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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 help2012-01-23
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    Your edit is exactly what I have so far. Unfortunately, I don't know how to proceed.2012-01-23
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    Maybe you can show that $y\mapsto y^4g(y)$ is zero almost everywhere?2012-01-23
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    Would that require measure theory? We haven't learned that yet.2012-01-23
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    This is unfortunate since the completeness property of order statistic is based on such integrals. Which makes me rather curious to see the details of the definition of completeness your instructor gave you. Would you provide these?2012-01-23
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    Let X1; ... ;Xn denote a random sample from a probability distribution with unknown parameter theta. Then a statistic U = g(X1;...;Xn) is said to be complete if for any function h, E(h(U)|theta) = 0 for all theta implies that h(U) = 0 with probability one.2012-01-24
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    *Any function h* of what kind? Any **measurable** function, right? Hence the definition requires to know how a measurable function such as h and an integral such as E(h(U)) are defined. Thus: see my last comment.2012-01-24
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    @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.2012-03-23
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    @guy: Thanks for the input. Sorry about your ordeal.2012-03-23