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The inverse function of the standard normal cdf is

f[x_] := InverseCDF[NormalDistribution[0, 1], x]

I also have

g[x_] := InverseErf[x]

These two should be essentially the same but the domain of f[x] is [0,1] while the domain of g[x] is [-1,1]. How exactly are the two connected? It is not f[x] = g[2x-1].

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    $\begingroup$ It should be g[x_]:=-Sqrt[2] InverseErf[1 - 2 x] $\endgroup$
    – JimB
    Apr 11, 2023 at 21:44
  • $\begingroup$ Thank you - I can accept it if you write it as an answer! Also, it is [2x -1] not [1-2x] $\endgroup$ Apr 11, 2023 at 22:39
  • $\begingroup$ Not quite true. If you use 2x-1 you'll need to remove the minus sign. However, now that you mention it, Sqrt[2] InverseErf[2 x - 1] is a nicer form (with "nicer form" being in the eye of the beholder. $\endgroup$
    – JimB
    Apr 11, 2023 at 23:02

1 Answer 1

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I use 2 unverified (but I think true) replacements:

FullSimplify[FunctionExpand[InverseCDF[NormalDistribution[0, 1], x]],
  Assumptions -> 0 <= x <= 1] /. InverseErfc[s_] -> InverseErf[1 - s]
(* -Sqrt[2] InverseErf[1 - 2 x] *)

This can be made into a nicer form as you mentioned:

% /. InverseErf[z_] -> -InverseErf[-z]
(* Sqrt[2] InverseErf[-1 + 2 x] *)
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