Mathematical Statistics Lesson of the Day – Complete Statistics

The set-up for today’s post mirrors my earlier Statistics Lesson of the Day on sufficient statistics.

Suppose that you collected data

\mathbf{X} = X_1, X_2, ..., X_n

in order to estimate a parameter \theta.  Let f_\theta(x) be the probability density function (PDF)* for X_1, X_2, ..., X_n.

Let

t = T(\mathbf{X})

be a statistic based on \mathbf{X}.

If

E_\theta \{g[T(\mathbf{X})]\} = 0, \ \ \forall \ \theta,

implies that

P \{g[T(\mathbf{X})]\} = 0] = 1,

then T(\mathbf{X}) is said to be complete.  To deconstruct this esoteric mathematical statement,

  1. let g(t) be a measurable function
  2. if you want to use g[T(\mathbf{X})] to form an unbiased estimator of the zero function,
  3. and if the only such function is almost surely equal to the zero function,
  4. then T(\mathbf{X}) is a complete statistic.

I will discuss the intuition behind this bizarre definition in a later Statistics Lesson of the Day.

*This above definition holds for discrete and continuous random variables.

One Response to Mathematical Statistics Lesson of the Day – Complete Statistics

  1. Samchappelle says:

    I’m glad someone else thinks it is a bizarre definition. I’ve never fully understood the motivation behind it – looking forward to the lesson!

Your thoughtful comments are much appreciated!

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: