# Likelihood ratio minimal sufficient

Consider a family of density functions $f(x|\theta)$ where the parameter space for $\theta$ is finite, that is, $\theta \in \{\theta_1, \cdots, \theta_p\}$. Assume that $\theta_p$ is such that $f(x|\theta_p) > 0$ for all $x$ on the support. Now consider the statistic $T(\mathbf{X}) = (T_1(\mathbf{X}), \cdots, T_p(\mathbf{X})) = \left(\frac{f(\mathbf{X}|\theta_1)}{f(\mathbf{X}|\theta_p)}, \cdots, \frac{f(\mathbf{X}|\theta_p)}{f(\mathbf{X}|\theta_p)} \right)$. Show that it is a minimal sufficient statistic of $\theta$.

I'm assuming I need to somehow apply the factorization theorem here to first show it is sufficient, then somehow show it is minimal. Is that the right approach? If not, how would I start?

Consider $$f(x|\theta)$$ for $$\theta \in \{\theta_1, \cdots, \theta_p\}$$. It can be written as \begin{align*} f(x|\theta)&=\frac{f(x|\theta)}{f(x|\theta_p)}\times f(x|\theta_p)\\ &=f(x|\theta_p)\,\exp\{\log[{f(x|\theta)}\big/{f(x|\theta_p)}]\} \end{align*} Since $$\theta \in \{\theta_1, \cdots, \theta_p\}$$, $$\theta$$ is one of the $$\theta_i$$'s, meaning $$\sum_{i=1}^p \mathbb{I}_{\theta=\theta_i}=\sum_{i=1}^p \mathbb{I}_{\theta_i}(\theta)=1$$ Thus $$\log[{f(x|\theta)}\big/{f(x|\theta_p)}]=\sum_{i=1}^p \mathbb{I}_{\theta_i}(\theta)\log[{f(x|\theta_i)}\big/{f(x|\theta_p)}]=\sum_{i=1}^p \mathbb{I}_{\theta_i}(\theta)\log[T_i(x)]$$and $$f(x|\theta)=\underbrace{f(x|\theta_p)}_{h(x)}\,\exp\underbrace{\left\{\sum_{i=1}^p \mathbb{I}_{\theta_i}(\theta)\log[T_i(x)]\right\}}_{R(\theta)^\text{T}S(x)}$$As a consequence, the density $$f(x|\theta)$$ belongs to an exponential family with natural parameter $$R(\theta)$$ and sufficient statistic $$S(x)$$. Since the components of $$R(\theta)$$ are linearly independent, the statistic $$S(x)$$ [and hence the statistic $$T(x)$$] are minimal. I remember seeing something like that in a paper by Don Fraser. (And this one as well.)

• Xi'an, could you remember which paper of Fraser you were referring to in the post? The link is dead and I thought I would edit it with the archived version, which unfortunately hasn't been done either. Nov 1, 2023 at 6:06