\pi}\:exp\{-\frac{(x_1^2+x_2^2)}{2}+\frac\theta 2(x_1+ x_1+2x_2)-\theta^2\}$$
$$f_{\theta}(x)=\frac{1}{2\pi}\:exp\{-\frac{(x_1^2+x_2^2)}{2}+\frac\theta 2(x_1+T)-\theta^2\}$$
Then I'm stuck right here, I'm a little bit confused by the Fischer-Neyman factorization theorem, because I don't know if I can define
$$g_\theta(T(x))=exp\{\frac\theta 2 (x_1+T)-\theta^2\}$$
and
$$h(x)=\frac1{2\pi} exp\{-\frac{(x_1^2+x_2^2)}{2}\}$$
which is obviously a proof that T is sufficient.
 A: If $T = X_1 + 2X_2$ were sufficient, then the conditional distribution of $(X_1, X_2)$ given $T$ would be independent of $\theta$. In particular, the conditional density of $X_1$ given $T = t$ should be independent of $\theta$. So let's compute this conditional density. Since $X_1$ and $X_2$ are independent, it follows that
$$\begin{pmatrix}X_1 \\ X_2\end{pmatrix} \sim \mathcal{N}\left(\begin{pmatrix}\theta \\ \theta\end{pmatrix}, I_2\right),$$
where $I_2$ denotes the $2 \times 2$ identity matrix. It then follows that
$$\begin{pmatrix}X_1 \\ T\end{pmatrix} = \begin{pmatrix}X_1 \\ X_1 + 2X_2\end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 1 & 2\end{pmatrix}\begin{pmatrix}X_1 \\ X_2\end{pmatrix} \sim \mathcal{N}\left(\begin{pmatrix}\theta \\ 3\theta\end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 1 & 5\end{pmatrix}\right).$$
Therefore, by the conditional distribution theorem for multivariate normal distribution, we have given $T = t$, the conditional distribution of $X_1$ is 
$$\mathcal{N}\left(\theta + 1 \times \frac{1}{5}(t - 3\theta), 1 - \frac{1}{5}\right) = \mathcal{N}\left(\frac{1}{5}t + \frac{2}{5}\theta, \frac{4}{5}\right),$$
which is NOT independent of $\theta$. Therefore $T$ is not sufficient. 
