2.1 Spine construction

In this section we extend our probability space by introducing the spine process. A more detailed description of this procedure may be found for example in [Reference Hardy and Harris9].

The spine of the branching process, which we shall denote by $\xi$ , is the infinite line of descent of particles chosen uniformly at random from all possible lines of descent. It is constructed in the following way. The initial particle $\emptyset$ of the branching process begins the spine. When the initial particle dies and is replaced with a random number of new particles, one of them is chosen uniformly at random to continue the spine. This procedure is then repeated recursively: whenever the particle in the spine dies, one of its children is chosen uniformly at random to continue the spine. We may then write the spine as $\xi = \{ \xi^{(0)}, \xi^{(1)}, \xi^{(2)}, \ldots\}$ , where $\xi^{(n)}$ is the label of the spine particle in the nth generation and $\xi^{(0)} = \emptyset$ .

We let $\tilde{\mathbb{P}}$ denote the probability measure under which the branching process is defined together with the spine. Hence $\mathbb{P} = \tilde{\mathbb{P}}|_{\mathcal{F}_\infty}$ . We let $\tilde{\mathbb{E}}$ be the expectation corresponding to $\tilde{\mathbb{P}}$ .

Below we introduce some new notation in relation with the spine process.

For any $t \ge 0$ we write ${{\mathop{\rm node}\nolimits} _t}(\xi )$ for the unique particle $u \in N_t \cap \xi$ . That is, ${{\mathop{\rm node}\nolimits} _t}(\xi )$ is the label of the spine particle at time t.

For any $t \ge 0$ we write $\xi_t$ for $X^u_t$ , where u is the unique particle in $N_t \cap \xi$ . Then $\xi_t$ is the spatial position of the spine particle at time t. It is not hard to check that the process ${(\xi_t)_{t \ge 0}}$ is a Brownian motion under $\tilde{\mathbb{P}}$ . We let ${(\tilde{L}_t)_{t \ge 0}}$ denote its local time at the origin.

For any ${t \ge 0}$ we write $n_t$ for the unique n such that $\xi^{(n)} \in N_t$ . Then ${(n_t)_{t \ge 0}}$ is the counting process of the number of branching events that have occurred along the path of the spine by time t. We denote the sequence of times of these branching events by $S_n$ and the number of particles produced at each such branching event by $A_n$ , $n \geq 1$ .

Moreover, we would like to distinguish branching events along the spine that occurred due to catalytic branching from those that occurred due to homogeneous branching. In order to do so, we denote the branching times along the spine that took place when the spine was at the origin by $S_n^0$ , and the number of particles produced at these times by $A_n^0$ , $n \geq 1$ . Similarly, we denote the branching times along the spine when it was not at the origin by $S_n'$ , and the number of particles produced at these times by $A_n'$ , $n \geq 1$ . We also denote the counting processes for ${(S_n^0)_{n \geq 1}}$ and ${(S_n')_{n \geq 0}}$ by ${(n_t^0)_{t \ge 0}}$ and ${(n_t')_{t \ge 0}}$ , respectively.

Observe that conditional on the path of the spine ${(\xi_t)_{t \ge 0}}$ , ${(n_t^0)_{t \ge 0}}$ and ${(n_t')_{t \ge 0}}$ are independent (inhomogeneous in the first case) Poisson processes (or Cox processes) with jump rates $\beta_0 \delta_0({\cdot})$ and $\beta$ , respectively, so that

$$\tilde{\mathbb{P}}( n_t^0 = k \!{\mid}\! (\xi_s)_{0 \leq s \leq t}) = {(\beta_0 \tilde{L}_t)^k}/{k!} \,\mathrm{e}^{- \beta_0 \tilde{L}_t}$$

and

$$\tilde{\mathbb{P}}( n_t' = k \!{\mid}\! (\xi_s)_{0 \leq s \leq t}) = {(\beta t)^k}/{k!} \,\mathrm{e}^{- \beta t}.$$

Finally, it is convenient to define several filtrations of the now extended probability space in order to take various conditional expectations.

2.2. Many-to-One Lemma and applications

The proof of the following result with a detailed discussion can be found for example in [Reference Hardy and Harris9] or [Reference Harris and Roberts11].

