Damien Horton

2022-07-23

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 statistic $T(\overrightarrow{X})$ is said to be complete for the distribution of $\overrightarrow{X}$ if, for every misurable function $g$, ${\mathbb{E}}_{\theta}[g(T)]=0\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}\theta \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{P}_{\theta}(g(T)=0)=1\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}\theta $

The statistic $T(\overrightarrow{X})$ is said to be complete for the distribution of $\overrightarrow{X}$ if, for every misurable function $g$, ${\mathbb{E}}_{\theta}[g(T)]=0\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}\theta \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{P}_{\theta}(g(T)=0)=1\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}\theta $

ab8s1k28q

Beginner2022-07-24Added 17 answers

Let ${T}_{1}=\sum _{i}{X}_{i}$ and ${T}_{2}=\sum _{i}{X}_{i}^{2}$ and suppose $P(2{T}_{1}^{2}-(n+1){T}_{2}=0)=1$. Then ${T}_{2}={\displaystyle \frac{2}{n+1}}{T}_{1}^{2}$ a.s$(P)$.

However by the triangle inequality,

${T}_{1}^{2}\le {T}_{2}={\displaystyle \frac{2}{n+1}}{T}_{1}^{2}\phantom{\rule{1em}{0ex}}a.s(P),$

which is a contradiction for all $n>2$.

However by the triangle inequality,

${T}_{1}^{2}\le {T}_{2}={\displaystyle \frac{2}{n+1}}{T}_{1}^{2}\phantom{\rule{1em}{0ex}}a.s(P),$

which is a contradiction for all $n>2$.

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