Sufficient Statistic for a family of distributions consisting of Poisson family and Bernoulli family

Question

Suppose $(X_1, . . . ,X_n)$ is an i.i.d. sample from the distribution $f_{\theta,k}(x)$, where $\theta \in (0, 1)$ and $k = 1, 2$. Assume that $$f_{\theta, k}(x)=\begin{cases} \text{Poisson($\theta)$}, &\text{if $k=1$}.\\ \\ \text{Bernoulli($\theta$)}, & \text{if $k=2$}. \end{cases}$$. Check if $T=\sum_{i=1}^nX_i$ is a sufficient statistic for this family. If not, then find a sufficient statistic for this family. $$$$My Attempt to the solutions is as follows : I found that $$\mathbb{P}(X_1=x_1, ...,X_n=x_n|T=t) =\begin{cases} \frac{n!}{x_1!x_2!....x_n!}(\frac{1}{n})^t & \text{if} &\sum_{i=1}^nx_i=t, &X_1, ...., X_n \sim \text{Poisson}(\theta) \\ \\\frac{1}{n \choose t} &\text{if} &\sum_{i=1}^nx_i=t, &X_1, ...., X_n \sim \text{Bernoulli}(\theta) \end{cases}$$ So $T$ is not sufficient for this family. $$$$Now we can write the joint density as $$f_{\theta, k}(x_1, ...., x_n)=\frac{e^{-n\theta}(\theta)^{\sum_{i=1}^nx_i}}{\prod_{i=1}^n(x_i)!}\textbf{1}(k=1)+(\theta)^{\sum_{i=1}^nx_i}(1-\theta)^{n-\sum_{i=1}^nx_i}\textbf{1}(0 \leq x_{(1)} \leq x_{(n)} \leq 1)\textbf{1}(k=2)$$ The indicator $\textbf{1}(0 \leq x_{(1)} \leq x_{(n)} \leq 1)$ is because the support in our case is $\chi=\mathbb{N} \cup 0$. So by the Factorization Theorem we get that $T(X_1, ...., X_n)=(\sum_{i=1}^nX_i, \prod_{i=1}^n(X_i)!, X_{(1)}, X_{(n)})$ is a sufficient statistic for this family as we can take $g_{\theta, k}(T(x_1, ...., x_n))$ equal to the density and $h(x_1, ...., x_n)=1$. $$$$Now to find the minimal sufficient statistic we consider the ratio $$\frac{f_{\theta, k}(x_1, ...., x_n)}{f_{\theta, k}(y_1, ...., y_n)}=\frac{\frac{e^{-n\theta}(\theta)^{\sum_{i=1}^nx_i}}{\prod_{i=1}^n(x_i)!}\textbf{1}(k=1)+(\theta)^{\sum_{i=1}^nx_i}(1-\theta)^{n-\sum_{i=1}^nx_i}\textbf{1}(0 \leq x_{(1)} \leq x_{(n)} \leq 1)\textbf{1}(k=2)}{\frac{e^{-n\theta}(\theta)^{\sum_{i=1}^ny_i}}{\prod_{i=1}^n(y_i)!}\textbf{1}(k=1)+(\theta)^{\sum_{i=1}^ny_i}(1-\theta)^{n-\sum_{i=1}^ny_i}\textbf{1}(0 \leq y_{(1)} \leq y_{(n)} \leq 1)\textbf{1}(k=2)}$$ Clearly if $T(x_1, ...., x_n)=T(y_1, ...., y_n)$ then the ratio is equal to $1$ which is a constant function of $\theta, k$. Now suppose that this ratio is a constant function of $\theta, k$. If we take $k=1$, then the ratio is $$\frac{f_{\theta, k=1}(x_1, ...., x_n)}{f_{\theta, k=1}(y_1, ...., y_n)}=\frac{\frac{e^{-n\theta}(\theta)^{\sum_{i=1}^nx_i}}{\prod_{i=1}^n(x_i)!}}{\frac{e^{-n\theta}(\theta)^{\sum_{i=1}^ny_i}}{\prod_{i=1}^n(y_i)!}}$$ which is a constant function of $\theta$ iff $\sum_{i=1}^nx_i=\sum_{i=1}^ny_i$ and that constant value is $\frac{\prod_{i=1}^n(y_i)!}{\prod_{i=1}^n(x_i)!}$. $$$$Now if we take $k=2$, then the ratio is $$\frac{f_{\theta, k}(x_1, ...., x_n)}{f_{\theta, k}(y_1, ...., y_n)}=\frac{(\theta)^{\sum_{i=1}^nx_i}(1-\theta)^{n-\sum_{i=1}^nx_i}\textbf{1}(0 \leq x_{(1)} \leq x_{(n)} \leq 1)}{(\theta)^{\sum_{i=1}^ny_i}(1-\theta)^{n-\sum_{i=1}^ny_i}\textbf{1}(0 \leq y_{(1)} \leq y_{(n)} \leq 1)}$$ Now if $x_{(n)}>y_{(n)}$ then $x_{(n)}=1, y_{(n)}=0$ and so $y_1, ...., y_n=0 \implies \sum_{i=1}^ny_i=0 \implies \sum_{i=1}^nx_i=0$ which is not possible as $x_{(n)}=1$. So we must have $x_{(n)}=y_{(n)}$ and in a similar way we can prove that $x_{(1)}=y_{(1)}$ and this implies that the constant value of the ratio is $1$ and hence $\prod_{i=1}^n(x_i)!=\prod_{i=1}^n(y_i)!$ and so $T(x_1, ...., x_n)=T(y_1, ...., y_n)$. So, $T(X_1, ...., X_n)=(\sum_{i=1}^nX_i, \prod_{i=1}^n(X_i)!, X_{(1)}, X_{(n)})$ is a minimal sufficient statistic for this family.

$$$$Is my solution correct?

Answer 1

Strong hints (rewritten in light of the Q&A):

1. If $\max x_i = 0$, then $P(x_i = 0) = 1$ regardless of $k$ and $\theta$,
2. If $\max x_i = 1$, then $P(x_i = 1; T) = T/n$, regardless of $k$ and $\theta$.
3. If $\max x_i > 1$ we know it's Poisson, so $k=1$, and $T$ is sufficient for the Poisson parameter.