Sufficiency

Recall Exponential Family

Sufficiency §

def. let statistic $T=T(\vec{X})$; and $X_i$ depends on unknown parameter $\theta$. Then $T$ is a sufficient statistic for $\theta$ if $\mathbb{P}[\vec{X}=\vec{x}|T=t]$ does not depend on $\theta$.

Info

If the distribution of $X_i$’s depends on unknown parameter $\theta$, but the distribution of $X_i$ given $T=t$ does not depend on $\theta$, it must be the case that all information about $\theta$ is in $T$. Once the value of $T$ is given, then $\theta$ becomes irrelevant in determining the value of $X$.

Info

There’s only one minimally sufficient statistic, and all other sufficient staistics are a function of this.

thm. Fischer-Neymann factorization. $T$ is a sufficient statistic iff:

i.e. the joint density function can be factored into a a function of $u,v$ where:

• $u(\vec{X})$ does not depend on $\theta$
• $v(T(\vec{X}),\theta )$ depends on $\theta$, and depend on $\vec{X}$ only through statistic $T$ ⇒ Intuitively, this menas when

Info

Very convenient for determining if statistic is sufficient or not.

lem 1. let ${\hat{\theta }}_1$ an estimator for $\theta$ and $T$ a sufficient statistic for $\theta$

then ${\hat{\theta }}_2=\mathbb{E}[{\hat{\theta }}_1|T]$ is also a sufficient estimator.

thm. Rao-Blackwell theorem. Continuing from above,

$\mathbb{MSE}[{\hat{\theta }}_2]\le \mathbb{MSE}[{\hat{\theta }}_1]$

Info

i.e. if you throw more data into a statistic, it often becomes a better statistic.

• In the exponential family of distributions, you can mostly do one iteration of Rao-Blackwell algorithm to get a pretty good estimator.
  • Bias of $\hat{\theta }$ and Bias of Rao-Blackwellized $\mathbb{E}[\hat{\theta }|T=t]$ is the same
• If $T$ is a minimally sufficient statistic, $\mathbb{E}[\hat{\theta }|T=t]$ is the best you will do