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代码：（是的，我知道它效率低下并且通常很糟糕。只是想知道为什么它不起作用。）数学可能是错误的，但很可能是编码。你能帮忙的话，我会很高兴。

public class NormalDistribution {

static void normalDistributionDecode(double u, double sd, double x) {
    double z = ((x-u)/sd);
    System.out.println("The point "+x+" on a normal distribution "
     +"with standard deviation "+sd+" and mean "+u+" translates"
     +" to point "+z+" on a normal distribution with mean 0 and"
     +" standard distribution 1");
}

static void standardNDAlgorithm(double x) {

    double pi    = 3.1415926535;
    double e     = 2.7182818284;
    double fx    = (Math.pow(e, (-0.5*Math.pow(x , 2))))/(2*pi);
    long end     = -10;
    double fend  = (Math.pow(e, (-0.5*Math.pow(end , 2))))/(2*pi);
    double a = x-0.25;
    double b = x-0.75;
    double c = x-1.0;
    double d = x-1.25;
    double E = x-1.5;
    double f = x-1.75;
    double g = x-2.2;
    double h = x-2.25;
    double i = x-2.5;
    double j = x-2.75;
    double k = x-3.0;
    double l = x-3.25;
    double m = x-3.5;
    double n = x-3.75;
    double o = x-4.0;
    double p = x-4.25;
    double q = x-4.5;
    double r = x-4.75;
    double s = x-5.0;
    double t = x-5.25;
    double u = x-5.5;
    double v = x-5.75;
    double w = x-6.0;
    double y = x-6.25;
    double z = x-6.5;
    double cw= 0.25;
    double f1 = (Math.pow(e, (-0.5*Math.pow(a , 2))))/(2*pi);
    double f2 = (Math.pow(e, (-0.5*Math.pow(b , 2))))/(2*pi);
    double f3 = (Math.pow(e, (-0.5*Math.pow(c , 2))))/(2*pi);
    double f4 = (Math.pow(e, (-0.5*Math.pow(d , 2))))/(2*pi);
    double f5 = (Math.pow(e, (-0.5*Math.pow(E , 2))))/(2*pi);
    double f6 = (Math.pow(e, (-0.5*Math.pow(f , 2))))/(2*pi);
    double f7 = (Math.pow(e, (-0.5*Math.pow(g , 2))))/(2*pi);       
    double f8 = (Math.pow(e, (-0.5*Math.pow(h , 2))))/(2*pi);
    double f9 = (Math.pow(e, (-0.5*Math.pow(i , 2))))/(2*pi);
    double f10= (Math.pow(e, (-0.5*Math.pow(j , 2))))/(2*pi);
    double f11= (Math.pow(e, (-0.5*Math.pow(k , 2))))/(2*pi);
    double f12= (Math.pow(e, (-0.5*Math.pow(l , 2))))/(2*pi);
    double f13= (Math.pow(e, (-0.5*Math.pow(m , 2))))/(2*pi);
    double f14= (Math.pow(e, (-0.5*Math.pow(n , 2))))/(2*pi);
    double f15= (Math.pow(e, (-0.5*Math.pow(o , 2))))/(2*pi);
    double f16= (Math.pow(e, (-0.5*Math.pow(p , 2))))/(2*pi);
    double f17= (Math.pow(e, (-0.5*Math.pow(q , 2))))/(2*pi);
    double f18= (Math.pow(e, (-0.5*Math.pow(r , 2))))/(2*pi);
    double f19= (Math.pow(e, (-0.5*Math.pow(s , 2))))/(2*pi);
    double f20= (Math.pow(e, (-0.5*Math.pow(t , 2))))/(2*pi);
    double f21= (Math.pow(e, (-0.5*Math.pow(u , 2))))/(2*pi);
    double f22= (Math.pow(e, (-0.5*Math.pow(v , 2))))/(2*pi);
    double f23= (Math.pow(e, (-0.5*Math.pow(w , 2))))/(2*pi);
    double f24= (Math.pow(e, (-0.5*Math.pow(y , 2))))/(2*pi);
    double f25= (Math.pow(e, (-0.5*Math.pow(z , 2))))/(2*pi);       
    double integfx= 0.5*cw*(fend+fx+2*(f1+f2+f3+f4+f5+f6+f7+f8+f9+f10
              +f11+f12+f13+f14+f15+f16+f17+f18+f19+f20+f21+f22+f23+f24+f25));
    System.out.println(integfx);
    System.out.println(fx);
    System.out.println(f1);
}
public static void main(String[] args) {
normalDistributionDecode(0, 1, 10);
standardNDAlgorithm(10);
}
}

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1 回答 1

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您似乎正在尝试通过辛普森规则计算概率密度函数积分的近似值。有不同的方法可以做到这一点，但这是一种可以接受的方法。

我发现您用于计算每个点的密度的公式存在一些问题。

(1) 分母应该是sd * sqrt(2 * pi)。

(2) 每个点 a、b、c 等都应该类似于(n.nn - u)/sdn.nn 为 0.25、0.75 等。

可能还有其他问题。

于 2013-06-20T00:02:21.193 回答