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.


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