エクセル関数
・誤差関数の逆関数
・\(\Large \displaystyle ERFINV(x) = NORMINV \left\{ \frac{1+x}{2}, 0, \frac{1}{\sqrt{2}} \right\} \)
まず，
・NORM.DIST(x，μ，σ，TRUE) ： 正規分布（平均μ, 標準偏差σ）の累積分布関数
\(\Large \displaystyle \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{- \infty}^x exp \left\{ - \frac{(t - \mu)^2}{2 \sigma^2} \right\} dt \)
ここで，
\(\Large \displaystyle \mu = 0, \ \sigma = \frac{1}{ \sqrt{2}} \)
とすると，
\(\Large \displaystyle NORM.DIST (x, 0, \frac{1}{ \sqrt{2}}, TRUE) = \frac{1}{\sqrt{ \pi}} \int_{- \infty}^x exp \left( - t^2 \right) dt \)
\(\Large \displaystyle = \frac{1}{\sqrt{ \pi}} \int_{- \infty}^0 exp \left( - t^2 \right) dt + \frac{1}{\sqrt{ \pi}} \int_{0}^x exp \left( - t^2 \right) dt\)
\(\Large \displaystyle = -\frac{1}{\sqrt{ \pi}} \int_0^{- \infty} exp \left( - t^2 \right) dt + \frac{1}{\sqrt{ \pi}} \int_{0}^x exp \left( - t^2 \right) dt\)
これは，\(\Large \displaystyle ERF(x) = \frac{2}{\sqrt{\pi}} \int_{0}^x exp \left( - t^2 \right) dt , \ ERF(- \infty) = -1 \)
から，
\(\Large \displaystyle = -\frac{-1}{2} + \frac{ERF(x)}{2} \)
\(\Large \displaystyle = \frac{1}{2} \left(1 + ERF(x) \right) \)
つまり，\(\Large \displaystyle 1 + ERF(x) = 2 \cdot NORM.DIST (x, 0, \frac{1}{ \sqrt{2}}, TRUE) \)
との関係を導き出すことができます，まだここまでは逆関数が出てきていません．
ここで，
\(\Large \displaystyle W \equiv NORM.DIST (x, 0, \frac{1}{ \sqrt{2}}, TRUE) \)
\(\Large \displaystyle Z \equiv ERF \left( x \right) \)
とおくと，WとZは，
\(\Large \displaystyle W = \frac{1 +Z}{2} \)
次にそれぞれの逆関数を求めてみましょう．
\(\Large \displaystyle NORMSDIST^{-1}(W, 0, \frac{1}{ \sqrt{2}}) = x \)
\(\Large \displaystyle ERF^{-1}(Z) = x \)
となりますので，
\(\Large \displaystyle ERF^{-1}(Z) = \frac{1}{\sqrt{2}} NORMSDIST^{-1}(W, 0, \frac{1}{ \sqrt{2}},) \)
\(\Large \displaystyle = \frac{1}{\sqrt{2}} NORMSDIST^{-1}\left\{\frac{1 +Z}{2} ,0, \frac{1}{ \sqrt{2}} \right\} \)
となって，
\(\Large \displaystyle ERFINV(x) = \frac{1}{\sqrt{2}} NORMSDIST^{-1} \left\{\frac{1 +x}{2} ,0, \frac{1}{ \sqrt{2}} \right\} \)
つまり，
\(\Large \displaystyle \color{red}{ERFINV(x) = NORMINV \left\{ \frac{1+x}{2}, 0, \frac{1}{\sqrt{2}} \right\}} \)
を導き出すことができました．