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Let $X_1, \ldots, X_n$ be i.i.d. sampled from the distribution $$ f(x; \theta) = \theta x^{\theta - 1}, $$ where $x \in (0, 1)$ and $\theta > 0$.

Show that $$ T(x_{1}, \ldots, x_{n}) = - \frac{1}{n} \sum_{i = 1}^n \ln(x_i) $$ is a Unique Minimum Variance Unbiased Estimator (UMVUE) of $\dfrac{1}{\theta} $.

I know that $g(T) = E[h(x) \mid T(x)]$ is UMVUE if $h(x)$ is an unbiased estimator and $T(x)$ is a complete sufficient statistic.

I tried guessing that $\dfrac{1}{x}$ would be a UE of $\dfrac{1}{\theta}$ using $E\left( \dfrac{k}{x} \right) = \int_0^1 \left( \dfrac{k}{x} \right) \cdot \theta \cdot x^{\theta-1} \ dx$, which gives $\dfrac{k \theta}{\theta - 1}$, so $\dfrac{\theta-1}{x}$ is UE of $\theta$.

Here, we can apply invariance property to see that $\dfrac{x}{\theta - 1}$ is UE of $\dfrac{1}{\theta}$.

I also know that the MLE of $\dfrac{1}{\theta}$ is $-\dfrac{1}{n} \sum_{i = 1}^n \ln(x_i)$.

However, I'm having trouble with actually calculating $g(T)$.

The Pointer
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kalybu
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  • Probably the reason someone voted to close this question is that it is phrased in language suitable for assigning homework. Generally that should not be done here. – Michael Hardy Apr 20 '15 at 15:22
  • Hi kalybu. Technically speaking, the estimator is a (measurable) function $ T: \mathbb{R}{> 0}^{n} \to \mathbb{R} $ such that $$ {\textbf{E}{\theta}}[T \circ (X_{1},\ldots,X_{n})] = \frac{1}{\theta}. $$ – Berrick Caleb Fillmore Apr 20 '15 at 15:32
  • You would need to show that the product $\prod_{i=1}^n X_i$, or equivalently the sum of the logarithms $\sum_{i=1}^n \log X_i$ is a sufficient statistic. That can be done via Fisher's factorization criterion. You need to show that it is complete, i.e. it admits no nontrivial unbiased estimator of zero. And you need to show that $T$ is unbiased for $1/\theta$. That means showing that $\int_0^1 (-\log x) \theta x^{\theta-1},dx=1/\theta$. Integration by parts should handle that. Can you say specifically what you have questions about in all this? ${}\qquad{}$ – Michael Hardy Apr 20 '15 at 15:33
  • I think what I'm really having trouble with is showing uniqueness. I have added some more to the original question to show where I am. – kalybu Apr 20 '15 at 17:48
  • Okay, second try. I see that -(1/n)sum(ln(x)) is MLE and UE. I still don't know how to show uniqueness though. – kalybu Apr 20 '15 at 18:10
  • https://math.stackexchange.com/a/3030815/321264, https://math.stackexchange.com/q/3464853/321264 – StubbornAtom Oct 28 '21 at 14:39

1 Answers1

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$\newcommand{\E}{\operatorname{E}}$ \begin{align} \E(-\log X) & = \int_0^1 (-\log x) \Big(\theta x^{\theta-1}\,dx\Big) = \int u\,dv = uv - \int v \, du \\[8pt] & = \left.(-\log x)x^\theta\vphantom{\frac11}\,\right|_0^1 - \int_0^1 x^\theta\Big( \frac{-dx} x \Big) \\[8pt] & = \int_0^1 x^{\theta-1} \,dx \quad(\text{L'Hopital's rule showed that the first term is 0.}) \\[8pt] & = \left.\frac {x^\theta}\theta\right|_0^1 = \frac 1 \theta. \end{align} That takes care of unbiasedness.

The joint density is $$ f(x_1,\ldots,x_n) = \theta^n (x_1\cdots x_n)^{\theta-1} \cdot 1 $$ where the "$1$" is a factor that does not depend on $\theta$ and in this case does not depend on $x_1,\ldots,x_n$, but dependence on those would not upset the following conclusion: the product, and therefore the sum of the logarithms, is sufficient.

You need to show that this sufficient statistic admits no nontrivial unbiased estimators of zero, i.e. there is no nonzero function $g(x_1,\ldots,x_n)$, not depending on $\theta$, for which $$ \int_0^1\cdots\int_0^1 g(x_1,\ldots,x_n)\theta^n(x_1\cdots x_n)^{\theta-1}\,dx_1\cdots dx_n = 0\text{ for all values of $\theta>0$}. $$ (You can divide both sides of that by $\theta^n$ and it's a little bit simpler.)

Maybe I'll be back later to deal with this integral${}\,\ldots\ldots$

Michael Hardy
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