Showing that likelihood function is sufficient but not minimal sufficient

[Usagi]

Let $f(x|\theta)$ be a family of densities where the parameter space is finite, i.e., $\theta \in \Theta = \{\theta_1, \cdots, \theta_p\}$. Now consider the likelihood function statistic, defined to be $T(\mathbf{X}) = (f_{\theta}(\mathbf{X}))_{\theta \in \Theta} = (T_1(\mathbf{X}), \cdots, T_p(\mathbf{X})) = (f_{\theta_1}(\mathbf{X}), \cdots, f_{\theta_p}(\mathbf{X}))$. Show that $T(\mathbf{X})$ is a sufficient statistic for $\theta$ but not a minimal sufficient statistic.

How do I go about proving this?

 

[Valkarie]

Any help is appreciated. Proof:To show that $T(\mathbf{X})$ is a sufficient statistic for $\theta$, we must show that the conditional distribution of $\mathbf{X}$ given $T(\mathbf{X})$ does not depend on $\theta$. That is, for any $\theta_1, \theta_2 \in \Theta$, we must show that $$P(\mathbf{X} | T(\mathbf{X}), \theta_1) = P(\mathbf{X} | T(\mathbf{X}), \theta_2). $$Since $T(\mathbf{X})$ is defined to be the collection of densities $f_{\theta}(\mathbf{X})$ evaluated at each value of $\theta$, it follows that $$P(\mathbf{X} | T(\mathbf{X}), \theta_1) = \prod_{i=1}^p f_{\theta_1}(x_i) = \prod_{i=1}^p f_{\theta_2}(x_i) = P(\mathbf{X} | T(\mathbf{X}), \theta_2). $$Therefore, $T(\mathbf{X})$ is a sufficient statistic for $\theta$.To show that $T(\mathbf{X})$ is not a minimal sufficient statistic, we must show that there exists a smaller statistic $S(\mathbf{X})$ such that $S(\mathbf{X})$ is also sufficient for $\theta$. To do this, let $S(\mathbf{X}) = (T_1(\mathbf{X}), \cdots, T_k(\mathbf{X}))$ where $k < p$. Since $S(\mathbf{X})$ consists of a subset of the densities in $T(\mathbf{X})$, it follows that $$P(\mathbf{X} | S(\mathbf{X}), \theta_1) = \prod_{i=1}^k f_{\theta_1}(x_i) = \prod_{i=1}