Let $X_1, \dots, X_n$ denote a random sample from the PDF

$$f_{\varphi}(x)= \begin{cases} \varphi x^{\varphi - 1} &\text{if}\, 0 < x < 1, \varphi > 0\\ 0 &\text{otherwise} \end{cases}$$

This density function is a member of the one-parameter exponential family.

Let $A_i = - \log(X_i)$, where $A_i$ has an exponential distribution. This means that $2\varphi \sum_{i = 1}^n A_i$ has a $\chi^2$ distribution with $2n$ degrees of freedom. Furthermore, I have that

$$ E\left[ \left( 2\varphi \sum_{i = 1}^n A_i \right)^{-1} \right] = \dfrac{1}{2n - 2} $$


$$\text{Var} \left( \dfrac{1}{2 \varphi \sum_{i = 1}^n A_i} \right) = \dfrac{1}{4(n - 1)^2(n - 2)}$$

Apparently, using this information, we can conclude that $\dfrac{n - 1}{\sum_{i = 1}^n A_i} = \dfrac{n - 1}{- \sum_{i = 1}^n \log(X_i)}$ is a UMVUE for $\varphi$. However, I am not able to follow this reasoning for how they calculated the UMVUE (nor for how they concluded that this is a UMVUE). If I had to guess, it seems to me that they might have done some kind of bias correction at $\dfrac{n - 1}{\sum_{i = 1}^n A_i} = \dfrac{n - 1}{- \sum_{i = 1}^n \log(X_i)}$, but I'm honestly not sure. What is the reasoning used here for why $\dfrac{n - 1}{\sum_{i = 1}^n A_i} = \dfrac{n - 1}{- \sum_{i = 1}^n \log(X_i)}$ is a UMVUE for $\varphi$? I'd greatly appreciate it if someone would please take the time to fill in the gaps and explain this reasoning so that I may understand it.


1 Answer 1


Note that the density can be written $$f(x \mid \varphi) = \varphi \frac 1 x e^{\varphi \log x}$$

with the constraints on $x,\varphi$ as given. By the factorization theorem, $\log x $ is a sufficient statistic for $\varphi$. For a sample, we have the likelihood to be $$f(\textbf x \mid \varphi) = \varphi^n\frac 1 {x^n}e^{\varphi\sum_{i=1}^n\log x}$$

and $\sum_i \log x$ is sufficient. Because the exponential family is full rank (not curved), then $\sum_i\log x$ is complete as well (could someone verify this? I'm not 100% sure on this part). Anyway, since we have a CSS (complete sufficient statistic) by the Lehmann-Scheffe Theorem any function of the CSS whose expected value is $\varphi$ will be a UMVU of $\varphi$.

YOu have written that $$\begin{split}E\left[\left(-2\varphi\sum_i \log X_i\right)^{-1}\right]&=\frac 1 {2n-2}\\ \mathbb E\left[\frac{n-1}{-\sum_i \log X_i}\right]=\varphi\end{split}$$

  • $\begingroup$ Thanks again for the answer! I researched the part you were unsure about, and I think you are correct! See definition 5, theorem 4, and theorem 5 of eml.berkeley.edu/~mcfadden/e240a_sp01/sufficiency.pdf But it says that this depends on whether "the $\theta_i$ are linearly independent and the $T_i$ are also linearly independent." Is this true for our case? And what is meant by "full rank" in this context? I'm familiar with rank in the matrix context en.wikipedia.org/wiki/Rank_(linear_algebra) $\endgroup$ Apr 17, 2021 at 5:29
  • 1
    $\begingroup$ I found here web.stat.tamu.edu/~debdeep/complete.pdf in Theorem 1 and similarly here www2.stat.duke.edu/~pdh10/Teaching/581/LectureNotes/expofam.pdf states "If the parameter space for an exponential family contains an s- dimensional open set, then it is called full rank. An exponential family that is not full rank is generally called a curved exponential family, as typically the parameter space is a curve in Rs of dimension less than s." Since we are working in $\mathbb R^1$ and $\varphi\in(0,\infty)$ which is an open set in $\mathbb R$ we can conclude that it is full rank/CSS. $\endgroup$
    – Vons
    Apr 17, 2021 at 6:16
  • $\begingroup$ that makes sense. Thanks for taking the time to clarify. $\endgroup$ Apr 17, 2021 at 6:18

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