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!


Erfc[x_] := 2*(1 - f[x*Sqrt[2]])
InverseErfc[x_] := -ff[x/2]/Sqrt[2];

and then I use

FullSimplify[1/2 (1 - p0) Erfc[(-((Sqrt[n] (μ0 - μ1))/σ) + 
 Sqrt[2] InverseErfc[
   2 (1 + 1/
       2 (-2 + 
         Erfc[(Sqrt[2] n (μ0 - μ1)^2 - 
           Sqrt[2] σ^2 Log[(-1 + p0)^2/p0^2])/(
          4 Sqrt[n] (μ0 - μ1) σ)]))])/Sqrt[2]] + 1/2 p0 (2 - 
 Erfc[(Sqrt[2] n (μ0 - μ1)^2 - 
   Sqrt[2] 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 $ff$.

  • 3
    $\begingroup$ Since CDF[NormalDistribution[]] includes Erfc, you're now defining f in terms of something which itself is now defined in terms of f. This can only lead to problems. Maybe use a replacement rule like {Erfc -> (2 f[-Sqrt[2] #] &)}? $\endgroup$ Mar 10, 2015 at 1:34
  • $\begingroup$ I would like to point you to a similar question I wrote about a different function Dot. I hope it will help: mathematica.stackexchange.com/questions/63147/… $\endgroup$
    – QuantumDot
    Jun 9, 2015 at 9:25

1 Answer 1


One complete overkill solution to the problem might be to use the Notation Package.

  • $\begingroup$ page does not exist. More explicitely? $\endgroup$ Mar 10, 2015 at 4:37
  • $\begingroup$ Sorry - try now. I'm new to stackexchange markup. $\endgroup$
    – iqchef
    Mar 11, 2015 at 5:32

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