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Let $f_X$ be a joint density function that comes from an $s$-parameter exponential family with sufficient statistics $(T_1, T_2, \dots, T_s)$ so that the density $f_X$ can be expressed as

$$f_{X|\theta}(x) = h(x) \exp \left(\sum_{i=1}^s T_i(x)\eta_i(\theta) - A(\theta) \right)$$

I have two questions:

  1. Expressed in this form, is it correct to say the statistics $T_1,\dots,T_s$ are jointly sufficient, as opposed to independently sufficient?

  2. Based on them being jointly sufficient, is it correct to say any unbiased estimator $\tau (X)$ such that $E[\tau(X)|T_i,T_j] = \theta_k$, for some $i,j,k$, must be UMVUE of $\theta_k \in \theta$?

I'm trying to understand the difference between being sufficient, and jointly sufficient

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    $\begingroup$ 1. Yes , and 2. You'll need joint completeness of $(T_i,T_j)$ alongside sufficiency for the conditional mean to be UMVUE. $\endgroup$ – StubbornAtom Sep 20 '18 at 2:06
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    $\begingroup$ The first part is easy to solve. You can express $\sum_{i=1}^s T_i(x) \eta_i(\theta)$ as $\eta^T(\theta) T(x)$ where $T(x) = [T_1(x),\ldots,T_s(x)]^T$ and $\eta(\theta) = [\eta_1(\theta),\ldots,\eta_s(\theta)]^T$. Thus, $T(x)$ is sufficient which implies jointly sufficient in this case. $\endgroup$ – Maxtron Sep 20 '18 at 2:07
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For exponential families, under the conditions that (a) the components $T_i(\cdot)$ are linearly independent [minimal representation of the exponential family] and (b) the parameter space Θ contains an open set in $\mathbb{R}^s$, the statistic $T(X)$ is also complete, which implies that the estimator $$\mathbb{E}_\theta[\tau(X)|T(X)]$$is UMVUE for estimating $\mathbb{E}_\theta[\tau(X)]$ by Lehmann-Scheffe. In the special case when$$\mathbb{E}_\theta[\tau(X)|T(X)]=\mathbb{E}_\theta[\tau(X)|T_i(X),T_j(X)]$$the later is (obviously) complete, but there is no reason in general for a sub-vector of $T(X)$ to be sufficient, hence for the expression $\mathbb{E}_\theta[\tau(X)|T_i(X),T_j(X)]$ to be independent from $\theta$.

As a remark, I never use the term jointly sufficient in my courses. As I find it unhelpful. A vector statistic $S(X)$ is sufficient or not, as a whole, and it may be that some non-bijective transforms of $S(X)$ are also sufficient, in which case $S$ is not minimal (and not complete). Similarly, I am keeping away from defining sufficiency for some component of the vector $\theta$, as this leads to paradoxes, as shown by e.g. Basu (1977).

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    $\begingroup$ Wow looks like those are presentation slides of your lectures. Neat! $\endgroup$ – StubbornAtom Sep 20 '18 at 5:57

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