C version of expected Normal order statistic program
C version of Royston's original FORTRAN program is located here.
/*-Algorithm AS 177
* Expected Normal Order Statistics (Exact and Approximate),
* by J.P. Royston, 1982.
* Applied Statistics, 31(2):161-165.
*
* Translation to C by James Darrell McCauley, mccauley@ecn.purdue.edu.
*
* The functions nscor1() and nscor2() calculate the expected values of
* normal order statistics in exact or approximate form, respectively.
*
*/
#define NSTEP 721
#define H 0.025
#include <math.h>
#include <stdio.h>
#include "local_proto.h"
/* Local function prototypes */
static double alnfac(int j);
static double correc(int i, int n);
/* exact calculation of normal scores */
void nscor1(double s[], int n, int n2, double work[], int *ifault)
{
double ani, c, c1, d, scor;
int i, j;
*ifault = 3;
if (n2 != n / 2)
return;
*ifault = 1;
if (n <= 1)
return;
*ifault = 0;
if (n > 2000)
*ifault = 2;
/* calculate the natural log of factorial(n) */
c1 = alnfac(n);
d = c1 - log((double)n);
/* accumulate ordinates for calculation of integral for rankits */
for (i = 0; i < n2; ++i) {
ani = (double)n - i - 1;
c = c1 - d;
for (scor = 0.0, j = 0; j < NSTEP; ++j)
scor += work[0 * NSTEP + j] *
exp(work[1 * NSTEP + j] + work[2 * NSTEP + j] * i
+ work[3 * NSTEP + j] * ani + c);
s[i] = scor * H;
d += log((double)(i + 1.0) / ani);
}
return;
}
void init(double work[])
{
double xstart = -9.0, pi2 = -0.918938533, xx;
int i;
xx = xstart;
/* set up arrays for calculation of integral */
for (i = 0; i < NSTEP; ++i) {
work[0 * NSTEP + i] = xx;
work[1 * NSTEP + i] = pi2 - xx * xx * 0.5;
work[2 * NSTEP + i] = log(alnorm(xx, 1));
work[3 * NSTEP + i] = log(alnorm(xx, 0));
xx = xstart + H * (i + 1.0);
}
return;
}
/*-Algorithm AS 177.2 Appl. Statist. (1982) Vol.31, No.2
* Natural logarithm of factorial for non-negative argument
*/
static double alnfac(int j)
{
static double r[7] = { 0.0, 0.0, 0.69314718056, 1.79175946923, 3.17805383035, 4.78749174278, 6.57925121101
};
double w, z;
if (j == 1)
return (double)1.0;
else if (j <= 7)
return r[j];
w = (double)j + 1;
z = 1.0 / (w * w);
return (w - 0.5) * log(w) - w + 0.918938522305 +
(((4.0 - 3.0 * z) * z - 14.0) * z + 420.0) / (5040.0 * w);
}
/*-Algorithm AS 177.3 Appl. Statist. (1982) Vol.31, No.2
* Approximation for Rankits
*/
void nscor2(double s[], int n, int n2, int *ifault)
{
static double eps[4] = { 0.419885, 0.450536, 0.456936, 0.468488 };
static double dl1[4] = { 0.112063, 0.121770, 0.239299, 0.215159 };
static double dl2[4] = { 0.080122, 0.111348, -0.211867, -0.115049 };
static double gam[4] = { 0.474798, 0.469051, 0.208597, 0.259784 };
static double lam[4] = { 0.282765, 0.304856, 0.407708, 0.414093 };
static double bb = -0.283833, d = -0.106136, b1 = 0.5641896;
double e1, e2, l1;
int i, k;
*ifault = 3;
if (n2 != n / 2)
return;
*ifault = 1;
if (n <= 1)
return;
*ifault = 0;
if (n > 2000)
*ifault = 2;
s[0] = b1;
if (n == 2)
return;
/* calculate normal areas for 3 largest rankits */
k = (n2 < 3) ? n2 : 3;
for (i = 0; i < k; ++i) {
e1 = (1.0 + i - eps[i]) / (n + gam[i]);
e2 = pow(e1, lam[i]);
s[i] = e1 + e2 * (dl1[i] + e2 * dl2[i]) / n - correc(1 + i, n);
}
if (n2 != k) {
/* calculate normal areas for remaining rankits */
for (i = 3; i < n2; ++i) {
l1 = lam[3] + bb / (1.0 + i + d);
e1 = (1.0 + i - eps[3]) / (n + gam[3]);
e2 = pow(e1, l1);
s[i] = e1 + e2 * (dl1[3] + e2 * dl2[3]) / n - correc(1 + i, n);
}
}
/* convert normal tail areas to normal deviates */
for (i = 0; i < n2; ++i)
s[i] = -ppnd16(s[i]);
return;
}
/*-Algorithm AS 177.4 Appl. Statist. (1982) Vol.31, No.2
* Calculates correction for tail area of noraml distribution
* corresponding to ith largest rankit in sample size n.
*/
static double correc(int i, int n)
{
static double c1[7] = { 9.5, 28.7, 1.9, 0.0, -7.0, -6.2, -1.6 };
static double c2[7] = { -6.195e3, -9.569e3, -6.728e3, -17.614e3, -8.278e3, -3.570e3, 1.075e3
};
static double c3[7] = { 9.338e4, 1.7516e5, 4.1040e5, 2.157e6, 2.376e6, 2.065e6, 2.065e6
};
static double mic = 1.0e-6, c14 = 1.9e-5;
double an;
if (i * n == 4)
return c14;
if (i < 1 || i > 7)
return 0.0;
else if (i != 4 && n > 20)
return 0.0;
else if (i == 4 && n > 40)
return 0.0;
/* else */
an = 1.0 / (double)(n * n);
return (c1[i - 1] + an * (c2[i - 1] + an * c3[i - 1])) * mic;
}