# College Math Teaching

## September 21, 2012

### A an example to demonstrate the concept of Sufficient Statistics

A statistic $U(Y_1, Y_2, ...Y_n)$ is said to be sufficient for $\hat{\theta}$ if the conditional distribution $f(Y_1, Y_2,...Y_n,|U, \theta) = f(Y_1, Y_2,...Y_n,|U)$, that is, doesn’t depend on $\theta$. Intuitively, we mean that a given statistic provides as much information as possible about $\theta$; there isn’t a way to “crunch” the observations in a way to yield more data.

Of course, this is equivalent to the likelihood function factoring into a function of $\theta$ and $U$ alone and a function of the $Y_i$ alone.

Though the problems can be assigned to get the students to practice using the likelihood function factorization method, I think it is important to provide an example which easily shows what sort of statistic would NOT be sufficient for a parameter.

Here is one example that I found useful:

let $Y_1, Y_2, ...Y_n$ come from a uniform distribution on $[-\theta, \theta]$.
Now ask the class: is there any way that $\bar{Y}$ could be sufficient for $\theta$? It is easy to see that $\bar{Y}$ will converge to 0 as $n$ goes to infinity.

It is also easy to see that the likelihood function is $(\frac{1}{2\theta})^n H_{-\theta, \theta}(|Y|_{(n)}$ where $H_{[a,b]}$ is the standard Heavyside function on the interval $[a,b]$ (equal to one on the support set $[a,b]$ and zero elsewhere) and $|Y|_{(n)}$ is the $Y_i$ of maximum magnitude (or the $n'th$ order statistic for the absolute values of the observations).

So one can easily see an example of a sufficient statistic as well.