Finding UMVUE of a family of continuous random variables Let $X$ has probability density function $f_X(x;\theta) = a(\theta)b(x)I_{(0, \theta)}(x)$ (where $a(\theta)$ and $b(x)$ are nonnegative). I have to find the UMVUE of $\theta$ or show that one doesn't exist. In a different part of the question, I showed that $Y^{(n)}$ (the maximum of the $X_1, X_2,...,X_n$) is a complete sufficient statistic. I want to use Lehmann-Scheffe's theorem to find the UMVUE of $\theta$. I usually do this by first finding $E(Y^{(n)})$ then trying to find a function of $Y^{(n)}$ with expectation equal to $\theta$. But I can't even find $E(Y^{(n)})$ in this question.
 A: Note that
$$\int_0^\theta a(\theta)b(x)\,dx=1\implies \int_0^\theta b(x)\,dx=\frac{1}{a(\theta)}$$
Distribution function of $T=\max\{X_1,X_2,\ldots,X_n\}$ is
\begin{align}
F_T(t)&=P(X_1\le t,X_2\le t,\ldots,X_n\le t)
\\&=(P(X_1\le t))^n
\\&=(a(\theta))^n \left(\int_0^t b(x)\,dx\right)^n\quad,\,0<t<\theta
\end{align}
Density of $T$ is therefore
\begin{align}
f_T(t)&=n(a(\theta))^n\left(\int_0^t b(x)\,dx\right)^{n-1}b(t)\,\mathbf1_{0<t<\theta}
\end{align}
I do not know how to simplify the expression for $E(T)$. 
But suppose an unbiased estimator of $\theta$ based on $T$ does exist. And let $h(T)$ be this estimator.
Then for all $\theta>0$, 
\begin{align}
&\qquad\quad E(h(T))=\theta
\\&\implies n(a(\theta))^n \int_0^\theta h(t)b(t)\left(\int_0^t b(x)\,dx\right)^{n-1}\,dt=\theta
\\&\implies\int_0^\theta h(t)b(t)\left(\int_0^t b(x)\,dx\right)^{n-1}\,dt=\frac{\theta}{n(a(\theta))^n}
\end{align}
Differentiating the above equation wrt $\theta$, we have for all $\theta>0$,
\begin{align}
h(\theta)b(\theta)\left(\underbrace{\int_0^\theta b(x)\,dx}_{=1/a(\theta)}\right)^{n-1}&=\frac{a(\theta)-n\theta a'(\theta)}{n(a(\theta))^{n+1}}
\\\\\implies h(\theta)&=\quad ?
\end{align}
Then by Lehmann-Scheffe, $h(T)$ would give you the UMVUE of $\theta$ assuming it already exists.
Now you can verify whether $E(h(T))$ actually exists or not. If it exists, no further problems.
