Hyperbolic Secant Distribution#

Related to the logistic distribution and used in lifetime analysis. Standard form is (defined over all $x$ )

\begin{array}{rcl} f (x) & = & \frac{1}{π} s e c h (x) \\ F (x) & = & \frac{2}{π} \arctan (e^{x}) \\ G (q) & = & \log (\tan (\frac{π}{2} q)) \end{array}

$\begin{eqnarray*} f\left(x\right) & = & \frac{1}{\pi}\mathrm{sech}\left(x\right)\\ F\left(x\right) & = & \frac{2}{\pi}\arctan\left(e^{x}\right)\\ G\left(q\right) & = & \log\left(\tan\left(\frac{\pi}{2}q\right)\right)\end{eqnarray*}$

M (t) = \sec (\frac{π}{2} t)

$M\left(t\right)=\sec\left(\frac{\pi}{2}t\right)$

\begin{array}{rcl} μ_{n}^{'} & = & \frac{1 + {(- 1)}^{n}}{2 π 2^{2 n}} n! [ζ (n + 1, \frac{1}{4}) - ζ (n + 1, \frac{3}{4})] \\ = & {\begin{array}{cc} 0 & n odd \\ C_{n / 2} \frac{π^{n}}{2^{n}} & n even \end{array} \end{array}

$\begin{eqnarray*} \mu_{n}^{\prime} & = & \frac{1+\left(-1\right)^{n}}{2\pi2^{2n}}n!\left[\zeta\left(n+1,\frac{1}{4}\right)-\zeta\left(n+1,\frac{3}{4}\right)\right]\\ & = & \left\{ \begin{array}{cc} 0 & n \text{ odd}\\ C_{n/2}\frac{\pi^{n}}{2^{n}} & n \text{ even} \end{array} \right.\end{eqnarray*}$

where $C_{m}$ is an integer given by

\begin{array}{rcl} C_{m} & = & \frac{(2 m)! [ζ (2 m + 1, \frac{1}{4}) - ζ (2 m + 1, \frac{3}{4})]}{π^{2 m + 1} 2^{2 m}} \\ = & 4 {(- 1)}^{m - 1} \frac{16^{m}}{2 m + 1} B_{2 m + 1} (\frac{1}{4}) \end{array}

$\begin{eqnarray*} C_{m} & = & \frac{\left(2m\right)!\left[\zeta\left(2m+1,\frac{1}{4}\right)-\zeta\left(2m+1,\frac{3}{4}\right)\right]}{\pi^{2m+1}2^{2m}}\\ & = & 4\left(-1\right)^{m-1}\frac{16^{m}}{2m+1}B_{2m+1}\left(\frac{1}{4}\right)\end{eqnarray*}$

where $B_{2m+1}\left(\frac{1}{4}\right)$ is the Bernoulli polynomial of order $2m+1$ evaluated at $1/4.$ Thus

\begin{aligned} μ_{n}^{'} = {\begin{array}{cc} 0 & n odd \\ 4 {(- 1)}^{n / 2 - 1} \frac{{(2 π)}^{n}}{n + 1} B_{n + 1} (\frac{1}{4}) & n even \end{array} \end{aligned}

$\begin{split}\mu_{n}^{\prime}=\left\{ \begin{array}{cc} 0 & n \text{ odd}\\ 4\left(-1\right)^{n/2-1}\frac{\left(2\pi\right)^{n}}{n+1}B_{n+1}\left(\frac{1}{4}\right) & n \text{ even} \end{array} \right.\end{split}$

\begin{array}{rcl} m_{d} = m_{n} = μ & = & 0 \\ μ_{2} & = & \frac{π^{2}}{4} \\ γ_{1} & = & 0 \\ γ_{2} & = & 2 \end{array}

$\begin{eqnarray*} m_{d}=m_{n}=\mu & = & 0\\ \mu_{2} & = & \frac{\pi^{2}}{4}\\ \gamma_{1} & = & 0\\ \gamma_{2} & = & 2\end{eqnarray*}$

h [X] = \log (2 π) .

$h\left[X\right]=\log\left(2\pi\right).$

Implementation: scipy.stats.hypsecant