I have a problem understanding a calculation in this paper (another form of the theorem an be found here at equation 11). For those who want to read the paper, I have difficulties with formula 2.14 in particular with the first passage in which the Authors moves from the derivative to that expression with sums.
For those who don't want to read the paper, I give a summary in the following: they take into account a Markov Jump process moving between state $\sigma$ with rates $k\left(\sigma,\sigma'\right)$ with $r\left(\sigma\right)=\sum_{\sigma'}k\left(\sigma,\sigma'\right)$
The generator is defined as: \begin{equation} Lf\left(\sigma\right)=\sum_{\sigma'}k\left(\sigma,\sigma'\right)\left[f\left(\sigma'\right)-f\left(\sigma\right)\right] \end{equation}
$\mu\left(\sigma,t\right)$ is probability distribution of $\sigma$ at time $t$ so that $\left\langle f\right\rangle_{\mu\left(t\right)}=\sum_{\sigma}\mu\left(\sigma,t\right)f\left(\sigma\right)$. We then have: \begin{equation} \frac{d}{dt}\left\langle f\right\rangle_{\mu\left(t\right)}=\left\langle L f\right\rangle_{\mu\left(t\right)} \end{equation} If the function $f$ is taken as a Kronecker delta at $\sigma$ we have that: \begin{equation} \frac{\partial\mu\left(\sigma,t\right)}{\partial t}=\sum_{\sigma'}k\left(\sigma',\sigma\right)\mu\left(\sigma',t\right)-r\left(\sigma\right)\mu\left(\sigma,t\right)=L^*\mu\left(\sigma,t\right) \end{equation}
Now they define this quantity: \begin{equation} g\left(\sigma,t\right)=\mathbb{E}_\sigma\left[e^{-\lambda W\left(t\right)}\right] \end{equation} where
-$\mathbb{E}_{\sigma}$ is the expectation value conditioned on the system being in state $σ$ at time $t=0$ (I think this is a different average with respect to the previous one but they must be somehow related)
-$W\left(t,\left\{\sigma_s,0\leqslant s\leqslant t\right\}\right)=\int_{0}^{t}\sum_{\sigma,\sigma'}\omega_{\sigma,\sigma'}\left(s\right)ds$ is an artificial quantity they define in which: $\sigma_s$ is a trajectory of the jump process and $\omega_{\sigma,\sigma'}\left(s\right)$ is a sequence of $δ$-functions, located exactly at those times $s$ when $\sigma_s$ jumps from $\sigma$ to $\sigma'$ with weight $\omega\left(\sigma,\sigma'\right)$.
Now here comes my problem: they say that, \begin{alignat}{2} \frac{d}{dt}g\left(\sigma,t\right)=&\sum_{\sigma'}k\left(\sigma,\sigma'\right)e^{-\lambda \omega\left(\sigma,\sigma'\right)}g\left(\sigma',t\right)-r\left(\sigma\right)g\left(\sigma,t\right)\\ =&\sum_{\sigma'}k\left(\sigma,\sigma'\right)^{1-\lambda}k\left(\sigma,\sigma'\right)^{\lambda}g\left(\sigma',t\right)-r\left(\sigma\right)g\left(\sigma,t\right)\\ =&L_{\lambda}g\left(\sigma,t\right)\\ \end{alignat} I didn't understand the definition of $\mathbb{E}_\sigma$ together with the first passage of the $\frac{d}{dt}g\left(\sigma,t\right)$.
I hope to have stated things in a clear and complete way! Also references are appreciated!
Thanks in advance!!