Let X a random sample with density function $$f(x|\theta)=\frac{\theta}{x^2}I_{(\theta,\infty)}(x)\hspace{0.5cm}\theta\in\mathbb{R}^+$$
Find the UMVUE for $h(\theta)=\theta$
$$f_\theta(\vec{x})=\theta^nI_{(0,x_{(1)})}(\theta)\prod_{i=1}^n\frac {1}{x_i^2}$$
Then $$T(X)=X_{(1)}$$ is sufficient for $\theta$
I am trying to prove that T is a complete statistic $$F_\theta(x)\,dx= \int_\theta^x f_\theta(x)\,dx = \int_\theta^x \frac{\theta}{x^2}\,dx=\left[-\frac{\theta}{x}\right]_\theta^x=\frac{x-\theta}{x}$$
Then $$F_{X_{(1)}}(y)=1-[1-F_\theta(y)]^n=1-\left[\frac{\theta}{y}\right]^n$$
Then $$f_{X_{(1)}}(y)=n\theta^ny^{-n-1}$$ Let $g(t)$ $$E[g(t)]=\int_\theta^{\infty}g(t)f_{X_{(1)}}=\int_\theta^{\infty}g(t)n\theta^nt^{-n-1}\,dt=n\theta^n\int_\theta^\infty g(t)t^{-n-1}\,dt $$
Then
$$0=n\theta^n\int_\theta^\infty g(t)t^{-n-1}\,dt$$
$$0={n^2\theta^{n-1}\int_\theta^\infty g(t)y^{-n-1}\,dt}+n\theta^n\frac{d}{d\theta}\left[\int_\theta^\infty g(t)t^{-n-1}\,dt\right]$$
It follows that $\int_0^\theta g(t)t^{{-n-1}}\;dt=0$ Then $$n\theta^n\frac{d}{d\theta}\left[\int_\theta^\infty g(t)t^{-n-1}\,dt\right]=0$$
I am stuck in this point ... I thougt T was complete but now I am not sure, can you help me please Thanks
It's almost done because $\frac{d}{d\theta}\left[\int_\theta^\infty g(t)t^{-n-1}\,dt\right]=0$, then $g(\theta)\theta^{-n-1}=0$, and since $\theta>0$, it follows that $g(\theta)=0$ for all $\theta$, so $g\equiv 0$