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

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?

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.
• If you look over my hints, you'll see no mention of $X_{(1)}$ or the product term. Commented Aug 24, 2023 at 18:34
• Rather than continue to look at your math, it might help more if you looked at a different approach and tried to understand that. Look up the Fisher Factorization Theorem, and observe the likelihood above. Can you rewrite it so there is one likelihood function regardless of $k$? (Yes). What does the Fisher Factorization Theorem tell you the sufficient statistics are? Commented Aug 24, 2023 at 20:06
• The Factorisation theorem is about the density function not the likelihood function, but you have removed the $\prod_{i=1}^n(x_i)!$ term in your answer. And why there is a $n \choose T$ term, the joint density of Bernoulli($\theta$) random variables is $\theta ^{\sum_{i=1}^nx_i}(1-\theta)^{n-\sum_{i=1}^nx_i}$ Commented Aug 25, 2023 at 5:40