2
$\begingroup$

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?

$\endgroup$

1 Answer 1

0
$\begingroup$

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.
$\endgroup$
13
  • $\begingroup$ Is my solution correct? $\endgroup$
    – user671269
    Aug 24, 2023 at 18:23
  • 1
    $\begingroup$ If you look over my hints, you'll see no mention of $X_{(1)}$ or the product term. $\endgroup$
    – jbowman
    Aug 24, 2023 at 18:34
  • $\begingroup$ Then what is wrong in my solution, can you please explain? Also the conditional distributions of X1, ...., Xn given T=t is different for both the families, so how can T be a sufficient statistics? $\endgroup$
    – user671269
    Aug 24, 2023 at 18:57
  • 1
    $\begingroup$ 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? $\endgroup$
    – jbowman
    Aug 24, 2023 at 20:06
  • $\begingroup$ 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 $\theta ^{\sum_{i=1}^nx_i}(1-\theta)^{n-\sum_{i=1}^nx_i}$ $\endgroup$
    – user671269
    Aug 25, 2023 at 5:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.