JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact support@jstor.org. This content downloaded from 128.235.251.160 on Sun, 22 Mar 2015 15:29:22 UTC All use subject to JSTOR Terms and Conditionsrestricted to satisfy the conditions pij > q (i = 1, . . ., k -1) with respective observed values xij (j = 1, . . ., k -1) and xik +y. The dimension of the new parameter space now decreases 1 unit to k. If Xlk 1 /Mik , > (Xik +Y)/(mik +n), then 9=Q51,. . **Pk,4),wherePik = 4 = (Xik +Y)/(Mik +n) and ij =Xi /mi for the otherjs. If (xik+y)/(mlk+n) > Xikl/mikl, use Lemma 2.1 again and reduce the dimension of the parameter space 1 more unit by setting pik_, = Pik = q and restricting pij > q (j = 1, . . ., k -2). If necessary, repeat the procedure to reduce the parameter space until the MLE 9 of 0 restricted to 0R is obtained. The maximum number of reductions will occur if reductions must be made to a one-dimensional space. In this case,This method for finding the restricted MLE of 0 can be considered a method of pooling the violating treatment sample of the smallest nonrestricted MLE with the control sample.The following algorithm can be used to find 9.Algorithm 2.1.Step 1. Compute Pi = ki/mi (i = 1, . . ., k) and q = y/n.Step 2. Compare q with the smallest of Pi. If q is less than or equal to this smallest Pi, the restricted MLE is obtained and is equal to the nonrestricted MLE.Step 3. Otherwise, pool the treatment sample with the smallest Pi with the control, and consider this treatment to be as effective as the control. Compute the new nonrestricted MLE q of q. Then go back to Step 2. Example 2.1. Consider five groups of patients, with each group containing 10 patients. The patients in the first four groups are under four different new treatments, respectively, and the patients in the last group are under the control treatment. Suppose that the numbers of patients having positive response are 3, 5, 4, 7, and 6, respectively. Then, the MLE's of the probabilities of positive response PI, P2, P3, P4, and q of these treatments under the restriction pi > q (i = 1, *. ., 4) through Algorithm 2.1 are given by Pi = 13/30, P2 = 5/10, p3 = 13/30, P4 = 7/10, and q= 13/30.The detailed steps in computing these are as follows.Step 1. j5 = 3/10, P2= 5/10, P3 = 4/10, P4= 7/10, and q = 6/10.Step 2. min=^3 =3/10< q=6/10. PoolingSample 1 to the control (p = q), q=9/20.Step 3. ml =nP = 4/10 < q= 9/20. Pooling Sample 3 to the control (pi =P3 = q), q = 13/30.Step 4. min =P2= 5/10 > q = 13/30. Then, pii=P3 = q = 13/30, Pj2= 5/10, andjP4 = 7/10.
APPENDIX: PROOF OF LEMMA 2.1The proof of this lemma is based on the fact that the likelihood function L is strictly increasing in each pi for pi < x1/mi (i = 1, . . ., k) and in q for q < y/n and is strictly decreasing in each pi for pi > xi/mi...