To compute the probability from a sequential test, you must specify a matrix that contains the boundaries. With the optional additional information concerning the time intervals and the target accuracy, or their default values, the SEQ subroutine returns the matrix that contains the probability and optionally returns the density from a sequential test evaluated at each given point of the boundary. Let denote the continuation set at each level . is defined to be the union at the th level of all the intervals bounded from below by the points with even indices and from above by the points with odd indices .

The SEQ subroutine computes, with , the densities

with

and

with the associated probability at any point at level to be

with

The notation denotes the vector of time intervals , and denotes the probability of continuation at the th level for a given domain , a given mean , and a given time vector . The variance at the level can be computed from .

It is important to understand the limitations that are imposed internally on the domain by the numerical method. Any element will always be limited within a symmetric interval with standardized values not to exceed . That is,

Given a domain , an optional time vector , and a probability level , the SEQSCALE subroutine finds the amount of scaling that would solve the problem

The result for the amount of scaling is returned as the second argument of the SEQSCALE subroutine, scale. Note that because of the complexity of the problem, the SEQSCALE subroutine will not attempt to scale a domain with multiple intervals of continuation.

For a significance level of , set .

Given a geometry , an optional time vector , and a power level , the SEQSHIFT subroutine finds the mean that solves .

Actually, a simple transformation of the variables in the sequential problem yields the following result:

where is given by .

Many options are available with the NLP family of optimization routines, which are described in Chapter 4, “Nonlinear Optimization Subroutines.”

Consider the following continuation intervals:

The following statements computes the probability from the sequential test at each boundary point specified in the geometry.

/* function to insert a into the kth column of m */
start table(m,a,k);
   if ncol(m) = 0 then   m = j(nrow(a),k,.);
   if nrow(m) < nrow(a) then  m = m// j(nrow(a)-nrow(m),ncol(m),.);
   if ncol(m) < k  then  m = m || j(nrow(m),k-ncol(m),.);
   m[1:nrow(a),k] = a;
finish;

call table(m, {-6,2}, 1);
call table(m, {-6,3}, 2);
call table(m, {-6,4,5,6}, 3);
call table(m, {-6,4}, 4);
call seq(prob,m) eps = 1.e-8 den="density";
print m, prob, density;

Figure 24.356 displays the values returned for , prob and den, respectively.

The probability at the level at the point is prob, while the density at the same point is density.

Consider the continuation intervals

Note that the continuation at level 2 can be effectively considered infinite, and it does not numerically affect the results of the computation at level 3. The following statements verify this by using the tscale parameter to compute this problem.

free m;
call table(m, {-20,2}, 1);
call table(m, {-20,20}, 2);
call table(m, {-3, 3}, 3);

/**************************************/
/* TSCALE has the default value of 1  */
/**************************************/
call seq(prob1,m) eps = 1.e-8 den="density";
print m[format=f5.]  prob1[format=e12.5];

call table(mm,{-20,2},1);
call table(mm,{-3,3},2);
/* You can use a 2-step separation between the levels */
/* while dropping the intermediate level at 2         */
tscale = { 2 };
call seq(prob2,mm) eps = 1.e-8 den="density" TSCALE=tscale;
print mm[format=f5.]  prob2[format=e12.5];

Figure 24.357 shows the values returned for the variables m, prob1, mm and prob2.

Some internal limitations are imposed on the geometry. Consider the three-level case with geometry in the preceding statements. Since the tscale variable is not specified, it is set to its default value, . The variance at the th level is for . The first level has a lower boundary point of , as represented by the value of . Since the absolute standardized value is larger than 8, this point is replaced internally by the value . Hence, the densities and the probabilities reported for the first level at this point are not for the given value ; instead, they are for the internal value of . For practical purposes, this limitation is not severe since the absolute error introduced is of the order of .

The computations performed by the first call of the SEQ subroutine can be simplified since the second level is large enough to be considered infinite. The matrix MM contains the first and third columns of the matrix M. However, in order to specify the two-step separation between the levels, you must specify tscale=2.

