离散拉普拉斯分布¶
在所有整数上定义 \(a>0\)
\BEGIN{eqnarray [}} p\left(k\right) & = & \tanh\left(\frac{{a}}{{2}}\right)e^{{-a\left|k\right|}},\\ F\left(x\right) & = & \left\{{ \begin{{array}}{{cc}} \frac{{e^{{a\left(\left\lfloor x\right\rfloor +1\right)}}}}{{e^{{a}}+1}} & \left\lfloor x\right\rfloor <0,\\ 1-\frac{{e^{{-a\left\lfloor x\right\rfloor }}}}{{e^{{a}}+1}} & \left\lfloor x\right\rfloor \geq0.\end{{array}}\right.\\ G\left(q\right) & = & \left\{{ \begin{{array}}{{cc}} \left\lceil \frac{{1}}{{a}}\log\left[q\left(e^{{a}}+1\right)\right]-1\right\rceil & q<\frac{{1}}{{1+e^{{-a}}}},\\ \left\lceil -\frac{{1}}{{a}}\log\left[\left(1-q\right)\left(1+e^{{a}}\right)\right]\right\rceil & q\geq\frac{{1}}{{1+e^{{-a}}}}.\end{{array}}\right.\end{{eqnarray] }
\BEGIN{eqnarray *}} M\left(t\right) & = & \tanh\left(\frac{{a}}{{2}}\right)\sum_{{k=-\infty}}^{{\infty}}e^{{tk}}e^{{-a\left|k\right|}}\\ & = & C\left(1+\sum_{{k=1}}^{{\infty}}e^{{-\left(t+a\right)k}}+\sum_{{1}}^{{\infty}}e^{{\left(t-a\right)k}}\right)\\ & = & \tanh\left(\frac{{a}}{{2}}\right)\left(1+\frac{{e^{{-\left(t+a\right)}}}}{{1-e^{{-\left(t+a\right)}}}}+\frac{{e^{{t-a}}}}{{1-e^{{t-a}}}}\right)\\ & = & \frac{{\tanh\left(\frac{{a}}{{2}}\right)\sinh a}}{{\cosh a-\cosh t}}.\end{{eqnarray* }
因此，
\[\mu_{n}^{\prime}=M^{\left(n\right)}\left(0\right)=\left [1+\left(-1\right)^{{n}}\right] \tExtrm{li}_{-n}\Left(e^{-a}\Right)\]
哪里 \(\textrm{{Li}}_{{-n}}\left(z\right)\) 是阶次的多对数函数 \(-n\) 评估时间 \(z.\)
\[H\Left [X\right] =-\log\left(\tanh\left(\frac{a}{2}\right)\right)+\frac{a}{\sinh a}\]