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Let $X =(X_1,\ldots, X_n), ~ X_i \mathrm{iid} \sim \mathcal N ( \theta, \theta^2), ~ \theta \in \Theta = \mathbb R \setminus \{0\}, ~ T(X)=(\sum_{i=1}^n X_i, \sum_{i=1}^n X_i^2)$.

I figured out that $T(X)$ is sufficient. To show that it's not complete I checked the function $g$ with $g(u,v) = 2u^2 - (n+1)v$ and noticed that $E_{\theta}[g(T(X))] = 0 ~ \forall \theta \in \Theta$ - but why is $P_{\theta}(g(T(X)) = 0) < 1$?

And how to show that $T(X)$ is minimal sufficient? I know that one can choose a subfamiliy $\mathcal P_0 \subset \mathcal P = \{f_{\theta} \mid \theta \in \Theta \}$ where $P$ is a family of densities with the same support and that's enough to show that $T$ is minimal sufficient for the subfamiliy. And if we have such a subfamily, $T^{*}(X) = \left(\frac{f_{\theta_1}(x)}{f_{\theta_0}(x)}, \ldots, \frac{f_{\theta_k}(x)}{f_{\theta_0}(x)}\right)$ is minimal sufficient (for $\Theta = \{\theta_0, \ldots, \theta_k\}$). But how to apply on the present case?

Michael Hardy
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  • In order that the probability that you ask about be equal to $1$, it would be necessary for $2\left(\sum_{i=1}^n X_i\right)^2 - (n+1)\sum_{i=1}^n(X_i^2)$ to be $0$ in all cases except on a set of probability $0$. But in fact it is true in almost no cases. ${}\qquad{}$ – Michael Hardy May 14 '15 at 14:15
  • $\ldots$ for example, suppose $(X_1,X_2,X_3)=(1,2,3)$. Then $2\left( \sum_{i=1}^n X_i\right)^2 - (n+1)\sum_{i=1}^n(X_i)^2 = 72 - 56\ne 0$. ${}\qquad{}$ – Michael Hardy May 14 '15 at 14:16
  • https://math.stackexchange.com/q/2667546/321264 – StubbornAtom May 14 '20 at 07:21

1 Answers1

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G(x) = (n+1)/2 ∑xi^2 – (∑xi )^2

E(G(x)) = 0 doesn't mean G(x)= 0

hence not complete

Vivek Singh
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