Lemma 1. (Many-to-One Lemma.) Let Y be a non-negative $\tilde{\mathcal{F}}_t$ -measurable random variable. It can be decomposed as

$$Y = \sum_{u \in N_t} Y(u) \mathbf{1}_{\{ {{\mathop{\rm node}\nolimits} _t}(\xi ) = u \}} ,$$

where for all $u \in N_t$ , Y(u) is $\mathcal{F}_t$ -measurable and then

$$\mathbb{E}^x \bigg( \sum_{u \in N_t} Y(u) \bigg) = \tilde{\mathbb{E}}^x ( Y \,\mathrm{e}^{\hat{\beta}_0 \tilde{L}_t + \hat{\beta} t}) {.}$$

In particular, if f is a non-negative functional such that $f((\xi_s)_{s \in [0,t]})$ is a $\mathcal{G}_t$ -measurable random variable, then

(14) $$\begin{equation} \mathbb{E}^x \bigg[ \sum_{u \in N_t} f ( (X^u_s)_{s \in [0,t]}) \bigg] = \tilde{\mathbb{E}}^x [ f((\xi_s)_{s \in [0,t]} ) \,\mathrm{e}^{\hat{\beta}_0 \tilde{L}_t + \hat{\beta} t}] . \end{equation}$$

Let us now apply (14) to prove equations (3) and (8) given as the motivation in the first section.

Proof of Proposition 1 and identity (8). Take $x \geq 0$ and $t \ge 0$ . Then

(15) $$\begin{align} \mathbb{E} |N_t^x| = \mathbb{E} \sum_{u \in N_t} \mathbf{1}_{\{X^u_t > x\}} = \tilde{\mathbb{E}} [ \mathbf{1}_{\{\xi_t > x\}} \,\mathrm{e}^{\hat{\beta}_0 \tilde{L_t} + \hat{\beta} t}] . \end{align}$$

We now make use of the joint density of $\xi_t$ and $\tilde{L}_t$ (which for example can be found in [Reference Karatzas and Shreve12]):

$$\tilde{\mathbb{P}} ( \xi_t \in \mathrm{d}y,\ \tilde{L}_t \in \mathrm{d} l) = {|y| + l}/{\sqrt{2 \pi t^3}} \exp \bigg\{ - {(|y| + l)^2}/{2t}\bigg\} \,\mathrm{d}y \,\mathrm{d}l{,} \quad y \in \mathbb{R},\ l \geq 0{,}$$

to complete the proof of (3). Then

$$\begin{align*} \tilde{\mathbb{E}} [ \mathbf{1}_{\{\xi_t > x\}} \,\mathrm{e}^{\hat{\beta}_0 \tilde{L_t} + \hat{\beta} t}] &= \mathrm{e}^{\hat{\beta} t} \int_0^\infty \int_x^{\infty} \,\mathrm{e}^{\hat{\beta}_0 l} {y + l}/{\sqrt{2 \pi t^3}} \exp \bigg\{ - {(y+l)^2}/{2t}\bigg\} \,\mathrm{d}y \,\mathrm{d}l\\ &= \mathrm{e}^{\hat{\beta} t} \int_0^\infty \,\mathrm{e}^{\hat{\beta}_0 l} {1}/{\sqrt{2 \pi t}} \exp \bigg\{ - {(x+l)^2}/{2t}\bigg\} \,\mathrm{d}l\\ &= \mathrm{e}^{\hat{\beta} t} \int_0^\infty {1}/{\sqrt{2 \pi t}} \exp \bigg\{ - {1}/{2t} ( l - (\hat{\beta}_0t - x))^2 + {\hat{\beta}_0^2}/{2}t - \hat{\beta}_0 x \bigg\} \,\mathrm{d}l\\ &= \mathrm{e}^{\hat{\beta}t + {\hat{\beta}_0^2}/{2}t - \hat{\beta}_0 x} \int_{- (\hat{\beta}_0 \sqrt{t} - \frac{x}{\sqrt{t}})}^\infty {1}/{\sqrt{2 \pi}} \,\mathrm{e}^{- \frac{z^2}{2}} \,\mathrm{d}z\\ &= \Phi \bigg( \hat{\beta}_0 \sqrt{t} - {x}/{\sqrt{t}}\bigg) \exp \bigg\{ {1}/{2} \hat{\beta}_0^2t -\hat{\beta}_0x + \hat{\beta}t \bigg\} . \end{align*}$$

For the expected total population we could have followed a similar calculation:

(16) $$\begin{equation} \mathbb{E}|N_t| = \tilde{\mathbb{E}} [ \mathrm{e}^{\hat{\beta}_0 \tilde{L_t} + \hat{\beta} t} ] = \cdots = 2 \Phi ( \hat{\beta}_0 \sqrt{t}) \exp \bigg\{ \bigg( {1}/{2} \hat{\beta}_0^2 + \hat{\beta}\bigg) t \bigg\}. \end{equation}$$

Let us now prove the upper bound for Theorem 1.

