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.
 A: 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.
