Fatigue Life (Birnbaum-Saunders) Distribution#

This distribution’s pdf is the average of the inverse-Gaussian $\left(\mu=1\right)$ and reciprocal inverse-Gaussian pdf $\left(\mu=1\right)$ . We follow the notation of JKB here with $\beta=S.$ There is one shape parameter $c>0$ , and the support is $x\geq0$ .

\begin{array}{rcl} f (x; c) & = & \frac{x + 1}{2 c \sqrt{2 π x^{3}}} \exp (- \frac{{(x - 1)}^{2}}{2 x c^{2}}) \\ F (x; c) & = & Φ (\frac{1}{c} (\sqrt{x} - \frac{1}{\sqrt{x}})) \\ G (q; c) & = & \frac{1}{4} {[c Φ^{- 1} (q) + \sqrt{c^{2} {(Φ^{- 1} (q))}^{2} + 4}]}^{2} \end{array}

$\begin{eqnarray*} f\left(x;c\right) & = & \frac{x+1}{2c\sqrt{2\pi x^{3}}}\exp\left(-\frac{\left(x-1\right)^{2}}{2xc^{2}}\right)\\ F\left(x;c\right) & = & \Phi\left(\frac{1}{c}\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)\right)\\ G\left(q;c\right) & = & \frac{1}{4}\left[c\Phi^{-1}\left(q\right)+\sqrt{c^{2}\left(\Phi^{-1}\left(q\right)\right)^{2}+4}\right]^{2}\end{eqnarray*}$

M (t) = c \sqrt{2 π} \exp (\frac{1}{c^{2}} (1 - \sqrt{1 - 2 c^{2} t})) (1 + \frac{1}{\sqrt{1 - 2 c^{2} t}})

$M\left(t\right)=c\sqrt{2\pi}\exp\left(\frac{1}{c^{2}}\left(1-\sqrt{1-2c^{2}t}\right)\right) \left(1+\frac{1}{\sqrt{1-2c^{2}t}}\right)$

\begin{array}{rcl} μ & = & \frac{c^{2}}{2} + 1 \\ μ_{2} & = & c^{2} (\frac{5}{4} c^{2} + 1) \\ γ_{1} & = & \frac{4 c \sqrt{11 c^{2} + 6}}{{(5 c^{2} + 4)}^{3 / 2}} \\ γ_{2} & = & \frac{6 c^{2} (93 c^{2} + 41)}{{(5 c^{2} + 4)}^{2}} \end{array}

$\begin{eqnarray*} \mu & = & \frac{c^{2}}{2}+1\\ \mu_{2} & = & c^{2}\left(\frac{5}{4}c^{2}+1\right)\\ \gamma_{1} & = & \frac{4c\sqrt{11c^{2}+6}}{\left(5c^{2}+4\right)^{3/2}}\\ \gamma_{2} & = & \frac{6c^{2}\left(93c^{2}+41\right)}{\left(5c^{2}+4\right)^{2}}\end{eqnarray*}$

Implementation: scipy.stats.fatiguelife