Proposition 2. (Upper bound for Theorem 1.)

(17) $$\begin{equation} \limsup_{t \to \infty} {1}/{t} \log |N_t| \leq {1}/{2} \hat{\beta}_0^2 + \hat{\beta} \quad \mathbb{P} \text{-a.s.} \end{equation}$$

Proof. Fix $\varepsilon > 0$ . Then, by the Markov inequality and (8),

$$\begin{align*} \mathbb{P} \bigg( \dfrac{1}{n} \log |N_n| > \dfrac{1}{2}\hat{\beta}_0^2 + \hat{\beta} + \varepsilon \bigg) = \mathbb{P} ( |N_n| > \mathrm{e}^{(\frac{1}{2} \hat{\beta}_0^2 + \hat{\beta} + \varepsilon)n} ) \leq \mathrm{e}^{-(\frac{1}{2} \hat{\beta}_0^2 + \hat{\beta} + \varepsilon)n} \mathbb{E}|N_n| < 2 \,\mathrm{e}^{- \varepsilon n}. \end{align*}$$

It follows from the Borel–Cantelli lemma that

$$\mathbb{P} \bigg( \bigg\{ \dfrac{1}{n} \log |N_n| > \dfrac{1}{2}\hat{\beta}_0^2 + \hat{\beta} + \varepsilon \bigg\} \text{ i.o.}\bigg) = 0$$

and thus

$$\limsup_{n \to \infty} \dfrac{1}{n} \log |N_n| \leq {1}/{2} \hat{\beta}_0^2 + \hat{\beta} + \varepsilon \quad \mathbb{P} \text{-a.s.}$$

By letting $\varepsilon \to 0$ we establish (17) with the limit taken over integer times. To get convergence over any real-valued sequence, we note that ${(|N_t|)_{t \ge 0}}$ is a non-decreasing process, and so for any $t > 0$

$$\dfrac{1}{t}\log |N_t| \leq \dfrac{\lceil t \rceil}{t} \dfrac{\log |N_{\lceil t \rceil}|}{\lceil t \rceil}{,}$$

and hence

$$\begin{align*} \limsup_{t \to \infty} {1}/{t} \log |N_t| \leq \limsup_{t \to \infty} \dfrac{\log |N_{\lceil t \rceil}|}{\lceil t \rceil} \leq {1}/{2} \hat{\beta}_0^2 + \hat{\beta} \quad \mathbb{P} \text{-a.s.}\\ \end{align*}$$

□

For upper bounds of Theorem 2 we need to adjust the previous argument because unlike ${(|N_t|)_{t \ge 0}}$ the process ${(|N_t^{\lambda t}|)_{t \ge 0}}$ is not monotone. We first establish the following result.

Proposition 3. For $\lambda > 0$ and $n \in \mathbb{N}\cup\{0\}$ , we define the following set of particles:

$$\hat{N}^{\lambda n}_n \,{:\!=}\, \Big\{ u \in N_{n+1} \colon \sup_{s \in [n, n+1]}X^u_s \geq \lambda n\Big\}.$$

Then

$$\limsup_{n \to \infty} \dfrac{1}{n} \log \mathbb{E} |\hat{N}^{\lambda n}_n| \leq \Delta_\lambda.$$

Note that for any $t \in [n, n+1]$ it is always true that $|N_t^{\lambda t}| \leq |\hat{N}^{\lambda n}_n|$ .

Proof of Proposition 3. By the Many-to-One Lemma we have

$$\begin{align*} \mathbb{E} |\hat{N}^{\lambda n}_n| &= \mathbb{E} \sum_{u \in N_{n+1}} \mathbf{1}_{\{\sup_{s \in [n, n+1]}X^u_s \geq \lambda n\}}\\ &= \tilde{\mathbb{E}} [ \mathbf{1}_{\{\sup_{s \in [n, n+1]}\xi_s \geq \lambda n\}} \,\mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)}]\\ &= \tilde{\mathbb{E}} [ \mathbf{1}_{\{\xi_{n+1} + \bar{\xi}_n \geq \lambda n\}} \,\mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)}], \end{align*}$$

where $\bar{\xi}_n \,{:\!=}\, \sup_{s \in [n, n+1]} (\xi_s - \xi_{n + 1})$ and $\bar{\xi}_n \stackrel{d}{=} \sup_{s \in [0,1]} \xi_s \stackrel{d}{=} |\mathcal{N}(0,1)|$ under $\tilde{\mathbb{P}}$ .

