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

[Albert Chan]

(11-28-2021 11:44 AM)deetee Wrote:  I found a similar result from Bowling et al (2009) with a maximal error of 1.4x10-4:

CDF = 1/(1+exp(-0.07056*z^3-1.5976*z))

Nice ! And, we can easily invert for z, without SOLVE

ln(1/CDF-1) = -0.07056*z^3 - 1.5976*z       // already a depressed cubic

With cubic and linear term having same sign, use identity: sinh(3θ) = 4*sinh(θ)^3 + 3*sinh(θ)

→ z ≈ 5.494 * sinh(asinh(-0.3418*ln(1/CDF-1))/3)

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We can use above approximation for the error function, and its inverse.

erfc(x) = 1 - erf(x) = cdf(-x*sqrt(2))*2 ≈ 2/(1+exp(0.2*x^3+2.259*x)

ierfc(x) = icdf(x/2)/-sqrt(2) ≈ 3.885 * sinh(asinh(0.3418*ln(2/x-1))/3)

--> ierf(x) ≈ 3.885 * sinh(asinh(0.6836*atanh(x))/3)