Are These Conjectures Regarding Sufficient Statistics True?

Question

I have these conjectures that I cannot quite prove (unless I impose another regularity condition of parameter-independent support for distribution, in which case, the conjectures are trivially true --- I think).

Would appreciate 1) confirmation (w/ proof ideally), or 2) rejection (w/ counter example ideally).

Conjectures:

Suppose $X_1, ..., X_n$ are IID scalar random variables with PDF $f(x; \theta_1, ..., \theta_k)$, where $k < n$ and $\theta_1, ..., \theta_k$ are well-identified without any constraints among them (I can make this technically precise if necessary). Suppose there are $s$ statistics, $T_1, ..., T_s$, each a scalar, that are jointly sufficient. Then,

A) $s \ge k$;

B) If $s = k$, then the sufficient statistics are minimal;

C) If $s = k$, then the sufficient statistics are complete.

End of Conjectures

Notes: If I further assume that the support of $f(x)$ is independent of $(\theta_1, ..., \theta_k)$, it is my understanding that there is a theorem that shows that $f(x)$ must be from the exponential family and all the conjectures above are true.

Answer 1

Nice conjectures you got there; shame if something were to happen to them

(A) is false for all $k>1$ for the simple reason that any vector of real numbers $T_1,...,T_s$ can be reduced to a single real number without loss of information (i.e., $\mathbb{R}^s$ is cardinally equivalent to $\mathbb{R}$; see e.g., here). For example (though there are many other mappings you could use), write the digits of the statistics when represented as decimal numbers as:

$$\begin{align} T_1 &= \cdots d_{1,3} \ d_{1,2} \ d_{1,1} \ d_{1,0} \cdot d_{1,-1} \ d_{1,-2} \ d_{1,-3} \cdots \\[6pt] T_2 &= \cdots d_{2,3} \ d_{2,2} \ d_{2,1} \ d_{2,0} \cdot d_{2,-1} \ d_{2,-2} \ d_{1,-3} \cdots \\[6pt] &\ \ \vdots \\[6pt] T_s &= \cdots d_{s,3} \ d_{s,2} \ d_{s,1} \ d_{s,0} \cdot d_{s,-1} \ d_{s,-2} \ d_{1,-3} \cdots \\[6pt] \end{align}$$

and then agglomerate these into a single real number:

$$T_* \equiv \cdots d_{1,1} \cdots d_{s,1} \ d_{1,0} \cdots d_{s,0} \cdot d_{1,-1} \cdots d_{s,-1} \ d_{1,-2} \cdots d_{s,-2} \cdots$$

It is simple to reverse this process to create a mapping $T_* \mapsto (T_1,...,T_s)$, which means that $T_* \in \mathbb{R}$ is also a sufficient statistic. Consequently, for any $k>1$ if $T_1,...,T_s$ is sufficient then we can formulate the sufficient statistic $T_*$ with length $s_* = 1 < k$.

(B) is false for all $k>1$ as a corollary of the above. Reduce the sufficient statistic to $T_*$ and then add useless statistics $T_{*2},...,T_{*k}$. This then gives you a vector $T_*,T_{*2},...,T_{*k}$ with length $k$ that is not minimal sufficient (since the latter elements contribute nothing to sufficiency). More generally, sufficiency of a statistic does not imply minimal sufficiency, even if the statistic has the same dimension as the parameter vector.

(C) is false by virtue of the fact that you can construct a sufficient non-complete statistic for the dimension $k=1$. For $k>1$ you can also apply the above to construct a sufficient statistic that is non-complete.

Answer 2

$U[0,\theta]$ is a counterexample to C. $T_1=\max_i X_i$ is sufficient for $\theta$ but it is not complete because $$E\left[\frac{n+1}{n}T_1\right]=\theta=E[2\bar X]$$

(I believe A and B are true.)