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

  • 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, 2015 at 19:23
  • $\begingroup$ Hm, I got it directly from an in class example .... I'll try to sort that out and repost. $\endgroup$ Sep 21, 2015 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. $\endgroup$ Sep 21, 2015 at 21:19

1 Answer 1


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.


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