Contents
Are complete sufficient statistics minimal?
A complete statistic is boundedly complete. If T is complete (or boundedly complete) and S = ψ(T) for a measurable ψ, then S is complete (or boundedly complete). It can be shown that a complete and sufficient statistic is minimal sufficient (Theorem 6.2. 28).
Why do we need sufficient statistic?
Sufficiency is ‘sought out’ because, along with other conditions (unbiasedness and completeness), it helps to identify estimators that have the smallest variance. The intuitive idea is that for purposes of estimating the parameter the sufficient statistic contains all relevant information.
What do you mean by sufficient statistics?
A sufficient statistic summarizes all of the information in a sample about a chosen parameter. For example, the sample mean, x̄, estimates the population mean, μ. x̄ is a sufficient statistic if it retains all of the information about the population mean that was contained in the original data points.
What is complete sufficiency?
Complete Sufficient Statistic It’s possible for a complete statistic to provide no information at all about θ. In order for complete statistics to be useful, they must also be a sufficient statistic; A sufficient statistic summarizes all of the information in a sample about a chosen parameter.
Why do we use minimal sufficient and complete statistics?
Minimal sufficient and complete statistics We introduced the notion of sufficient statistics in order to have a function of the data that contains all information about the parameter. However, a sufficient statistic does not have to be any simpler than the data itself.
Which is an example of a minimal sufficient statistic?
minimal sufficient statistic is unique in the sense that two statistics that are functions of each other can be treated as one statistic. For example, if T is minimal sufficient, then so is (T;eT), but no one is going to use (T;eT). If the range of X is Rk, then there exists a minimal sufficient statistic. Example 6.2.15.
How to figure out the statistic F ( x1 x2 xn )?
A statistic T = r(X1,X2,···,Xn) is sufficient if and only if the joint density can be factored as follows: f(x1,x2,···,xn|θ) = u(x1,x2,···,xn)v(r(x1,x2,···,xn),θ) (2) where u and v are non-negative functions.