Then, for any $\delta \in (0, \lambda)$ we can split the latter expectation as

$$\begin{align*} \tilde{\mathbb{E}} [ \mathbf{1}_{\{\xi_{n+1} + \bar{\xi}_n \geq \lambda n\}} \,\mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)}] &= \tilde{\mathbb{E}} [ \mathbf{1}_{\{\xi_{n+1} + \bar{\xi}_n \geq \lambda n\}} \,\mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)} \mathbf{1}_{\{|\xi_{n+1}| \leq (\lambda - \delta)n\}} ]\\&\quad\, + \tilde{\mathbb{E}} [ \mathbf{1}_{\{\xi_{n+1} + \bar{\xi}_n \geq \lambda n\}} \,\mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)} \mathbf{1}_{\{|\xi_{n+1}| > (\lambda - \delta)n\}}] {.} \end{align*}$$

We shall refer to the first term in the sum as $I_1$ and the second one as $I_2$ . First we show that the contribution of $I_1$ is negligibly small as it has a faster than exponential decay rate:

$$\begin{align*} I_1 &= \tilde{\mathbb{E}} [ \mathbf{1}_{\{\xi_{n+1} + \bar{\xi}_n \geq \lambda n\}} \,\mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)} \mathbf{1}_{\{|\xi_{n+1}| \leq (\lambda - \delta)n\}}]\\ &\leq \tilde{\mathbb{E}} [ \mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)} \mathbf{1}_{\{\bar{\xi}_n \geq \delta n\}} ]\\ &\leq ( \tilde{\mathbb{E}} [ \mathrm{e}^{2\hat{\beta}_0 \tilde{L}_{n+1} + 2\hat{\beta}(n+1)} ] )^{{1}/{2}} ( \tilde{\mathbb{P}} ( \bar{\xi}_n \geq \delta n ) )^{{1}/{2} } \end{align*}$$

using the Cauchy–Schwarz inequality in the last line. Then, as we know from (16),

$$\dfrac{1}{n} \log ( \tilde{\mathbb{E}} [ \mathrm{e}^{2\hat{\beta}_0 \tilde{L}_{n+1} + 2\hat{\beta}(n+1)} ] )^{{1}/{2} } \to \hat{\beta}_0^2 + \hat{\beta} \quad \text{as } n \to \infty{,}$$

while, since $\bar{\xi}_n \stackrel{d}{=} |\mathcal{N}(0,1)|$ ,

$$\dfrac{1}{n^2} \log ( \tilde{\mathbb{P}} ( \bar{\xi}_n \geq \delta n ) )^{{1}/{2} } \to \dfrac{-\delta^2}{4} \quad \text{as } n \to \infty.$$

Thus

(18) $$\begin{equation} \limsup_{n \to \infty} \dfrac{1}{n^2} \log I_1 \leq - \dfrac{\delta^2}{4}. \end{equation}$$

On the other hand,

$$\begin{align*} I_2 &= \tilde{\mathbb{E}} [ \mathbf{1}_{\{\xi_{n+1} + \bar{\xi}_n \geq \lambda n\}} \,\mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)} \mathbf{1}_{\{|\xi_{n+1}| > (\lambda - \delta)n\}} ]\\ &\leq \tilde{\mathbb{E}} [ \mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)} \mathbf{1}_{\{|\xi_{n+1}| > (\lambda - \delta)n\}}]\\ &= 2\tilde{\mathbb{E}} [ \mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)} \mathbf{1}_{\{\xi_{n+1} > (\lambda - \delta)n\}}]\\ &= 2\mathbb{E}|N_{n+1}^{(\lambda - \delta)n}|{,} \end{align*}$$

using symmetry in the third line and identity (15) in the fourth line. Thus, from (3) we can see (just as we did in (4)–(6)) that

(19) $$\begin{equation} \limsup_{n \to \infty} \dfrac{1}{n} \log I_2 \leq \Delta_{\lambda - \delta}. \label{eqn19} \end{equation}$$

From (18) and (19) we have that, for any $\delta \in (0, \lambda)$ ,

$$\limsup_{n \to \infty} \dfrac{1}{n} \log \mathbb{E} |\hat{N}^{\lambda n}_n| = \limsup_{n \to \infty} \dfrac{1}{n} \log(I_1 + I_2) \leq \Delta_{(\lambda-\delta)}.$$

Letting $\delta \to 0$ and using continuity and monotonicity of $\Delta_{\lambda}$ as a function of $\lambda$ , we obtain the sought result.

Proposition 3 can now be applied to prove the upper bounds for Theorem 2.