This example verifies some of the results published in Table 3 of Pocock (1982). That is, the following statements verify for the given domain that the significance level is and that the power is under the alternative hypothesis:

proc iml;
/* check whether the numbers yield 0.95 for the alpha level */
bm      ={-3.663  -2.884 -2.573 -2.375 -2.037,
          -2.988  -2.537 -2.407 -2.346 -2.156,
          -2.598  -2.390 -2.390 -2.390 -2.310,
          -2.446  -2.404 -2.404 -2.404 -2.396};

bplevel = { 0.5 0.25 0.1  0.05};
level   = 0.95; /* this the required alpha value  */
sigma   = diag(sqrt(1:5)); /* global sigma matrix */

do i = 1 to 4;
   m      = bm[i,];
   plevel = bplevel[i];
   geom   = (m//(-m))*sigma;

   /* Try the null hypothesis */
   call seq(prob,geom) eps = 1.e-10;
   palpha  = (prob[2,]-prob[1,])[5];

   /* Try the alternative hypothesis */
   call seqshift(prob,mean,geom,plevel);
   beta  = (prob[2,] -prob[1,])[5];
   p     = prob[3,]-prob[2,]+prob[1,];

   /* Number of patients per group   */
   tn    = 4*mean##2;
   maxn  = 5*tn;

   /* compute the average sample number */
   asn   = tn *( 5 - (4:0) * p`);
   summary = summary // ( palpha  || level || beta ||
             plevel || tn  || maxn ||asn);
end;
print summary[format=10.5];

Notice that the variables eps and tscale have been internally set to their default values. Figure 24.358 shows the computed values, which compare well with the values shown in Table 3 of Pocock (1982). Differences are of the order of .

This example shows how to verify the results in Table 1 of Wang and Tsiatis (1987). For a given , the following program finds that yields a symmetric continuation interval given by

with a given significance level of :

proc iml;
start func(delta,k) global(level);
   m       = ((1:k))##delta;
   mm      = (-m//m);
   /* meet the significance level by scaling */
   call seqscale(prob,scale,mm,level);
   return(scale);
finish;

/* alpha levels of 0.05 and 0.01 */
blevel        = {0.95 0.99};
do i = 1 to 2;
   level      = blevel[i];
   free summary;
   do delta   = 0 to .7 by .1;
      free row;
      do k=2 to 5;
         x    = func(delta,k);
         row  = row || x;
      end;
      summary = summary //row;
   end;
   print summary[format=10.5];
end;

Figure 24.359 shows the value of SUMMARY for the 0.95 and 0.99 levels. Notice that since eps and tscale are not specified, they are internally set to their default values.

This example verifies the results in Table 2 of Pocock (1977). The following program finds that yields a symmetric continuation interval given by

for five groups. The overall significance level is (the probability ), and the power is .

%let nl = 5;
proc iml;
start func(plevel) global(level,scale,mean,palpha,beta,tn,asn);
   m       = sqrt((1: &nl));
   mm      =  -m //m;
   /* meet the significance level by scaling */
   call seqscale(prob,scale,mm,level);
   palpha  = (prob[2,]-prob[1,])[&nl];
   mm      = mm *scale;
   /* meet the power condition    */
   call seqshift(prob,mean,mm,plevel);
   return(mean);
finish;

/* alpha = 0.95 */
level   = 0.95;
bplevel = { 0.5 .25 .1 0.05 0.01};
free summary;
do i = 1 to 5;
   summary = summary || func(bplevel[i]);
end;
print summary[format=10.5];

Figure 24.360 shows the results, which agree with Table 2 of Pocock (1977).

