Suppose we have to identically distributed Bernoulli$(p)$ random variables $X_1, X_2$ which are not necessarily independent. We would like to show that $X_1+X_2$ is a sufficient estimator for $p$.
To do this, I tried to compute $$ P(X_1=x_1, X_2=x_2 \mid X_1+X_2 = t) $$ Note that if $x_1+x_2 \neq t$ then the above evaluates to $0$. There remains 4 cases: \begin{align*} P(X_1=0, X_2=0 \mid X_1+X_2 = 0)\\ P(X_1=1, X_2=0 \mid X_1+X_2 = 1)\\ P(X_1=0, X_2=1 \mid X_1+X_2 = 1)\\ P(X_1=1, X_2=1 \mid X_1+X_2 = 2) \end{align*} Clearly, $$ P(X_1=0, X_2=0 \mid X_1+X_2 = 0) = P(X_1=1, X_2=1 \mid X_1+X_2 = 2) = 1 $$
How do I handle the 2 other cases?
I know that if $X_1, X_2$ are independent then $$ P(X_1=1, X_2=0 \mid X_1+X_2 = 1) = P(X_1=0, X_2=1 \mid X_1+X_2 = 1) = 1/2 $$ On the other hand, if one picks $X_1$ and $X_2=X_1$ then $$ P(X_1=1, X_2=0 \mid X_1+X_2 = 1) = P(X_1=0, X_2=1 \mid X_1+X_2 = 1) = 0 $$ The answer will therefore depend on something, but I am not sure how to formulate a general solution.
If $\mathbb P(X_1+X_2 = 1)=0$, then one need not to consider this conditional probability. In other case, $$ \mathbb P(X_1=1, X_2=0 \mid X_1+X_2 = 1) = \dfrac{\mathbb P(X_1=1, X_2=0, X_1+X_2 = 1)}{\mathbb P(X_1+X_2 = 1)}= \dfrac{\mathbb P(X_1=1, X_2=0)}{\mathbb P(X_1=1, X_2=0)+\mathbb P(X_1=0, X_2 =1)} $$ Prove that both probabilities in the denominator are coincide: $$ \mathbb P(X_1=0, X_2=1) + \mathbb P(X_1=1, X_2=1) = \mathbb P(X_2=1)=p, $$ $$ \mathbb P(X_1=1, X_2=0) + \mathbb P(X_1=1, X_2=1) = \mathbb P(X_1=1)=p, $$ and then $$ \mathbb P(X_1=0, X_2=1) = p - \mathbb P(X_1=1, X_2=1) = \mathbb P(X_1=1, X_2=0). $$ Therefore, the required conditional probability equals to $1/2$: $$ \mathbb P(X_1=1, X_2=0 \mid X_1+X_2 = 1) = \dfrac{\mathbb P(X_1=1, X_2=0)}{2\mathbb P(X_1=1, X_2=0)}=\frac12. $$ And $$ \mathbb P(X_1=0, X_2=1 \mid X_1+X_2 = 1) = \dfrac{\mathbb P(X_1=0, X_2=1)}{2\mathbb P(X_1=0, X_2=1)}=\frac12. $$ So, the sum is sufficient statistics.