Proposition 4. (Upper bounds for Theorem 2.)

If $\lambda < \lambda_{\mathrm{crit}}$ $(\Delta_\lambda > 0)$ , then

(20) $$\begin{equation} \limsup_{t \to \infty} {1}/{t} \log |N_t^{\lambda t}| \leq \Delta_\lambda \quad \mathbb{P} \text{-a.s.} \end{equation}$$

If $\lambda > \lambda_{\mathrm{crit}}$ $(\Delta_\lambda < 0)$ , then

(21) $$\begin{equation} \lim_{t \to \infty} |N_t^{\lambda t}| = 0 \quad \mathbb{P} \text{-a.s.} \end{equation}$$

and

(22) $$\begin{equation} \limsup_{t \to \infty} {1}/{t} \log \mathbb{P} (|N_t^{\lambda t}| > 0) \leq \Delta_{\lambda} . \end{equation}$$

Proof. For any $\lambda >0$ , let $\hat{N}_n^{\lambda n}$ be as in the previous proposition and fix $\varepsilon > 0$ . Then the Markov inequality gives

$$\mathbb{P} ( |\hat{N}_n^{\lambda n}| > \mathrm{e}^{(\Delta_\lambda + \varepsilon)n} ) \leq \mathrm{e}^{-(\Delta_\lambda + \varepsilon)n} \mathbb{E} |\hat{N}_n^{\lambda n}|{,}$$

and from Proposition 3 the right-hand side decays exponentially fast. Therefore, by the Borel–Cantelli lemma,

$$\mathbb{P} ( \{|\hat{N}_n^{\lambda n}| > \mathrm{e}^{(\Delta_\lambda + \varepsilon)n} \} \text{ i.o.}) = 0.$$

This is equivalent to saying that

$$|\hat{N}_n^{\lambda n}| \leq \mathrm{e}^{(\Delta_\lambda + \varepsilon)n} \quad \text{eventually } \mathbb{P} \text{-a.s.}$$

So $\mathbb{P}$ -almost surely, for all t sufficiently large,

$$|N_t^{\lambda t}| \leq |\hat{N}_{\lfloor t \rfloor}^{\lambda \lfloor t \rfloor}| \leq \mathrm{e}^{(\Delta_\lambda + \varepsilon)\lfloor t \rfloor}{.}$$

Then, if $\lambda > \lambda_{\mathrm{crit}}$ , we can take $\varepsilon$ sufficiently small that $\Delta_\lambda + \varepsilon < 0$ and hence $|N_t^{\lambda t}| < 1$ for t large enough, thus proving (21).

If $\lambda < \lambda_{\mathrm{crit}}$ , then we get

$$\limsup_{t \to \infty} {1}/{t} \log |N_t^{\lambda t}| \leq \Delta_\lambda + \varepsilon \quad \mathbb{P} \text{-a.s.}{,}$$

and letting $\varepsilon \to 0$ yields (20).

Finally, if $\lambda > \lambda_{\mathrm{crit}}$ then (22) follows from the Markov inequality and (5):

$$\begin{align*} \limsup_{t \to \infty} {1}/{t} \log \mathbb{P} (|N_t^{\lambda t}| > 0) \leq \limsup_{t \to \infty} {1}/{t} \log \mathbb{E} |N_t^{\lambda t}| = \Delta_{\lambda}.\\ \end{align*}$$

2.3. Additive martingales and applications

Recall that under the probability $\tilde{\mathbb{P}}$ the branching process together with the spine may be described as follows.

• The process starts with a single spine particle whose path ${(\xi_t)_{t \ge 0}}$ is distributed like a Brownian motion.

• At instantaneous rate $\beta_0 \delta_0({\cdot}) + \beta$ along its path, the spine particle splits into $A({\cdot})$ particles. If splitting took place at position x, then

  $$\begin{equation*} \tilde{\mathbb{P}}(A(x) = n) = \begin{cases} q_n& \text{if } x = 0, \\ p_n& \text{if } x \neq 0. \end{cases} \end{equation*}$$

• 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 copies of a branching process with branching rate $\beta_0 \delta_0({\cdot}) + \beta$ and offspring distribution $A({\cdot})$ as under $\mathbb{P}^x$ .

We shall now describe a family of martingale changes of measure that will put a certain bias on the motion of the spine particle as well as the birth rate and the offspring distribution along the path of the spine particle. Again, for a detailed discussion the reader is referred to [Reference Hardy and Harris9].

