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Take a random sample $X_1, X_2,\ldots X_n$ from the distribution $f(x;\theta)=1/\theta$ for $0\le x\le \theta$.

I need to show that $Y=\max(X_1,X_2,...,X_n)$ is complete.

Now, I know I should multiply the sample distribution of $Y$ and multiply it with a function of $Y$, then integrate over the range of $\theta$ and equate them to zero. But how do I get the sampling distribution of $Y$?

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For every positive $y\lt\theta$, the event $[Y\lt y]$ is the intersection of $n$ independent events $[X_k\lt y]$, each with probability $y/\theta$, hence $P_\theta(Y\lt y)=(y/\theta)^n$ and $Y$ has density $$ f_Y(y;\theta)=ny^{n-1}\theta^{-n}\mathbf 1_{0\lt y\lt\theta}. $$ Now, assume that $g$ is a measurable function such that $E_\theta(g(Y))=0$ for every $\theta\gt0$ and note that $E_\theta(g(Y))=n\theta^{-n}G(\theta)$, with $$ G(\theta)=\int_0^\theta y^{n-1}g(y) \, \mathrm dy. $$ In particular, $G$ is differentiable almost everywhere and $G'(\theta)=\theta^{n-1}g(\theta)$. If $G(\theta)=0$ for every $\theta\gt0$ then $G'=0$ hence $g=0$.

This proves that $Y$ is a complete statistic for the parameter $\theta$.

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    $\begingroup$ I was proving completeness, not sufficiency, that is why I require the sampling distribution of Y. Your working, doesn't help with that."Quite easily" is a relative term. Kindly show it instead. $\endgroup$ – Bree Mar 5 '14 at 11:41
  • $\begingroup$ "Quite easily" was not in the previous version of my answer. $\endgroup$ – Did Mar 5 '14 at 22:18
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The p.d.f of $Y=X_{(n)}$ is $$\delta_{\theta}(y)=\frac{n}{\theta^n}y^{n-1},y\in[0,\theta]$$ For any function $f$ satisfying $E_{\theta}\left[f(y)\right]=0,\forall\theta\in\mathbb{R}^+$, we have $$\int_0^\theta f(y)y^{n-1}dy=0,\forall\theta\in\mathbb{R}^+$$ Since $[0,\theta]$ generate the Borel $\sigma$-algebra of $\mathbb{R}^+$, we have $$\int_A f(y)y^{n-1}dy=0, \text{ for all Borel set }A\subset\mathbb{R}^+$$ Therefore $$f(y)y^{n-1}= 0, a.e.\iff f(y)\equiv0, a.e.$$ This means $Y=X_{(n)}$ is complete.

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