# Showing completeness

How would I show that the order statistic Y5 is complete for the parameter theta? I have the pdf of Y5 but I am unsure of the process on how to show completeness using the definition. Any help would be appreciated.

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Since you know the distribution of $Y_5$, you can use this distribution to write $\mathrm E_\theta(g(Y_5))$, for every $\theta$ and for a given measurable function $g$. Now your task is to prove that the condition that $\mathrm E_\theta(g(Y_5)) = 0$ for all $\theta$ implies that $\mathrm P_\theta(g(Y_5) = 0) = 1$ for all $\theta$. (I fail to see what is blocking you.)

Edit Slightly clarifying what you wrote in a comment, one gets $$\mathrm E_\theta(g(Y_5)) =\frac5{\theta^5}\int_0^\theta g(y)y^4\mathrm dy.$$ So, assuming the RHS is $0$ for every $\theta$, one could want to show that $g=0$ almost surely...

Second edit If one assumes furthermore that $g$ is continuous (a quite odd hypothesis in this context but see the comments), a proof based on first principles could start with the fact that $G(\theta)=0$ for every positive $\theta$, where $G(\theta)=\int\limits_0^\theta g(y)y^4\mathrm dy$.

Assume that $\theta_1\gt0$ is such that $g(\theta_1)\ne0$, for example $g(\theta_1)\geqslant2u$ with $u\gt0$. Then, by continuity of $g$ at $\theta_1$, there exists $v\gt0$ such that $|g(\theta)-g(\theta_1)|\leqslant u$ for every $\theta$ such that $|\theta-\theta_1|\leqslant v$. In particular $g\geqslant g(\theta_1)-u\geqslant u$ on $(\theta_1,\theta_1+v)$.

Thus $G(\theta_1+v)\geqslant G(\theta_1)+u\int\limits_{\theta_1}^{\theta_1+v}y^4\mathrm dy\gt G(\theta_1)$, hence $G(\theta_1)=G(\theta_1+v)=0$ is impossible, a contradiction.

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Are you suggesting I take the expectation of g(Y5) and then show that for g(y5) * pdf of y5 I must have g(y5) = 0? How would I show that it must be 0 for all theta? I have that the pdf of y5 is 5(y5)^4/theta^5. The expectation would be 5(y5)^4/theta^5 * g(Y5) evaluated from 0 to theta and all that would be equal to 0... I fail to see how I can proceed showing that g(y5) must be 0... Thanks so much for the help – icobes Jan 23 '12 at 2:07
Your edit is exactly what I have so far. Unfortunately, I don't know how to proceed. – icobes Jan 23 '12 at 2:20
Maybe you can show that $y\mapsto y^4g(y)$ is zero almost everywhere? – Did Jan 23 '12 at 2:26
Would that require measure theory? We haven't learned that yet. – icobes Jan 23 '12 at 2:41
@Didier If he is using (say) Casella and Berger, they seem to expect one to use the Fundamental Theorem of Calculus to do such problems, under the tacit assumption that $g$ is continuous. I always find this incredibly distasteful, and I can't get away from it because I tutor graduate students for our first year exam and "proving" completeness like this is a frequent exam problem. – guy Mar 23 '12 at 4:24