# Can this statistic be shown not to be sufficient for $\theta$?

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?

• 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. Commented May 1, 2020 at 6:21
• stats.stackexchange.com/q/253211/119261 Commented Dec 6, 2021 at 11:06

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