(15C)(DM15) - PDF/CDF and Inverse of a Normal Distribution

[Albert Chan]

Hi, deetee

With 0 < CDF < 1, 1/CDF - 1 > 0

Also, cdf/icdf using tanh/atanh is equivalent to original exp/ln

(1 + tanh(x))/2 = (sinh(x)+cosh(x)) / (2*cosh(x)) = e^x / (e^x + e^-x) = 1 / (1 + e^(-2x))

CDF ≈ 1/(1+exp(-0.07056*z^3-1.5976*z)) = (1 + tanh(0.03528*z^3+0.7988*z))/2

Similarly, for the inverse, ln(x) = 2*atanh((x-1)/(x+1))

z ≈ 5.494 * sinh(asinh(-0.3418*ln(1/CDF-1))/3) = 5.494 * sinh(asinh(-0.6836*atanh(1-2*CDF))/3)