Let {$X_1, X_2,...,X_n$} be a random sample from a population with the following pmf

$$f(x|\theta,p)= \left\{ \begin{array}{lcc} (1-p)p^{x-\theta} & x=\theta, \theta+1,... \\ \\ 0 &\mbox{ otherwise}, \\ \\ \end{array} \right.$$ where $\theta$ is an positive integer and $0<p<1$, both are unknown. Find a sufficient statistic for the parameter vector $(\theta,p)$.

My Approach

Using Neyman-Fisher Factorization $$ f_X(x|\theta, p) = \prod_{i=1}^n (1-p)p^{x_i-\theta} = (1-p)^{n} p^{\sum_{i=1}^n x_i-n\theta} I_{(x_i \in {\theta,\theta+1,... })}. $$ This can be written $$ f_X(x|\theta,p) = g(T_1(x),T_2(x);\theta,p)\, h(x) \, , $$ where $g(T_1(x),T_2(x))=p^{\sum_{i=1}(x_i-n\theta)}$ and $h(x) = (1-p)^n$, which demonstrates that $(X_{(1)},\sum_{i=1}^n X_i)$ is a sufficient statistic for the parameter $(\theta,p)$.

Is the above correct? What am I missing?

  • $\begingroup$ How come the probabilities for the two values of $x$ add up to $1-p^2$ and not $1$? $\endgroup$ – s0ulr3aper07 Jan 30 at 19:57
  • $\begingroup$ @s0ulr3aper07, honestly I don't understand your question. $\endgroup$ – Lady Jan 30 at 20:00
  • $\begingroup$ If $x$ is a discrete RV which can take only two values: $\theta$ and $\theta +1$, then $P(X=\theta)+P(X=\theta +1)$ should be $1$. Here, it is $1-p^2$. $\endgroup$ – s0ulr3aper07 Jan 30 at 20:07
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    $\begingroup$ @s0ulr3aper07, I apologise. I have made the necessary correction in the problem statement. $\endgroup$ – Lady Jan 30 at 20:14
  • $\begingroup$ i suggest , you request for minimal sufficient too and you should consider θ is an positive integer! (i think you did not use it).it cause change in minimal sufficient. see related link math.stackexchange.com/questions/3170632/… that $\theta$ is integer. $\endgroup$ – masoud Apr 1 at 22:21

Your result is correct, but the treatment of the indicators in the working is wrong, and you have made things messier than they need to be. It would be much simpler to write the likelihood function in terms of the sample mean $\bar{x} = \sum_i x_i /n$ and the sample minimum $x_{(1)}$. For all $\mathbf{x} \in \mathbb{N}^n$ we can write the likelihood function as:

$$L_\mathbf{x}(p, \theta) = (1-p)^n \cdot p^{n(\bar{x} - \theta)} \cdot \mathbb{I}(x_{(1)} \geqslant \theta).$$

Every term in this factorisation contains at least one of the parameters $p$ and $\theta$ so we cannot remove any of the terms to a function $h(\mathbf{x})$. Hence, the sufficient statistic for $(p, \theta)$ is $(\bar{x}, x_{(1)})$. If you prefer you can use the sample total to say equivalently that $(n \bar{x}, x_{(1)})$ is sufficient.

  • $\begingroup$ Does $\mathbb{I}(x_{(1)} \geqslant \theta)$ really provide a strong enough condition to build the likelihood function? For example, if $\theta \lt x_{(1)} \lt \theta +1$, then the likelihood function will still be positive when it should be zero. $\endgroup$ – s0ulr3aper07 Jan 31 at 8:25
  • $\begingroup$ You can't have $\theta < x_{(1)} < \theta+1$ because $\theta$ and $x_1,...,x_n$ have all been specified to be positive integers. $\endgroup$ – Ben Jan 31 at 9:11
  • $\begingroup$ I may be missing something here, but as per my understanding the question does not put any such restriction on the values that $x_1,\ldots,x_n$ can take. Could you show me which part you've inferred that from? $\endgroup$ – s0ulr3aper07 Jan 31 at 10:10
  • $\begingroup$ It specifies that $\theta$ is a positive integer and that the sampling density is only non-zero when $x = \theta, \theta+1, ...$. $\endgroup$ – Ben Jan 31 at 10:51

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