Let $X_i\big|_{i = 1...n}$ be random sample from the PMF:
$P(X_i = 0) = \frac{1-\theta}2;\;P(X_i = 1) = \frac12 ; P(X_i = 2) = \frac\theta2$ where $\theta\in(0,1)$. Find the minimal sufficient statistics.
Let $X = (X_1, \ldots, X_n) $ and $x = (x_1, \ldots, x_n)$, then
$P(X = x, \theta ) \\= \prod_{i=1}^n \big(\frac{1-\theta}2\big)^{I(x_i = 0)} \big(\frac{1}2\big)^{I(x_i = 1)} \big(\frac{\theta}2\big)^{I(x_i = 2)}\\ = \prod_{i=1}^n \big(\frac{1}2\big)^{I(x_i = 0) +I(x_i = 1) + I(x_i = 2)} (1-\theta)^{I(x_i = 0)} \;\theta^{I(x_i = 2)} \\ = \prod_{i=1}^n \frac12 (1-\theta)^{I(x_i = 0)} \;\theta^{I(x_i = 2)} \text{ because } I(x_i = 0) +I(x_i = 1) + I(x_i = 2) = 1\\ = \frac{1}{2^n}(1-\theta)^{\sum_{i = 1}^nI(x_i = 0)} \theta^{\sum_{i=1}^nI(x_i = 2)} $
Now, $$ \frac{P(X= x, \theta )}{P(X = y, \theta)} = (1-\theta)^{\sum_{i = 1}^nI(x_i = 0)- \sum_{i = 1}^nI(y_i = 0)} \;\theta^{\sum_{i=1}^nI(x_i = 2) - \sum_{i = 1}^nI(y_i = 2)}$$ is independent of $\theta$ iff $$\sum_{i = 1}^nI(x_i = 0)= \sum_{i = 1}^nI(y_i = 0)$$ and $$\sum_{i=1}^nI(x_i = 2) = \sum_{i = 1}^nI(y_i = 2)$$
Hence, $T(X) = (\sum_{i=1}^nI(X_i = 0), \sum_{i=1}^nI(X_i = 2))$ is minimal sufficient.
Is my attempt correct?