Let us consider a process of the form

$$\skew1\tilde{M}_t = \bigg[ \prod_{n=1}^{n_t} A_{n} \bigg] \,\mathrm{e}^{- \hat{\beta}t - \hat{\beta}_0 \tilde{L}_t} \skew1\tilde{M}^{(1)}_t = \bigg[ \prod_{n=1}^{n_t'} A_{n}' \times \prod_{n=1}^{n_t^0} A_{n}^0 \bigg] \,\mathrm{e}^{- \hat{\beta}t - \hat{\beta}_0 \tilde{L}_t} \skew1\tilde{M}^{(1)}_t,$$

where ${(A_n)_{n \geq 1}}$ , ${(A^0_n)_{n \geq 1}}$ , ${(A_n')_{n \geq 1}}$ , ${(n_t)_{t \ge 0}}$ , ${(n^0_t)_{t \ge 0}}$ , and ${(n_t')_{t \ge 0}}$ were defined in Subsection 2.1 and ${(\skew1\tilde{M}^{(1)})_{t \ge 0}}$ is a non-negative ${\tilde{\mathbb{P}}}$ -martingale of mean 1 with respect to the filtration ${(\mathcal{G}_t)_{t \ge 0}}$ . The effect of ${\skew1\tilde{M}^{(1)}}$ , if used as a change of measure martingale, is to put some drift on ${(\xi_t)_{t \ge 0}}$ .

For this particular paper we shall take ${\skew1\tilde{M}^{(1)}}$ to be either

$${\skew1\tilde{M}^{(1)}_t = \mathrm{e}^{\lambda \xi_t - {\lambda^2}/{2}t} , \quad t \ge 0}$$

or

$$\begin{align*} \skew1\tilde{M}^{(1)}_t = \mathrm{e}^{\lambda |\xi_t| - \lambda \tilde{L}_t - {\lambda^2}/{2}t} = \mathrm{e}^{\lambda \int_0^t \operatorname{sgn}(\xi_s) \,\mathrm{d}\xi_s - {\lambda^2}/{2}t} , \quad t \ge 0. \end{align*}$$

The first choice is the classical Girsanov martingale, which has the effect of adding constant drift ${\lambda}$ to ${(\xi_t)_{t \ge 0}}$ .

The second choice has the effect of adding instantaneous drift ${\lambda \operatorname{sgn}({\cdot})}$ to ${(\xi_t)_{t \ge 0}}$ , so that if ${\lambda < 0}$ then this is drift of magnitude ${|\lambda|}$ towards the origin, whereas ${\lambda> 0}$ then this is drift of magnitude ${\lambda}$ away from the origin. For more details one can refer to [Reference Bocharov and Harris1] or [2, ‘Brownian motion with alternating drift’, pp. 128–129]. Alternatively, the effect of this martingale can be seen as adding constant drift ${\lambda}$ to the Brownian motion ${(|\xi_t| - \tilde{L}_t)_{t \ge 0}}$ .

Recalling that ${{m = \smash{\sum_{n \geq 1}} n p_n}}$ and ${{m_0 = \smash{\sum_{n \geq 1}} n q_n}}$ are the means of the offspring distribution due to homogeneous and catalytic branching, we see that ${(\skew1\tilde{M})_{t \ge 0}}$ can be decomposed into a product of three martingales:

$${\tilde{M}_t = \skew1\tilde{M}^{(1)}_t \skew1\tilde{M}^{(2)}_t \skew1\tilde{M}^{(3)}_t{,}}$$

where

$$\tilde M_t^{(2)} = {m^{{n_{t'}}}}{\mkern 1mu} {{\rm{e}}^{ - \hat \beta t}} \times m_0^{n_t^0}{\mkern 1mu} {{\rm{e}}^{ - {{\hat \beta }_0}{{\tilde L}_t}}}$$

and

$$\skew1\tilde M_t^{(3)} = \prod\limits_{n = 1}^{{n_{t'}}} {\frac{{{A_{n'}}}}{m}} \times \prod\limits_{n = 1}^{n_t^0} {\frac{{A_n^0}}{{{m_0}}}} .$$

