# Sequentially defined probability density function and it's UMVUE estimator.

Let $f_{0}(x)$, $-\infty<x<\infty$ : probability density function(p.d.f.) and $$F_{0}(x)=\int_{-\infty}^{x}f_{0}(t)\ dt$$ be the corresponding distribution function.

And let $X_{1}$, $\cdots$, $X_{n}$ be a random sample from distribution function $F(x;\theta)=\left(F_{0}\left(x\right)\right)^{\theta}$.

I want to find UMVUE of $1/\theta$.

Consider $n=1$. We want an unbiased function of the complete sufficient statistic $\log F_0(X)$ or equivalently $F_0(X)$: $$E_\theta [h(F_0(X)] = 1/\theta$$ This translate to $\theta^{-1} = \int h(F_0(x)) d [F_0(x)^\theta] = \int_0^1 h(t) d t^\theta = h(1) - \int_0^1 t^\theta h'(t) dt$. It seems like Beta integrals might be helpful.
From the integral above, $E h(F_0(X)) = \int_0^1 \theta t^{\theta-1} h(t) dt$, we see that $F_0(X)$ has a distribution with density $\theta t^{\theta-1} 1\{ t \in (0,1)\}$. This is a Beta distribution with parameters $\theta$ and $1$.
Guessing $h(t) = \sum_i a_i t^{\alpha_i} (1-\theta)^{\beta_i-1}$, have $E h = \sum_i a_i \theta B(\theta + \alpha_i,\beta_i)$. Something like this might work (no guarantees).
• You mean I need to choose $h'$ as a beta distribution? Hmm.. any more suggestions? – kayak Dec 17 '16 at 6:43
• It is a bit tricky. Take $h(1) = 1$ and look at what the mean of a Beta distribution looks like. That mean also has an integral representation in terms of the density of the Beta. – passerby51 Dec 17 '16 at 7:54
• $\frac{\alpha}{\alpha+\beta}=\int_{0}^{1}\frac{x}{B(\alpha,\beta)}x^{\alpha-1}(1-x)^{\beta-1}\ dx$ So choose $\beta=1$, $\alpha=\theta-1$. Then $h'(t)=\frac{1}{B(\theta-1,1)}\frac{1}{t}$. Is that $h(x)=\frac{1}{B(\theta-1,1)}\int_{0}^{x}\frac{1}{t}\ dt=\infty$??? – kayak Dec 17 '16 at 8:50
• OK... that won't work. I missed the Beta factor. $h$ shouldn't depend on $\theta$. Another alternative equation is $\int_0^1 h(t) t^{\theta-1}dt = \theta^{-2}$. Not sure what the solution is, if it has any solution at all. – passerby51 Dec 17 '16 at 14:42