Loss Data Analysis

where Φ(x) is cumulative distribution function of an N(0, 1) random variable, therefore

${\left(-\alpha \right)}^n{e}^{{\alpha }^2/2}={\displaystyle \underset{ℝ}{\int }\frac{d^n{e}^{-ax}}{d{x}^n}\frac{e^{-x^2/2}}{\sqrt{2\pi }}dx={\left(-1\right)}^n}{\displaystyle \underset{ℝ}{\int }{e}^{-ax}\frac{d^{n+1}\Phi \left(x\right)}{d{x}^{n+1}}dx},$

where the last step follows from an integration by parts. That is,

${\varphi }_Z\left(\alpha \right)\approx {\displaystyle \underset{ℝ}{\int }{e}^{-\alpha x}\left(d\Phi \left(x\right)-\frac{a_3}{6}d{\Phi }^{\left(3\right)}\left(x\right)+\frac{a_4}{24}d{\Phi }^{\left(4\right)}\left(x\right)+\frac{a_3^2}{72}d{\Phi }^{\left(6\right)}\left(x\right)\right).}$

As both sides of the equation are Laplace transforms, it follows that

$P\left(Z\le x\right)\approx \Phi \left(x\right)-\frac{a_3}{6}{\Phi }^{\left(3\right)}\left(x\right)+\frac{a_4}{24}{\Phi }^{\left(4\right)}\left(x\right)+\frac{a_3^2}{72}{\Phi }^{\left(6\right)}\left(x\right).$

Now, undoing the standardization Z → (S − μS)/σS, we obtain

$\begin{array}{l}P\left(S\le x\right)=P\left(\frac{S-{\mu }_S}{{\sigma }_S}\le \frac{x-{\mu }_S}{{\sigma }_S}\right)\\ \approx \Phi \left(\frac{x-{\mu }_S}{{\sigma }_S}\right)-\frac{a_3}{6}{\Phi }^{\left(3\right)}\left(\frac{x-{\mu }_S}{{\sigma }_S}\right)+\frac{a_4}{24}{\Phi }^{\left(4\right)}\left(\frac{x-{\mu }_S}{{\sigma }_S}\right)+\frac{a_3^2}{72}{\Phi }^{\left(6\right)}\left(\frac{x-{\mu }_S}{{\sigma }_S}\right).\\ \left(5.11\right)\end{array}$

A very common ...