Solved – Basu’s Theorem Proof

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

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 $P(Vin A)$ 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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