Pairwise sufficiency. A statistic T is pairwise sufficient for P if it is sufficient for every pair of distributions in P.

(i) If P is countable and T is pairwise sufficient for P, then T is sufficient for P.

(ii) If P is a dominated family and T is pairwise sufficient for P, then T is sufficient for P.

[(i): Let P = {P0, P1,...}, and let A0 be the sufficient subfield induced by T.

Let λ = ciPi (ci > 0) be equivalent to P. For each j = 1, 2,... the probability measure λj that is proportional to (c0/n)P0 + cjPj is equivalent to {P0, Pj}.

Thus by pairwise sufficiency, the derivative fj = dP0/[(c0/n) dP0 + cj dPj ] is A0-measurable. Let Sj = {x : fj (x)=0} and S = n j=1 Sj . Then S ∈ A0, P0(S) = 0, and on X − S the derivative dP0/dn j=1 cjPj equals (n j=1 1/fj )

−1 which is A0-measurable. It then follows from