Exponential Distribution.

For the first case, where $ h(x)=x$, we let $ x_n$ follow the standard exponential distribution $ p_0(x)=\exp(-x)$ for $ x>0$ (Section 16.1.2). Then, for $ {\rm Re}(\lambda)<1$,

$\displaystyle e(u,\lambda)= \frac{1}{1-\lambda} \; e^{(\lambda-1)u}
\;\;\;$    for $\displaystyle \; u>0; \; \frac{1}{1-\lambda}$    for $\displaystyle \; u<0.
$

$\displaystyle c(u,\lambda)= \frac{1}{1-\lambda}
\left(1-e^{(\lambda-1)u}\right)
\;\;\;$    for $\displaystyle \; u>0; \; 0$    for $\displaystyle \; u<0.
$



Baggenstoss 2017-05-19