Mathematica uses complementary error function and its inverse as functions for example when integral of a Gaussian is taken. Therefore, all output expressions of Mathematica involve Erfc and InverseErfc, if needed.
My question is: If there is an easy way to manipulate mathematica so that it outputs CDF of standard normal Gaussian distribution for Erfc and inverse CDF of standard Gaussian for InverseErfc?
I used the following and it should be working BUT I am not able to verify this!
Unprotect[Erfc]; Unprotect[InverseErfc]; Erfc[x_] := 2*(1 - f[x*Sqrt]) InverseErfc[x_] := -ff[x/2]/Sqrt;
and then I use
FullSimplify[1/2 (1 - p0) Erfc[(-((Sqrt[n] (μ0 - μ1))/σ) + Sqrt InverseErfc[ 2 (1 + 1/ 2 (-2 + Erfc[(Sqrt n (μ0 - μ1)^2 - Sqrt σ^2 Log[(-1 + p0)^2/p0^2])/( 4 Sqrt[n] (μ0 - μ1) σ)]))])/Sqrt] + 1/2 p0 (2 - Erfc[(Sqrt n (μ0 - μ1)^2 - Sqrt σ^2 Log[(-1 + p0)^2/p0^2])/( 4 Sqrt[n] (μ0 - μ1) σ)])]
and I only guess that the output is correct. But I connot confirm it. When I define later on
f[x_] := CDF[NormalDistribution[0, 1], x]; ff[x_] := InverseCDF[NormalDistribution[0, 1], x];
I got problems. Is the way I follow sound? How can I justify that Mathematica indeed fullsimplified correctly, I mean in terms of $f$ and $ff$.