When used as the Radon–Nikodym derivative, ${(\tilde{M}^{(2)}_t)_{t \ge 0}}$ has the effect of changing the instantaneous jump rate of ${(n_t')_{t \ge 0}}$ from ${\beta}$ to ${m \beta}$ and the jump rate of ${(n_t^0)_{t \ge 0}}$ from ${\beta_0 \delta_0({\cdot})}$ to ${m_0 \beta_0 \delta_0({\cdot})}$ . The effect of ${(\skew1\tilde{M}^{(3)}_t)_{t \ge 0}}$ is to change the distribution of random variables ${{(A_n')_{n \geq 1}}}$ from ${{(p_k)_{k \geq 1}}}$ to ${{({k}/{m} p_k)_{k \geq 1}}}$ and the distribution of random variables ${{(A_n^0)_{n \geq 1}}}$ from ${{(q_k)_{k \geq 1}}}$ to ${({k}/{m_0} q_k)_{k \geq 1}}$ (while keeping them all independent). If we now define a new probability measure ${\tilde{\mathbb{Q}}}$ as

$$\begin{equation*} \dfrac{\mathrm{d} \tilde{\mathbb{Q}}}{\mathrm{d} \tilde{\mathbb{P}}} \bigg\vert_{\tilde{\mathcal{F}}_t} = \skew1\tilde{M}_t , \quad t \ge 0 {,} \end{equation*}$$

then for any events ${E_1 \in \mathcal{G}_t}$ , ${E_2 {\in} \sigma ( (n_s')_{0 \leq s \leq t}, (n_s^0)_{0 \leq s \leq t})}$ and ${E_3 {\in} \sigma (A_1', \ldots, A_{n_t'}';\ A_1^0, \ldots, A_{n_t^0}^0)}$ we have that

$$\begin{align*} \tilde{\mathbb{Q}}(E_1, E_2, E_3) & = \tilde{\mathbb{E}} ( \mathbf{1}_{E_1} \mathbf{1}_{E_2} \mathbf{1}_{E_3} \skew1\tilde{M}^{(1)}_t \skew1\tilde{M}^{(2)}_t \skew1\tilde{M}^{(3)}_t )\\ & = \tilde{\mathbb{E}} [ \mathbf{1}_{E_1} \skew1\tilde{M}^{(1)}_t \tilde{\mathbb{E}} ( \mathbf{1}_{E_2} \skew1\tilde{M}^{(2)}_t \tilde{\mathbb{E}} ( \mathbf{1}_{E_3} \tilde{M}^{(3)}_t \!{\mid}\! \sigma (\mathcal{G}_t, (n_s')_{0 \leq s \leq t}, (n_s^0)_{0 \leq s \leq t}) ) \!{\mid}\! \mathcal{G}_t ) ] \end{align*}$$

so it can be seen that the effects of ${\skew1\tilde{M}^{(1)}}$ , ${\skew1\tilde{M}^{(2)}}$ , and ${\tilde{M}^{(3)}}$ superimpose. Thus, under ${\tilde{\mathbb{Q}}}$ the branching process has the following description.

• The process starts with a single spine particle whose path ${(\xi_t)_{t \ge 0}}$ is distributed like a Brownian motion with drift imposed by ${\skew1\tilde{M}^{(1)}}$ .

• At instantaneous rate ${m_0 \beta_0 \delta_0({\cdot}) + m \beta}$ along its path the spine particle splits into ${A({\cdot})}$ particles. If splitting took place at position x, then

$$\begin{equation*} \tilde{\mathbb{Q}}(A(x) = n) = \begin{cases} \dfrac{n}{m_0}q_n& \text{if } x = 0, \\ \dfrac{n}{m}p_n& \text{if } x \neq 0. \end{cases} \end{equation*}$$

• 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 ${\mathbb{P}^x}$ .

Suppose now that, for all ${t \ge 0}$ , ${\skew1\tilde{M}^{(1)}_t}$ can be represented as

(23) $$\begin{equation} \skew1\tilde{M}^{(1)}_t = \sum_{u \in N_t} M^{(1)}_t(u) \mathbf{1}_{\{ {{\mathop{\rm node}\nolimits} _t}(\xi ) = u \}}, \end{equation}$$

where, for all ${u \in N_t}$ , ${M^{(1)}_t(u)}$ is ${\mathcal{F}_t}$ -measurable. For example, if ${\skew1\tilde{M}^{(1)}_t = \mathrm{e}^{\lambda |\xi_t| - \lambda \tilde{L}_t - {\lambda^2}/{2}t}}$ , then we get the required representation by taking ${M^{(1)}_t(u) = \mathrm{e}^{\lambda |X^u_t| - \lambda L^u_t - {\lambda^2}/{2}t}}$ . If we define

