Damien Horton

2022-07-23

Given a sample ${X}_{1},...,{X}_{n}\sim N\left(\theta ,{\theta }^{2}\right)$ show, using the definition of completeness, that the statistic $T=\left(\sum _{i}{X}_{i},\sum _{i}{X}_{i}^{2}\right)$ is not complete for $n\ge 2$. Use the fact that ${\mathbb{E}}_{\theta }\left[2\left(\sum _{i}{X}_{i}{\right)}^{2}-\left(n+1\right)\sum _{i}{X}_{i}^{2}\right]=0$
The statistic $T\left(\stackrel{\to }{X}\right)$ is said to be complete for the distribution of $\stackrel{\to }{X}$ if, for every misurable function $g$, ${\mathbb{E}}_{\theta }\left[g\left(T\right)\right]=0\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall }\theta \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{P}_{\theta }\left(g\left(T\right)=0\right)=1\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall }\theta$

ab8s1k28q

Let ${T}_{1}=\sum _{i}{X}_{i}$ and ${T}_{2}=\sum _{i}{X}_{i}^{2}$ and suppose $P\left(2{T}_{1}^{2}-\left(n+1\right){T}_{2}=0\right)=1$. Then ${T}_{2}=\frac{2}{n+1}{T}_{1}^{2}$ a.s$\left(P\right)$.
However by the triangle inequality,
${T}_{1}^{2}\le {T}_{2}=\frac{2}{n+1}{T}_{1}^{2}\phantom{\rule{1em}{0ex}}a.s\left(P\right),$
which is a contradiction for all $n>2$.

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