# 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. – BruceET May 1 '20 at 6:21

## 1 Answer

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