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Given a sample $X_{1},...,X_{n} \sim N(\theta,\theta^{2})$ show, using the definition of completeness, that the statistic $T=(\sum_{i}X_{i},\sum_i X_{i}^{2})$ is not complete for $n \ge 2$. Use the fact that $\mathbb{E}_{\theta}[2(\sum_{i}X_{i})^{2}-(n+1)\sum_i X_{i}^{2}]=0$

The definition of complete statistic given is:

The statistic $T(\vec{X})$ is said to be complete for the distribution of $\vec{X}$ if, for every misurable function $g$, $\mathbb{E}_{\theta}[g(T)]=0 \; \forall \theta \implies P_{\theta}(g(T)=0)=1 \; \forall \theta$

So it is sufficient to show that $P_{\theta}(2(\sum_{i}X_{i})^{2}-(n+1)\sum_i X_{i}^{2}=0)\neq1$ but i can't figure out how. Can anyone give me a clue?

NCh
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Galois
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1 Answers1

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Let $ T_1 = \sum_i X_i$ and $T_2 = \sum_i X_i^2 $ and suppose $P(2T_1^2-(n+1)T_2=0) = 1$. Then $T_2 = \dfrac{2}{n+1}T_1^2$ a.s(P).
However by the triangle inequality, $$ T_1^2 \leq T_2 = \dfrac{2}{n+1}T_1^2\quad a.s(P), $$ which is a contradiction for all $n>2$.

Car Loz
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