variable. It can be decomposed as $$Y = \sum_{u \in N_t} Y(u) \mathbf{1}_{\{ node_t(\xi) = u \}} \text{,}$$ where for all $u \in N_t$, $Y(u)$ is $\mathcal{F}_t$-measurable and then $$E^x \Big( \sum_{u \in N_t} Y(u) \Big) = \tilde{E}^x \Big( Y \mathrm{e}^{\hat{\beta}_0 \tilde{L}_t + \hat{\beta} t}\Big) \text{,}$$ In particular, if $f$ is a sufficiently nice functional then $$\label{many_to_one_eq}
E^x \Big[ \sum_{u \in N_t} f \Big( \big(X^u_s\big)_{s \in [0,t]}\Big) \Big] = \tilde{E}^x \Big[ f\Big(\big(\xi_s\big)_{s \in [0,t]} \Big) 
\mathrm{e}^{\hat{\beta}_0 \tilde{L}_t + \hat{\beta} t}\Big] \text{.}$$

Let us now apply to prove equations and given as the motivation in the first section.

Take $x \geq 0$ and $t \geq 0$. Then $$\begin{aligned}
\label{eq_enx1}
E |N_t^x| &=  E \sum_{u \in N_t} \mathbf{1}_{\{X^u_t > x\}}\nonumber\\
&= \tilde{E} \Big[ \mathbf{1}_{\{\xi_t > x\}} \mathrm{e}^{\hat{\beta}_0 \tilde{L_t} + \hat{\beta} t}\Big] \text{.}\end{aligned}$$ We now make use of the joint density of $\xi_t$ and $\tilde{L}_t$ (which for example can be found in [@KS84]): $$\tilde{P} \Big( \xi_t \in \mathrm{d}y, \ \tilde{L}_t \in \mathrm{d} l\Big) = 
\frac{|y| + l}{\sqrt{2 \pi t^3}} \exp \Big\{ - \frac{(|y| + l)^2}{2t}\Big\} \mathrm{d}y \mathrm{d}l \qquad y \in \mathbb{R}, \ l \geq 0$$ to complete the proof of . $$\begin{aligned}
\tilde{E} \Big[ \mathbf{1}_{\{\xi_t > x\}} \mathrm{e}^{\hat{\beta_0} \tilde{L_t} + \hat{\beta} t}\Big] &= 
