Can the dimension of a (potentially) sufficient statistic exceed the dimension of the parameter it estimates?

Question

I understand that if the dimension of a sufficient statistic exceeds that of the parameter it estimates, then that particular sufficient statistic won't be minimal. Now, in the following case, I define $M$ to be the sample median in the usual form with the order statistics. However, I am trying to determine whether $U = (\overline{X},S^2, M)$ is a sufficient statistic for $\theta = (\mu,\sigma)$ in the case where $X_1,\ldots,X_n \overset{\mathrm{iid}}{\sim} N(\mu,\sigma^2)$ with both $\mu,\sigma$ unknown. I know that using the Factorization Theorem and Likelihood function is the best way to proceed, but how do I involve $M$ in the likelihood function to determine whether $U$ is a sufficient statistic for $\theta$? That is, how can I get $M$ into the following equation: \begin{align*} L(\mu,\sigma) &= \prod_{i = 1}^{n} \frac{1}{\sqrt{2\pi}\sigma} \exp\left(\frac{-(x_i - \mu)^2}{2\sigma^2}\right) \end{align*} Now, this can be broken down into \begin{align} &\frac{1}{(2\pi\sigma^2)^n} \exp\left(\frac{-\sum_{i = 1}^{n}(x_i - \mu)^2}{2\sigma^2}\right)\\ &=\frac{1}{(2\pi\sigma^2)^n} \exp \left\{ \frac{-1}{2\sigma^2} \sum_{i = 1}^{n} [(x_i - \overline{x})^2 + (\overline{x}-\mu)^2 ]\right\} \newline \end{align} How do I incorporate $M$ here?

Answer 1

if the dimension of a sufficient statistic exceeds that of the parameter it estimates, then that particular sufficient statistic won't be minimal

Here is a counter-example. Take an iid sample $X_1,\ldots,X_n$ from a uniform distribution on the interval $(θ,θ+1)$, $\theta\in\mathbb R$. The likelihood is $$L(\theta|x_1,\ldots,x_n)=\prod_{i=1}^n \mathbb I_{θ<x_i<θ+1}=\mathbb I_{\max x_i−1<\theta<\min x_i}$$ Thus, the pair$$(\min x_i,\max x_i)\equiv$$is sufficient (to write down the likelihood) and minimal since two different values of $(\min x_i,\max x_i)$ obviously define a different likelihood function by changing its support. Thus, $T(X)=(X_{(1)},X_{(n)})$ is a minimal sufficient statistic for this model, while of dimension two. (It is not complete since $\mathbb E_\theta[X_{(1)}-X_{(n)}]$ is constant.)

As to answer the main question, the median did nothing in the right-side [of the likelihood]. That was the significant incident, to paraphrase Sherlock Holmes.