Sufficient statistics for $p$ for a random sample from $\text{Ber}(p)$ distribution Let $X_1,X_2, X_3$ be a random sample from Bernoulli distribution $B(p)$. Which of the following is sufficient statistic for $p$ ?
$(A)\ \ X_{1}^{2}+X_{2}^{2}+X_{3}^{2}$
$(B) \ \ X_1+2X_{2}+X_{3}$
$(C)\ \ 2X_1-X_{2}-X_{3}$
$(D) \ \ X_1+X_{2}$
$(E)\ \ 3X_1+2X_{2}-4X_{3}$
We know $T=\sum_i^3 X_i$ is sufficient statistic for $p$ via Neymann Factorization theorem. 
The reasoning for different options. 
$(A)$ It's not a one-one function of sufficient statistic $T$ (If it was $T^2$ , was it sufficient statistic then? My answer is yes because it is one to one function of sufficient statistic our random variable is positive am I right?).
$(B)X_1+X_{2}+X_{3}+X_{2}$ It contains original statistic therefor it is suffient statistic for $p$.
$(D)$ It doesn't include $X_3$ So it's not sufficient statistic for $p$
All other options include subtraction in it so I ruled out all of them.
I think my reasoning is not very good I lack some intuition behind finding sufficient statistic. Correct me here, please.  
 A: Since $T=X_1+X_2+X_3$ is a minimal sufficient statistic for $p$, implying that it is a function of every other sufficient statistic, option (D) is eliminated.
Now $X_i^2$ is a one-to-one function function of $X_i$ because $X_i\in\{0,1\}$ for each $i$. 
So $X_1^2+X_2^2+X_3^2$ is also a one-to-one function of $T$, implying that the former is sufficient for $p$. (Thanks to @Alex for pointing this out)
For the remaining options I create a table of the possible values that the statistics can take:
\begin{array}{|c|c|c|}
\hline (X_1,X_2,X_3)&T_1=X_1+2X_2+X_3&T_2=2X_1-X_2-X_3&T_3=3X_1+2X_2-4X_3\\ \hline(0,0,0)&0&0&0\\ 
\hline(0,0,1)&1&-1&-4\\ 
\hline(0,1,0)&2&-1&2\\ 
\hline (0,1,1)&3&-2&-2\\
\hline (1,0,0)&1&2&3\\
\hline (1,0,1)&2&1&-1\\
\hline (1,1,0)&3&1&5\\
\hline (1,1,1)&4&0&1\\
\hline
\end{array}
I am looking whether $T_1,T_2,T_3$ are one-to-one functions of the sample $(X_1,X_2,X_3)$ or not. If they are, then they are sufficient statistics. If you can see right away that the only bijection is $$T_3:\{0,1\}^3\to\{-4,-2,\ldots,3,5\}$$, then you are done. Because a bijection with the sample implies that observing $T_3(X_1,X_2,X_3)$ and observing $(X_1,X_2,X_3)$ are equivalent. In other words, $T_3$ is a sufficient statistic for $p$.
Alternatively, if you consider the case $T_1=2$, you will find that the conditional distribution $P(\{X_1,X_2,X_3\}\mid T_1)$ depends on $p$. Similar argument holds for $T_2$ if you consider the case $T_2=0$. Also it is apparent from the table above that $P(\{X_1,X_2,X_3\}\mid T_3)$ is independent of $p$ for all possible values of $T_3$, because $T_3$ takes 8 distinct values corresponding to 8 different tuples. 
I used the following threads for reference:


*

*Verification of sufficiency of a linear combination of the sample $(X_i)_{i\ge1}$ where $X_i\stackrel{\text{i.i.d}}\sim\text{Ber}(\theta)$

*Sufficiency or Insufficiency
