I have the following problem
Let $f(x ; \theta)=g(\theta) h(x), a(\theta) \leqslant x \leqslant b(\theta)$ with $a(\theta)$ decreases and $b(\theta)$ increases with $\theta$ and $g(\theta)$ differentiable with random sample $X_{1}, \ldots, X_{n}$ and $X_{(1)}<\ldots<X_{(n)}$ the order statistics. Consider the statistic $T=\max \left(a^{-1}\left(X_{(1)}\right), b^{-1}\left(X_{(n)}\right)\right)$
I have to prove that the MVUE of $\theta$ is $\hat{\theta}=T-\frac{g(T)}{n g^{\prime}(T)}$ and thus by the same notion, I am trying to prove
$$E(T) = \theta + E(\frac{g(T)}{n g^{\prime}(T)})$$
I know that $$P_{\theta}(T \leq t) = (g(\theta))^{n}\left(\int_{a(t)}^{b(t)} h(x) d x\right)^{n}$$
and $$ f_{T}(t)=n(g(\theta))^{n}\left(\int_{a(t)}^{b(t)} h(x) d x\right)^{n-1}\left[h(b(t)) b^{\prime}(t)-h(a(t)) a^{\prime}(t)\right] \mathbf{1}_{0<t<\theta} $$
Using $ \int u d v=u v-\int v d u $ I am trying to make $u v = \theta$ and
$$ -\int v d u = E\left(\frac{g(T)}{n g^{\prime}(T)}\right) $$
but i am not sure how to get this following result. Here is what I have so far:
$$ E(T) = \int_{a(\theta)}^{b(\theta)} \biggl(t n(g(\theta))^{n}\left(\int_{a(t)}^{b(t)} h(x) d x\right)^{n-1}\left[h(b(t)) b^{\prime}(t)-h(a(t)) a^{\prime}(t)\right] \mathbf{1}_{0<t<\theta} \biggr)dt$$
$$ u = \operatorname{tn}(g(\theta))^{n}\left(\int_{a(t)}^{b(t)} h(x) d x\right)^{n-1} $$
$$du = ng(\theta)^n \biggl[\int_{a(t)}^{b(t)} h(x) d x + s\biggl((h(b(t))b'(t) - h(a(t))a'(t)\biggr)\biggr]dt$$
$$dv = h(b(t)) b^{\prime}(t)-h(a(t)) a^{\prime}(t)$$ $$v = \int_{a(\theta)}^{b(\theta)} h(b(t)) b^{\prime}(t)-h(a(t)) a^{\prime}(t)dt$$
I was wondering if my line of attack is correct since doing $v$ causes the following:
$$v = \int_{a(\theta)}^{b(\theta)} h(b(t)) b^{\prime}(t)dt - \int_{a(\theta)}^{b(\theta} h(a(t)) a^{\prime}(t)dt$$ but doing the substitution $r = b(t), dr = b'(t)dt$ and $s = a(t), ds = a'(t)dt$ and $\int_{a(\theta)}^{b(\theta)} h(x) d x=\frac{1}{g(\theta)}$, v should be 0 so i am stumped. I was hoping for some hint as to create $uv = \theta$
Thank you for your time