Parameters |
$$µ \in \mathbb{R}, $$ $$-\infty < min < +\infty , $$ $$\sigma>0 , $$ $$ 0 < P_{90} < P_{10} <1 . $$ |
Support | $$x \in \mathbb{R} \quad and \quad p \in \:]0,1 [$$ |
$$ PDF = \frac{1} {x \frac{ln(P_{10})-ln(P_{90})}{\sqrt2 (E_2-E_1)} \sqrt{2 \pi}} e^{- \left\{ \frac{ \left[ ln(\bar{x})- \left( ln(P_{90})-E_1 \frac{ln(P_{90})-ln(P_{10})}{E_2-E_1} \right) \right] ^2}{2 \times \left[ \frac {ln(P_{90})-ln(P_{10})}{ \sqrt2(E_2-E_1)} \right] ^2 } \right\} } $$ $$ with \qquad \bar{x} , \qquad the \: normalized \: x \: value $$ $$ with \qquad E_1= erf^{-1} (-0.8) \quad and \quad E_2=erf^{-1} (0.8) $$ |
|
CDF |
$$ CDF = \frac{1}{2} \left\{ 1 + er f \left[ \frac {ln(\bar{x})- \left( ln (P_{90}) – E_1 \frac {ln(P_{90})-ln(P_{10})}{(E_2-E_1)} \right)} {\frac {ln(P_{90})-ln(P_{10})}{(E_2-E_1)}} \right]\right\} $$ $$ with \qquad \bar{x} , \qquad the \: normalized \: x \: value $$ $$ with \qquad E_1= erf^{-1} (-0.8) \quad and \quad E_2=erf^{-1} (0.8) $$ |
CDF-1 |
$$ CDF^{-1} = min + e^{ \left\{ \frac {ln(P_{10}) – ln(P_{90})} {(E_2-E_1)} \times er f (2 \times p – 1 ) + \left( ln(P_{90})- E_1 \frac {ln(P_{90}) – ln(P_{10})} {(E_2-E_1)} \right) \right\}} $$ $$ with \qquad E_1= erf^{-1} (-0.8) \quad and \quad E_2=erf^{-1} (0.8) $$ |