This problem comes from Casella and Berger, who do not rigorously demonstrate (in their solution key) that the statistic is not sufficient.

Let $X_1,\dots,X_n$ be a random sample from a population with PDF $f(x|\theta)=\theta x^{\theta-1} \cdot 1_{\{x\in(0,1)\}}$ for $\theta>0$. Show that $\sum_i X_i$ is not sufficient for $\theta$.

If you write out the PDF $p(\vec{x}|\theta)$ of the random sample, it is clear by the factorization theorem that $\prod_i X_i$ or $\sum_i \log(X_i)$ are sufficient for $\theta$; the PDF $p$ also suggests $\sum_i X_i$ is not sufficient for $\theta$. However, to rigorously demonstrate this, we need to analyze $p(\vec{x},\theta)/q(T(\vec{x},\theta)$, where $p$ is the distribution of sample $\vec{x}$, $q$ is the distribution of the statistic $T(\vec{X})=\sum_i X_i$.

But finding the distribution of $T(X)=\sum_i X_i$ seems impractical. I have observed that $f$ is the PDF of a Beta($\theta$,1) distribution, but checking online, the distribution of the sum of Beta random variables doesn't appear to have a closed form. Are there any alternative routes (e.g., showing there is no factorization involving $T(\vec{X})$)? Did C&B leave out a full explanation that $\sum_i X_i$ is not sufficient because none actually exist?

  • 5
    $\begingroup$ This is a beta distribution with shape parameter $\alpha$ unknown and shape parameter $\beta = 1.$ If you will look at the (very long) Wikipedia article under 'maximum likelihood estimation' you will see that estimation is messy, conclude that the sample mean is not a sufficient statistic, and see that the MLE depends approximately on the geometric mean of the data. Also, knowing the name of the distribution might help you google additional helpful pages. $\endgroup$
    – BruceET
    Commented May 1, 2020 at 6:21
  • $\begingroup$ stats.stackexchange.com/q/253211/119261 $\endgroup$ Commented Dec 6, 2021 at 11:06

1 Answer 1


Lets we want to prove $U=\sum X_i$ is not a sufficient statistic.

1) Find a minimal sufficient ($T=\prod X_i$)

2)Show that the minimal sufficient is not a function of $U$

3)Compare with the fact that a minimal sufficient statistic is a function of any sufficient statistic. So conclude $U$ is not a sufficint statistic.

Note that $T$ is a function of $U$ if $U(a_1)=U(a_2 )$ $\Rightarrow T(a_1)=T(a_2)$. So it is enough to find two points $a_1$ and $a_2$ that $U(a_1)= U(a_2)$ but $T(a_1)\neq T(a_2)$ , and hence $T$ is not a function of $U$ and hence $U$ is not a sufficient statistic.

On the other hand let $T$ is a minimal sufficient statistic. $U$ is not a sufficient statistic if there exist two points $a_1,a_2$ such that

$U(a_1)=U(a_2)$ but $T(a_1)\neq T(a_2)$

see this and this.

For $n=2$

$a_1=(\frac{1}{2} , \frac{1}{2})$ ,$a_2=(\frac{1}{4} , \frac{3}{4})$

$U(a_1)=U(a_2)=1$ but $\frac{1}{4}=T(a_1)\neq T(a_2)=\frac{3}{16}$.


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