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) IfP is a dominated family and T is pairwise sufficient forP, then T is sufficient for P.

[(i): Let

P = {P0, P1,...}, and let A0 be the sufficient subfield induced by T . Let λ =

ci Pi (ci > 0) be equivalent to P. For each j = 1, 2,... the probability measure

λj that is proportional to (c0/n)P0 + c j Pj is equivalent to {P0, Pj}. Thus by pairwise sufficiency, the derivative f j = d P0/[(c0/n) d P0 + c j d Pj] is A0-measurable. Let Sj = {x : f j(x) = 0} and S = n j=1 Sj . Then S ∈ A0, P0(S) = 0, and on X − S the derivative d P0/d n

j=1 c j Pj equals (

n j=1 1/ f j)−1 which is A0-measurable. It then follows from