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Inverse of Cumulative Normal Distribution Function with parameters

I want to implement equivalent of matlab icdf function in C++, I have already found this useful post: https://www.johndcook.com/blog/cpp_phi_inverse/. But I want it with optional mu and sigma parameters as in matlab.

What I am supposed to change?

like image 800
MerklT Avatar asked Mar 01 '19 12:03

MerklT


1 Answers

Inspired from https://gist.github.com/kmpm/1211922/6b7fcd0155b23c3dc71e6f4969f2c48785371292:

double inverse_of_normal_cdf(const double p, const double mu, const double sigma)
{
    if (p <= 0.0 || p >= 1.0)
    {
        std::stringstream os;
        os << "Invalid input argument (" << p
            << "); must be larger than 0 but less than 1.";
        throw std::invalid_argument(os.str());
    }

    double r, val;

    const double q = p - 0.5;

    if (std::abs(q) <= .425) {
        r = .180625 - q * q;
        val =
            q * (((((((r * 2509.0809287301226727 +
                33430.575583588128105) * r + 67265.770927008700853) * r +
                45921.953931549871457) * r + 13731.693765509461125) * r +
                1971.5909503065514427) * r + 133.14166789178437745) * r +
                3.387132872796366608)
            / (((((((r * 5226.495278852854561 +
                28729.085735721942674) * r + 39307.89580009271061) * r +
                21213.794301586595867) * r + 5394.1960214247511077) * r +
                687.1870074920579083) * r + 42.313330701600911252) * r + 1);
    }
    else {
        if (q > 0) {
            r = 1 - p;
        }
        else {
            r = p;
        }

        r = std::sqrt(-std::log(r));

        if (r <= 5) 
        {
            r += -1.6;
            val = (((((((r * 7.7454501427834140764e-4 +
                .0227238449892691845833) * r + .24178072517745061177) *
                r + 1.27045825245236838258) * r +
                3.64784832476320460504) * r + 5.7694972214606914055) *
                r + 4.6303378461565452959) * r +
                1.42343711074968357734)
                / (((((((r *
                    1.05075007164441684324e-9 + 5.475938084995344946e-4) *
                    r + .0151986665636164571966) * r +
                    .14810397642748007459) * r + .68976733498510000455) *
                    r + 1.6763848301838038494) * r +
                    2.05319162663775882187) * r + 1);
        }
        else { /* very close to  0 or 1 */
            r += -5;
            val = (((((((r * 2.01033439929228813265e-7 +
                2.71155556874348757815e-5) * r +
                .0012426609473880784386) * r + .026532189526576123093) *
                r + .29656057182850489123) * r +
                1.7848265399172913358) * r + 5.4637849111641143699) *
                r + 6.6579046435011037772)
                / (((((((r *
                    2.04426310338993978564e-15 + 1.4215117583164458887e-7) *
                    r + 1.8463183175100546818e-5) * r +
                    7.868691311456132591e-4) * r + .0148753612908506148525)
                    * r + .13692988092273580531) * r +
                    .59983220655588793769) * r + 1);
        }

        if (q < 0.0) {
            val = -val;
        }
    }

    return mu + sigma * val;
}
like image 124
Anne Bierhoff Avatar answered Oct 31 '22 18:10

Anne Bierhoff