The following theorem can be used to demonstrate that a statistic is minimal sufficient:

Let $f(X|\theta)$ be the pmf or pdf of a sample X. Suppose $\exists$ a function $T(X)$ such that, for every two samples X and Y:

$\frac{f(X|\theta)}{f(Y|\theta)} \amalg \theta$ iff $T(X) = T(Y)$.

Then $T(X)$ is a minimal suffienct statistic for $\theta$.

(Reference: Casella & Berger Theorem 6.2.13.)

I'm confused about how this works when the densities contain indicator functions. Here is an example that I constructed in order to confuse myself:

Let $X_{i}$ and $Y_{i} \sim Unif(0,\theta)$. Let $I$ be the indicator function. Then:

$$ \frac{f(X|\theta)}{f(Y|\theta)} = \frac{ I( max(X) < \theta ) \cdot I( min(X) > 0 ) }{ I( max(Y) < \theta ) \cdot I( min(Y) > 0 )}$$

We'll forget about the ratio of the second indicator functions because it's already independent of $\theta$. So the question becomes, "When is the ratio of the first indicator terms independent of $\theta$?" The ratio will be 1 if $I( max(X) < \theta ) = I( max(Y) < \theta ) = 1$. If $I( max(X) < \theta ) = I( max(Y) < \theta ) = 0$, I'm hazy on what to do because the denominator is 0, but at least the ratio shouldn't depend on $\theta$. Okay -- so $max(X)$ is a candidate MSS, and it fulfills the "if" direction.

Let's look at the "only if" direction. If $max(X) \ne max(Y)$, then the ratio is either $1/0$ or $0/1$. To my mind, these are independent of $\theta$ regardless of any function of $X$ or $Y$. But this conclusion doesn't make any sense; therefore the premise that I understand indicator functions and minimal sufficiency is wrong.

I'd love some guidance to set my understanding straight.


1 Answer 1


Someone described to me a clever proof sketch that I will flesh out for any interested parties.

First, let's fix the problem about having 0 in the denominator by constructing an equivalent condition:

$$ \frac{f(X|\theta)}{f(Y|\theta)} = \frac{ I( max(X) < \theta ) }{ I( max(Y) < \theta )} $$

is equivalent to:

$$ \exists \; g(X,Y) \amalg \theta \; \text{such that:}$$ $$ \; {I( max(X) < \theta ) = g(X,Y) \cdot I( max(Y) < \theta )} \; \forall \; \theta \; \; \; \; \; \; (*)$$

Take $max(X)$ as the candidate MSS. For the "if" direction, simply note that when $max(X) = max(Y)$, $(*)$ holds regardless of $\theta$, if we let $g(X,Y) = 1$.

For the "only if" direction, we need to show that $(*)$ implies $max(X) = max(Y)$.

We will prove the contrapositive: $max(X) \ne max(Y)$ implies $(*)$ does not hold, i.e., $g(X,Y)$ depends on $\theta$. WLOG, let $max(X) < max(Y)$.

Now there are three cases:

(1) $\theta < max(X) < max(Y)$. Then both indicators are 0, so $g(X,Y)$ can be anything and $(*)$ will hold.

(2) $max(X) < \theta < max(Y)$. Then the indicators don't match, so $g(X,Y)$ must evaluate to 0 to fulfill $(*)$.

(3) $max(X) < max(Y) < \theta$. Then the indicators are both 1, so $g(X,Y)$ must evaluate to 1 to fulfill $(*)$.

Therefore, $max(X) \ne max(Y)$ implies that $(*)$ does not hold, since $g(X,Y)$ would be a piecewise function dependent on $\theta$. And we're done.

  • $\begingroup$ Can you explain how you know that there must exist a function $g(x,y)$ that is independent of $\theta$? $\endgroup$ Aug 17, 2018 at 22:04
  • $\begingroup$ We don't know that it exists, but rather we're translating the ratio that defines minimal sufficiency into an equivalent condition, which gives properties that the function would have to have, if it existed. $\endgroup$
    – half-pass
    Aug 19, 2018 at 18:13

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