5
$\begingroup$

I am having trouble with the proof of Basu's theorem... specifically, I'm not sure about the $\theta$s in the expectations below:

Let $T$ be a complete sufficient statistic. Let $V$ be an ancillary statistic. Let $A$ be an event in the sample space.

Basu's theorem states that $V$ and $T$ are independent. We need to show:

$\mathbb{P}( V \in A | T )$ $=$ $\mathbb{P}(V \in A)$

So, $\mathbb{P}(V \in A)$ $=$ $\mathbb{E}[I(V \in A)]$

$=$ $\mathbb{E}_{\theta}[(I(V \in A)]$ (Question: Why is $\theta$ here if we're talking about an ancillary statistic?)

=$\mathbb{E}_{\theta}\mathbb{E}_{\theta}[I(V \in A|T]$

$=$ $\mathbb{E}_{\theta}\mathbb{E}[I(V \in A|T]$ (Question: I understand that the $\theta$ disappears from the second expectation here since T is a sufficient statistic?)

From this we conclude $\mathbb{E}_{\theta}(g(t)$ $-$ $\mathbb{P}(V \in A)$ $=$ $0$ for all $\theta$ in the sample space. (Queston: Why is $g(t)$ subtracted from $\mathbb{P}(V \in A)$ here? Why are we concluding from the above that the expectation is 0?

Thus $\mathbb{E}_{\theta}(I(V \in A)|T)$ $=$ $\mathbb{P}(V \in A|T)$ $=$ $\mathbb{P}(V \in A)$

$\endgroup$
  • 2
    $\begingroup$ "From this we conclude &tc..." is incorrect: If you define the function $g$ by $g(t)=\mathbb{E}[I(V \in A|T=t]$, then $\mathbb{E}[g(T)]=\mathbb{P}(V \in A)$ and therefore $\mathbb{E}[g(T)-\mathbb{P}(V \in A)]=0$ for all $\theta$'s, which is a contradiction with $T$ being complete unless $g(t)=\mathbb{P}(V \in A)$ everywhere. $\endgroup$ – Xi'an Sep 21 '15 at 19:23
  • $\begingroup$ Hm, I got it directly from an in class example .... I'll try to sort that out and repost. $\endgroup$ – LotsofQuestions Sep 21 '15 at 19:53
  • $\begingroup$ The first $\theta$ shouldn't be there for the reason you identified. This also explains the 'disappearance' of $\theta$ later on. $\endgroup$ – rightskewed Sep 21 '15 at 21:19
0
$\begingroup$

The first questions were already answered in the comments. You could add that the inner expectation in

$E_{\theta}E_{\theta}[I_{V \in A}|T]$

is taken "with a fixed T = t", giving a function $g(t)=E_{\theta}[I_{V \in A}|T=t]$, as Xi'an said. The second $E_{\theta}$ then takes the expectation of g(t) (so you vary t now).

From this we conclude $E_{\theta}[g(t) − P(V \in A)] = 0$ for all θ in the sample space. (Queston: Why is $g(t)$ subtracted from $P(V\in A)$ here? Why are we concluding from the above that the expectation is 0?

This is using the definition of a complete statistic:

If you have a function h(T) that

  • 1) does not depend on the parameter $\theta$ directly, but only on T (as it is written, "h(T)", and
  • 2) for which $E_{\theta}(h(T)) = 0$ for whatever $\theta$ you pick,

then $h(t)$ is itself zero almost everywhere (or: $P_{\theta}(h(T) = 0) = 1$), again for any value of $\theta$.

In the proof, T is complete, and the function h of T is $h(T) = g(T) − P(V \in A) = g(T) - c$. We need that V is ancillary because else, h would not be purely a function of T, but some $h(\theta, T)$.

"From this we conclude ...": your first five lines of formulas say that $E_{\theta}(I_{V \in A}) = E_{\theta}[g(T)]$,so $ E_{\theta}[h(T)]$ is their difference and zero, so the second point is fulfilled, too, so the conclusion follows.


I know it's over two years late. I just figured this out for myself (or so I think). My problem was overlooking the first requirement. Hope it makes sense.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.