0, \\
    \frac{n}{m}p_n& \text{if } x \neq 0. 
    \end{array} \right.$$

-   Uniformly at random one of the new particles is selected to continue the spine and thus to stochastically repeat the behaviour of the initial particle starting from $x$.

-   The remaining $A(x) - 1$ particles initiate independent unbiased copies of a branching process with branching rate $\beta_0 \delta_0(\cdot) + \beta$ and offspring distribution $A(\cdot)$ as under $P^x$.

Suppose now that for all $t \geq 0$ $\tilde{M}^{(1)}_t$ can be represented as $$\tilde{M}^{(1)}_t = \sum_{u \in N_t} M^{(1)}_t(u) \mathbf{1}_{\{ node_t(\xi) = u \}},$$ where for all $u \in N_t$, $M^{(1)}_t(u)$ is $\mathcal{F}_t$-measurable. E.g. if $\tilde{M}^{(1)}_t = \mathrm{e}^{\lambda |\xi_t| - \lambda \tilde{L}_t - \frac{\lambda^2}{2}t}$ then we get the required representation by taking $M^{(1)}_t(u) = \mathrm{e}^{\lambda |X^u_t| - \lambda L^u_t - \frac{\lambda^2}{2}t}$. If we define $$M_t := \sum_{u \in N_t} M^{(1)}_t(u) \mathrm{e}^{- \hat{\beta}_0 L_t^u - \hat{\beta}t} 
\qquad \text{ , } t \geq 0$$ it can be checked that $(M_t)_{t \geq 0}$ is a unit-mean $P$-martingale such that $$M_t = \tilde{E} \Big( \tilde{M}_t \big\vert \mathcal{F}_t\Big)$$ and that if $Q := \tilde{Q} \big\vert_{\mathcal{F}_\infty}$ then $$\frac{\mathrm{d}Q}{\mathrm{d}P} \Big\vert_{\mathcal{F}_t} = M_t \text{ , } \qquad t \geq 0.$$

Since $(M_t)_{t \geq 0}$ is always a non-negative $P$-martingale, it 