Sufficient statistic of Bernoulli Trial Let $X_1$ and $X_2$ be iid random variables from a $Bernoulli(p)$ distribution. Verify if the statistic $X_1+2X_2$ is sufficient for $p$.
I calculated and found out $X_1+X_2$ as a sufficient statistic for $p$.
Is this enough to rule out the possibility of $X1+2X2$ as a sufficient statistic?
Is there a better way to show that explicitly?
Addendum: I want to check for $X_1+2X_2$, sorry for the mistake. It is now edited.
 A: Let $T=X_1+2X_2$ , $S=X_1+X_2$. We know $S$ is a minimal sufficient statistics.
$\{T=0\}=\{ (0,0)\}$
$\{T=1\}=\{ (1,0)\}$
$\{T=2\}=\{ (0,1)\}$
$\{T=3\}=\{ (1,1)\}$
$\sigma(T)=\sigma\bigg( \color{red}\{(0,0)\color{red}\} ,\color{red}\{(1,0)\color{red}\} , \color{red}\{(0,1)\color{red}\},\color{red}\{(1,1)\color{red}\} \bigg)$
$\sigma(S)=\sigma\bigg( \color{red}\{(0,0)\color{red}\} ,\color{red}\{(1,0), (0,1)\color{red}\} , \color{red}\{(1,1)\color{red}\} \bigg)$
where $\sigma(T)$ denotes the sigma generated by T and
$\sigma(S)$ denotes the sigma generated by S.
Since $\sigma(S)\subset \sigma(T)$ (the information in $T$ is more than $S$) ,$S$ is a minimal sufficient statistic and $S$ is a function of $T$ ,hence $T$ is a sufficient statistic(But not a minimal one). We can also compare it with $\sigma(X_1,X_2)$
and find $\sigma(X_1,X_2)=\sigma(T)$ ($T$ and $(X_1,X_2)$ have a same information) and obtain that $T$ is a sufficient statistics.
We can also use the definition of a sufficient statistics as follows:
\begin{eqnarray}
P(X_1=x_1,X_2=x_2|T=t)=
\left\{
\begin{array}{cc}
*1 & t=0 \\
*2 & t=1 \\
*3 & t=2 \\
*4 & t=3
\end{array}
\right.
\end{eqnarray}
and find *1,*2,*3 and *4. For example(*1)
\begin{eqnarray}
P(X_1=x_1,X_2=x_2|T=0)=
\left\{
\begin{array}{cc}
1 & x_1=0,x_2=0 \\
0 & O.W. 
\end{array}
\right.
\end{eqnarray}
and in all cases it does not depend of the parameter.
A: Since $T \equiv X_1+X_2$ is a sufficient statistic, the question boils down to whether or not you can recover the value of this sufficient statistic from the alternative statistic $T_* \equiv X_1 + 2 X_2$.  Formally, is there any function that maps $T_*$ to $T$?  The other answer by Masoud gives you the information you need to construct such a mapping, so use this to have a go constructing a function of this kind.  You can then appeal directly to the Fisher-Neyman factorisation to show that the latter statistic is also sufficient.
