remain motionless with a probability $1-dt/{\tau^\prime}$. Given that the length of each step is the random variate $\xi$, this sheme generates a compound Poisson process, $P_q(t)$; cf. [@KardarI Sec. 2.3]. Its cumulant-generating function is $$\kappa(\tilde{P}_q) = \frac{[\mu(\tilde\xi)-1] t}{{\tau^\prime}}\vline_{\tilde\xi=\tilde{P}_q},$$ which together with the approximant given by [Eq. (\[eq:MuPade2\])]{} yields $$\kappa(\tilde{P}_q) \approx \frac{(1-q) B \tilde{P}_q t}{\tau(1-B \tilde{P}_q)}.$$ The dimensionless parameter $(1-q)$ can be further absorbed into the constant $\tau = {\tau^\prime}/(1-q)$, and we thus obtain the compound Poisson process $P(t)$ with exponentially distributed intensity $${\label{eq:pois}}
    \kappa(\tilde{P}) = \frac{B \tilde{P} t}{(1-B x) \tau},$$ cf. [Ref. [@PRE2016II]]{}.

Another way to derive [Eq. (\[eq:pois\])]{} is to seek directly a Padé approximant for the cumulant generating function $\kappa(\tilde\xi) \approx \kappa_{[1/1]}(\tilde\xi)$. Like for $\mu_{[1/1]}$ in [Eq. (\[eq:MuPade1\])]{}, we begin with a most general form $${\label{eq:KaPade1}}
    \kappa_{[1/1]}(\tilde\xi) = \frac{a_{1,1} + a_{1,2} \tilde\xi}{a_{2,1} + a_{2,2} \tilde\xi},$$ where $a_{1,1}$, $a_{1,2}$, etc. are constants. Since a cumulant-generating function must vanish at $\tilde\xi=0$, as required by the normalization of probability densities, we then have $a_{1,1} = 0$. With $B = -a_{2,2}/a_{2,1}$ and $a_2=-a_{1,2} / (B a_{2,2})$, [Eq. (\[eq:KaPade1\])]{} becomes $${\label{eq:KaPade2}}
    \kappa_{[1/1]}(\tilde\xi) = \frac{a_2 B \tilde\xi}{1 - B \tilde\xi}.$$

Now we can take a continuous limit in the manner we did for the 