This example illustrates how to find the optimal boundary of the -class of Wang and Tsiatis (1987). The -class boundary has the form

The -class boundary is optimal if it minimizes the average sample number while satisfying the required significance level and the required power . You can use the following program to verify some of the results published in Table 2 and Table 3 of Wang and Tsiatis (1987):

%let nl = 5;
proc iml;
start func(delta) global(level,plevel,mean,
                         scale,alpha,beta,tn,asn);
   m       = ((1: &nl))##delta;
   mm      = (-m//m);
   /* meet the significance level */
   call seqscale(prob,scale,mm,level);
   alpha   = (prob[2,]-prob[1,])[&nl];
   mm      = mm *scale;
   /* meet the power condition    */
   call seqshift(prob,mean,mm,plevel);
   beta    = (prob[2,]-prob[1,])[&nl];
   /* compute the average sample number */
   p       = prob[3,]-prob[2,]+prob[1];
   tn      = 4*mean##2; /* number per group */
   asn     = tn *( &nl - p *((&nl-1):0)`);
   return(asn);
finish;

/* set up the global variables needed by func */
level   = 0.95;
plevel  = 0.01;

/* set up options used to control the optimization routine */
opt     = {0 1 0 1 6};
tc      = repeat(.,1,12);
tc[1]   = 100;
tc[7]   = 1.e-4;
par     = { 1.e-13 . 1.e-10 . . .} || . || epsd;

/* provide the initial guess and call nlpdd */
delta   = 0.5;
ods select IterStop ConvergenceStatus;
call nlpdd(rc,rx,"func",delta) opt=opt tc=tc par=par;

Figure 24.361 displays the results. The optimal function value of 34.88 agrees with the entry in Table 2 of Wang and Tsiatis (1987) for five groups, , and . Notice that the variables eps and tscale are internally set to their default values. For more information about the NLPDD subroutine, see the section NLPDD Call. For details about the opt, tc, and par arguments in the NLPDD call, see the section Options Vector, the section Termination Criteria, and the section Control Parameters Vector, respectively.

You can replicate other values in Table 2 of Wang and Tsiatis (1987) by changing the values of the variables NL and PLEVEL. You can obtain values from Table 3 by changing the value of the variable LEVEL to 0.99 and specifying NL and PLEVEL accordingly.

This example illustrates how to find the boundaries that minimize ASN given the required significance level and the required power. It replicates some of the results published in Table 3 of Pocock (1982). The program computes the domain that

The last two nonlinear conditions on the optimization process can be incorporated as a penalty applied on the error in these nonlinear conditions. The following program does the computations for a power of 0.9.

%let nl=5;
proc iml;
start func(m) global(level,plevel,sigma,epss,
               geometry,stgeom,gscale,mean,alpha,beta,tn,asn);
   m       =  abs(m);
   mm      = ( -m  // m)*sigma;
   /* meet the significance level */
   call seqscale(prob,gscale,mm,level) iguess=gscale eps=epss;
   stgeom  = gscale*m;
   geometry= mm*gscale;
   alpha   = (prob[2,]-prob[1,])[&nl];
   /* meet the power condition    */
   call seqshift(prob,mean,geometry,plevel) iguess=mean eps=epss;
   beta    = (prob[2,]-prob[1,])[&nl];
   p       = prob[3,] - prob[2,]+prob[1,];
   /* compute the average sample number */
   tn      = 4*mean##2; /* number per group */
   asn     = tn *( &nl - p *((&nl-1):0)`);
   return(asn);
finish;

/* set up the global variables needed by func */
epss    = 1.e-8;
epso    = 1.e-5;
level   = 9.50000E-01;
plevel  = 0.05;
sigma   = diag(sqrt(1:5));

/* set up options used to control the optimization routine */
opt     = {0 2 0 1 6};
tc      = repeat(.,1,12);
tc[1]   = 100;
tc[7]   = 1.e-4;
par     = { 1.e-13 . 1.e-10 . . .} || . || epso;

/* provide the constraint matrix to ensure monotonically
   increasing significance levels               */
con     = { .  .  .  .  .    .  . ,
            .  .  .  .  .    .  . ,
            1 -1  .  .  .    1  0 ,
            .  1 -1  .  .    1  0 ,
            .  .  1 -1  .    1  0 ,
            .  .  .  1 -1    1  0 };

/* provide the initial guess and call nlpdd */
m       = { 1  1  1  1  1  };
call nlpdd(rc,rx,"func",m) opt=opt blc = con tc=tc par=par;
print stgeom;

Although eps has been set to eps=, tscale has been internally set to its default value. You can choose to run the program with and without the specification of the keyword IGUESS to see the effect on the execution time.

Notice the following about the optimization process: