636 PROBLEMS AND SOLUTIONS

An OC Curve Inequality

Problem 86-20", by P. A. ROEDIGER at/d J. G. MARDO (U.S. Army Armament,Munitions and Chemical Command, Dover, NJ).Let

where n > c > 0 and 0 < q < 1. Prove or disprove that

OC(n,c,q)<[OC(n,c, ql/m)] forallm> 1.

In the terminology of lot-by-lot sampling inspection by attributes, e.g., per MIL-STD-105D, the Operating Characteristic (OC) curve defines the probability of ac-cepting a lot whose true fraction effective is q, when the criterion is to accept if andonly if (n- c) or more effectives are found in a random n-sample. When m >quality characteristics are distinguished, having effect rates q., lot quality is describedby the profile ( (ql, q2, qm) and, generally, the accept/reject criteria are suchthat probability of acceptance has the form

PA() It OC(n, c, q,).i=1

Since total lot quality q is the product of the q;’s, one is naturally interested inPA(OIq), for a given q. The two sides of the proposed inequality can be shown to beoptimal PA values, under this constraint. The difficulty is deciding which is the maxand which is the min.

Solution by A. A. JAGERS (Universiteit Twente, Enschede, the Netherlands).For fixed n and c let h(u)= log {OC (n, c, e-")} with u _-> 0. Then h(0)= 0,

h(u) < 0 for u > 0, and the inequality can be written as m-h(u) < h(m-u). It sufficesto prove that h is concave, or that h"(u) < 0 for u > 0. To do so, we note that

c!(n c )! fxn)

OC (n, c, e-u)_ t--( t) dt

e-(-’( e-’) ds

:;f(ul

(a result which expresses a familiar relation between the binomial and beta distribu-tions and the order statistics for a sample from an exponential distribution). It followsby a simple computation that h"(u) has the same sign as

g(u) := (np c)f(u) q-p/ l,

with q e-u and p q. Now

g’(u)=nqf(u)-q-c/lpand

f(u >p e-"-)S ds

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PROBLEMS AND SOLUTIONS 637

Thus g’(u) > 0. Since g(u) 0 as u --> , we see that g(u) < 0 for all u > 0. The samemay be said for h"(u), so h is concave.

Also solved by W. BOHM (Wirtschaftsuniversitt, Wien).

Problem 85-20 (Dec. 1986, p. 574).

There were two misprints in the published solution. Line 6 should read

Gxy r(xy)r-(xr q. yr)-3[(1 r)(xr+l -}- yr+) "t" (1 + r)(xy h- yxr)] >0,

and line 9 should read

g(t) := G(x(t), y(t)), 0 <- <- 1,

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