How to prove or disprove that $T(X_{1},X_{2}) = X_{1} + X_{2}$ is a sufficient statistic

Question

Let $X_{1},X_{2},\ldots,X_{n}$ be random sample from a population whose distribution is given by $X\sim\text{Bernoulli}(\theta)$, $0 < \theta < 1$.

a. Show that $T(x) = \displaystyle\sum_{i=1}^{n}X_{i}$ is sufficient for $\theta$.

b. In the same context, consider that $T = X_{1} + X_{2}$. What is the distribution of $T$? Is $T$ a sufficient statistic?

MY ATTEMPT

a. To begin with, let us determine the likelihood function for this sample \begin{align*} L(\textbf{x}|\theta) = \prod_{i=1}^{n}\theta^{x_{i}}(1-\theta)^{1-x_{i}} = \theta^{\sum x_{i}}(1-\theta)^{n - \sum x_{i}} = h(\textbf{x})g_{\theta}(T(\textbf{x})) \end{align*}

Therefore, according to the factorization theorem, $T(\textbf{x})$ is sufficient for $\theta$.

b. As it is known, the sum of independent Bernoulli random variables is a Binomial random variable. Therefore $T = X_{1} + X_{2} \sim \text{Binomial}(2,\theta)$.

Then I get stuck. Can someone help me out? Thanks in advance!

Answer 1

All you need to show is that $$P(X_1=x_1,X_2=x_2,...X_n=x_n|X_1+X_2 = t )$$ depends on $\theta$ to prove that $T=X_1+X_2$ is NOT a sufficient statistic.