c中的反累積分佈函數

Question

我正在搜索一個函數或一些代碼，它返回c中給定值的INVERSE累積正態分佈。所以如果我輸入0.5，我得到0，0.157給我-1 aso。

有沒有一種方法可以在c中實現？

Answer 1

這應該做的伎倆。這是Objective-C的代碼，但應該很容易轉換爲c。我用它進行統計計算，它工作得很好。

- (double)getInverseCDFValue:(double)p { 

double a1 = -39.69683028665376; 
double a2 = 220.9460984245205; 
double a3 = -275.9285104469687; 
double a4 = 138.3577518672690; 
double a5 =-30.66479806614716; 
double a6 = 2.506628277459239; 

double b1 = -54.47609879822406; 
double b2 = 161.5858368580409; 
double b3 = -155.6989798598866; 
double b4 = 66.80131188771972; 
double b5 = -13.28068155288572; 

double c1 = -0.007784894002430293; 
double c2 = -0.3223964580411365; 
double c3 = -2.400758277161838; 
double c4 = -2.549732539343734; 
double c5 = 4.374664141464968; 
double c6 = 2.938163982698783; 

double d1 = 0.007784695709041462; 
double d2 = 0.3224671290700398; 
double d3 = 2.445134137142996; 
double d4 = 3.754408661907416; 

//Define break-points. 

double p_low = 0.02425; 
double p_high = 1 - p_low; 
long double q, r, e, u; 
long double x = 0.0; 


//Rational approximation for lower region. 

if (0 < p && p < p_low) { 
    q = sqrt(-2*log(p)); 
    x = (((((c1*q+c2)*q+c3)*q+c4)*q+c5)*q+c6)/((((d1*q+d2)*q+d3)*q+d4)*q+1); 
} 

//Rational approximation for central region. 

if (p_low <= p && p <= p_high) { 
    q = p - 0.5; 
    r = q*q; 
    x = (((((a1*r+a2)*r+a3)*r+a4)*r+a5)*r+a6)*q/(((((b1*r+b2)*r+b3)*r+b4)*r+b5)*r+1); 
} 

//Rational approximation for upper region. 

if (p_high < p && p < 1) { 
    q = sqrt(-2*log(1-p)); 
    x = -(((((c1*q+c2)*q+c3)*q+c4)*q+c5)*q+c6)/((((d1*q+d2)*q+d3)*q+d4)*q+1); 
} 


//Pseudo-code algorithm for refinement 

if((0 < p)&&(p < 1)){ 
    e = 0.5 * erfc(-x/sqrt(2)) - p; 
    u = e * sqrt(2*M_PI) * exp(x*x/2); 
    x = x - u/(1 + x*u/2); 
} 


iCFDValue = x; 
return iCFDValue; 
}