Sufficient Statistic for non-exponential family distribution

Question

Question: Let $X_1,X_2,\ldots,X_n$ be an iid sample from $N(\theta , 4 \theta^2 )$. I want to show that this model is not a member of the exponential family and to find a sufficient statistic for $\theta$

Attempt: \begin{align*} f(~\underline{x}~;\theta) &= \prod_{i=1}^n \frac{1}{\sqrt{8 \pi \theta^2}} \exp\left(\frac{-1}{8 \theta^2} \sum_{i=1}^n (x_i - \theta)^2\right)\\ &=\exp \left(\ln(8\pi \theta^2)^{-n/2}- \frac{1}{8 \theta^2}\sum_{i=1}^n x_i^2 + \frac{1}{4 \theta} \sum_{i=1}^n x_i - \frac{n}{8}\right) \end{align*}

So clearly this is not a member of the exponential family as it is the representation of a two dimensional exponential family, but we only have one parameter.

I am struggling to find a sufficient statistic however, can I have a two dimensional statistic if I am estimating one parameter?

Update

So after doing a similar question I am fairly certain that a sufficient statistic is given by: $S=(S_1,S_2) =(\sum_{i=1}^n x_i^2,\sum_{i=1}^n x_i) $. So i guess my question just boils down to how can we have a two dimensional statistic to estimate one parameter, seems counter intuitive?

Also, I've learned that this is a member of the curved exponential family, a further generalization of the exponential family.

Answer 1

First this is an exponential family (as shown by the above excerpt from Brown, 1986) since the density writes down as $$\exp\{\Phi_1(\theta) S_1({\mathbf x})+\Phi_2(\theta) S_2({\mathbf x})-\Psi(\theta)\}$$against a particular dominating measure. That the two coefficients $\Phi_1(\theta)$ and $\Phi_2(\theta)$ are connected with a functional relation is not an issue: they also both depend deterministically on $\theta$. The fact that $\theta$ is one-dimensional and the family is two-dimensional is a case of curved exponential families (see excerpt below from Brown, 1986). This family can be extended to a (full) truly two-dimensional parameter space, of which the ${\cal N}(\theta, 4\theta²)$ is a special case. Or a curve like $\Psi_1=\Psi_2^2/2$ in the extended (full) parameter space. But curved exponential families are special cases of exponential families, not generalisations.

Another reason for this distribution to be from an exponential family is that there exists a sufficient statistic of dimension two, whatever the sample size $n$ is. By the Darmois-Pitman-Koopman lemma this can only occur in an exponential family.

For the same reason as before, there can be a sufficient statistic of dimension two and a parameter of dimension one and this is not a contradiction, as the same sufficient statistic of dimension two serves for the extended (full) exponential family with two parameters. Examples (or paradoxes) where this happens abound in the literature. See for instance Romano and Siegel (1987). As pointed out by Kjetil B Halvorsen, these "paradoxes" are generally connected with a lack of completeness.

Answer 2

There are different definitions of "sufficient statistic(s)" and "exponential family" by different authors. I wish the theoretical statisticians can agree on a single set of definitions to avoid confusing the world.

Most of authors do not require the dimensionality of sufficient statistics to match that of parameters. Examples: https://en.wikipedia.org/wiki/Sufficient_statistic and "The Theory of Point Estimation" by Lehmann and Casella.

A few authors do. Examples: The great Mr. Fisher himself, and "Econometrics: Statistical Foundations and Applications" by Dhrymes.

Under the "more common" definition of exponential family, OP's example is a curved exponential, where the number of "natural" parameters $s$, exceeds the number of "original" parameters $k$. When $s = k$, it is called full-rank. It should be noted that $s < k$ can never happen, if the original parameters are identified in any open $k$ dimensional rectangle in $R^k$,