filtration which contains all the information about all the particles’ motion and their genealogy. It doesn’t however have any information about the spine.

-   $\tilde{\mathcal{F}}_t := \sigma \big(\mathcal{F}_t, \ node_t(\xi)\big)$. Thus $\tilde{\mathcal{F}}$ has all the information about the branching process and all the information about the spine. This will be the largest filtration.

-   $\mathcal{G}_t := \sigma \big( \xi_s : 0 \leq s \leq t \big)$. This filtration only contains information about the path of the spine but it doesn’t know which particles make up the spine along its path at different times.

-   $\tilde{\mathcal{G}}_t := \sigma \big( \mathcal{G}_t, \ node_s(\xi) : 0 \leq s \leq t , \ A^0_n : n \leq n^0_t, \ A_m' : m \leq n_t'\big)$. This filtration has information about both the path of the spine, its genealogy and how many particles are born along the path of the spine. However it has no information about anything happenning off the spine.

We note that $\mathcal{F}_t \subset \tilde{\mathcal{F}}_t$ and $\mathcal{G}_t \subset \tilde{\mathcal{G}}_t \subset \tilde{\mathcal{F}}_t$.

Many-to-One lemma and applications
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The proof of the following result with a detailed discussion can be found for example in [@HH09] or [@HR].

\[many\_to\_one\_lemma\] Let $Y$ be a non-negative $\tilde{\mathcal{F}}_t$-measurable random 