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

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?

• Could you explain the conceptual meaning of estimating an element of a $d$-dimensional space by means of a statistic that is in a $c$-dimensional space with $c\ne d$? As far as your question goes, what definition of "sufficient statistic" are you using? The ones I know all immediately imply that when you adjoin additional statistics to a sufficient statistic, it's still sufficient: after all, since by definition the sufficient statistic supplies "all the relevant information about the sample," that must also be the case when you add in any number of other statistics.
– whuber
Nov 30, 2022 at 19:31
• Sure thing, the definition of sufficient statistic that was given in my class was "The statistic $U = g(Y_1,...,Y_n)$ is sufficient for $\theta$ if the conditional distribution of $Y_1,...,Y_n$ given $U$ does not depend on $\theta.$" Then, we used the usual factorization theorem as defined in Wackerly Mathematical Statistics (Theorem 9.4) to also determine whether statistics were sufficient or not. Nov 30, 2022 at 19:35
• Also, I think your response that "when you adjoin additional statistics to a sufficient statistic, it's still sufficient" hits the nail on the head. I think the dimensionality differences between $U$ and $\theta$ tripped me up when in reality it changes nothing in terms of sufficiency, all it changes is that it's not the minimal sufficiency statistic. Nov 30, 2022 at 19:39

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 $$L(\theta|x_1,\ldots,x_n)=\prod_{i=1}^n \mathbb I_{θ Thus, the pair$$(\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.)