Java实现Excel中的NORMSDIST函数和NORMSINV函数

由于工作中需要将Excel中的此两种函数转换成java函数，从而计算内部评级的资本占用率和资本占用金额。经过多方查阅资料和整理，总结出如下两个转换方法

标准正态分布累计函数NORMSDIST：

public static double NormSDist(double z) {
        // this guards against overflow
        if (z > 6)
            return 1;
        if (z < -6)
            return 0;

        double  gamma = 0.231641900, 
                a1 = 0.319381530, 
                a2 = -0.356563782, 
                a3 = 1.781477973, 
                a4 = -1.821255978, 
                a5 = 1.330274429;

        double x = Math.abs(z);
        double t = 1 / (1 + gamma * x);


        double n = 1
                - (1 / (Math.sqrt(2 * Math.PI)) * Math.exp(-z * z / 2))
                * (a1 * t + a2 * Math.pow(t, 2) + a3 * Math.pow(t, 3) + a4
                        * Math.pow(t, 4) + a5 * Math.pow(t, 5));
        if (z < 0)
            return 1.0 - n;

        return n;
    }

标准正态分布累计反函数NORMSINV：

public static double normsinv(double p) {
        double LOW = 0.02425;
        double HIGH = 0.97575;

        double a[] = { -3.969683028665376e+01, 2.209460984245205e+02,
                -2.759285104469687e+02, 1.383577518672690e+02,
                -3.066479806614716e+01, 2.506628277459239e+00 };

        double b[] = { -5.447609879822406e+01, 1.615858368580409e+02,
                -1.556989798598866e+02, 6.680131188771972e+01,
                -1.328068155288572e+01 };

        double c[] = { -7.784894002430293e-03, -3.223964580411365e-01,
                -2.400758277161838e+00, -2.549732539343734e+00,
                4.374664141464968e+00, 2.938163982698783e+00 };

        double d[] = { 7.784695709041462e-03, 3.224671290700398e-01,
                2.445134137142996e+00, 3.754408661907416e+00 };

        double q, r;

        if (p < LOW) {
            q = Math.sqrt(-2 * Math.log(p));
            return (((((c[0] * q + c[1]) * q + c[2]) * q + c[3]) * q + c[4])
                    * q + c[5])
                    / ((((d[0] * q + d[1]) * q + d[2]) * q + d[3]) * q + 1);
        } else if (p > HIGH) {
            q = Math.sqrt(-2 * Math.log(1 - p));
            return -(((((c[0] * q + c[1]) * q + c[2]) * q + c[3]) * q + c[4])
                    * q + c[5])
                    / ((((d[0] * q + d[1]) * q + d[2]) * q + d[3]) * q + 1);
        } else {
            q = p - 0.5;
            r = q * q;
            return (((((a[0] * r + a[1]) * r + a[2]) * r + a[3]) * r + a[4])
                    * r + a[5])
                    * q
                    / (((((b[0] * r + b[1]) * r + b[2]) * r + b[3]) * r + b[4])
                            * r + 1);
        }
    }

不足之处，敬请指出

 

转载于:https://www.cnblogs.com/wxfistudy/p/5898693.html