(24) $$\begin{equation} M_t \,{:\!=}\, \sum_{u \in N_t} M^{(1)}_t(u) \,\mathrm{e}^{- \hat{\beta}_0 L_t^u - \hat{\beta}t} , \quad t \ge 0 {,} \end{equation}$$

then ${(M_t)_{t \ge 0}}$ is a unit-mean ${\mathbb{P}}$ -martingale such that

$$M_t = \tilde{\mathbb{E}} ( \skew1\tilde{M}_t \!{\mid}\!\mathcal{F}_t ).$$

To see this, note that

$$\begin{align*} \tilde{\mathbb{E}} ( \skew1\tilde{M}_t \!{\mid}\! \mathcal{F}_t ) &= \tilde{\mathbb{E}} \bigg( \bigg[ \prod_{n=1}^{n_t} A_n \bigg] \,\mathrm{e}^{- \hat{\beta}t - \hat{\beta}_0 \tilde{L}_t} \skew1\tilde{M}^{(1)}_t \biggm\vert \mathcal{F}_t \bigg)\\&= \tilde{\mathbb{E}} \bigg( \sum_{u \in N_t} \bigg( \bigg[ \prod_{v<u} A_v \bigg] \,\mathrm{e}^{- \hat{\beta}t - \hat{\beta}_0 L^u_t} \tilde{M}^{(1)}_t(u) \mathbf{1}_{\{{{\mathop{\rm node}\nolimits} _t}(\xi ) = u\}} \bigg) \biggm\vert \mathcal{F}_t \bigg)\\ &= \sum_{u \in N_t} \bigg( \bigg[ \prod_{v<u} A_v \bigg] \,\mathrm{e}^{- \hat{\beta}t - \hat{\beta}_0 L^u_t} \skew1\tilde{M}^{(1)}_t(u) \tilde{\mathbb{P}} ( u \in \xi\!{\mid}\!\mathcal{F}_t ) \bigg)\\ &= \sum_{u \in N_t} \bigg( \bigg[ \prod_{v<u} A_v \bigg] \,\mathrm{e}^{- \hat{\beta}t - \hat{\beta}_0 L^u_t} \skew1\tilde{M}^{(1)}_t(u) \prod_{v<u} \dfrac{1}{A_v} \bigg)\\&= M_t. \end{align*}$$

For the martingale property of ${(M_t)_{t \ge 0}}$ , we then check that, for any ${s \leq t}$ and an event ${A \in \mathcal{F}_s}$ ,

$$\begin{align*} \mathbb{E}(M_t \mathbf{1}_A) & = \tilde{\mathbb{E}}(M_t \mathbf{1}_A) \\ & = \tilde{\mathbb{E}} ( \tilde{\mathbb{E}}(\skew1\tilde{M}_t \!{\mid}\! \mathcal{F}_t) \mathbf{1}_A) \\ & = \tilde{\mathbb{E}} ( \tilde{\mathbb{E}} (\skew1\tilde{M}_t \mathbf{1}_A \!{\mid}\! \mathcal{F}_t) ) \\ & = \tilde{\mathbb{E}} (\skew1\tilde{M}_t \mathbf{1}_A) \\ & = \tilde{\mathbb{E}} (\skew1\tilde{M}_s \mathbf{1}_A) \\ & = \tilde{\mathbb{E}} ( \tilde{\mathbb{E}} (\skew1\tilde{M}_s \mathbf{1}_A \!{\mid}\! \mathcal{F}_s) ) \\ & = \tilde{\mathbb{E}} ( \tilde{\mathbb{E}}(\skew1\tilde{M}_s \!{\mid}\! \mathcal{F}_s) \mathbf{1}_A)\\ & = \tilde{\mathbb{E}}(M_s \mathbf{1}_A) \\ & = \mathbb{E}(M_s \mathbf{1}_A). \end{align*}$$

If we now define ${\mathbb{Q} \,{:\!=}\, \tilde{\mathbb{Q}} \vert_{\mathcal{F}_\infty}}$ , then we would have that

$${\dfrac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{P}} \bigg\vert_{\mathcal{F}_t} = M_t , \quad t \ge 0 {,}}$$

since for any event ${A \in \mathcal{F}_t}$ ,

$$\begin{align*} \mathbb{Q}(A) = \tilde{\mathbb{Q}}(A) = \tilde{\mathbb{E}}(\mathbf{1}_A \skew1\tilde{M}_t) = \tilde{\mathbb{E}} (\tilde{\mathbb{E}}( \mathbf{1}_A \skew1\tilde{M}_t \!{\mid}\! \mathcal{F}_t) ) = \tilde{\mathbb{E}} ( \mathbf{1}_A \tilde{\mathbb{E}}(\skew1\tilde{M}_t \!{\mid}\! \mathcal{F}_t) ) = \tilde{\mathbb{E}}(\mathbf{1}_A M_t) = \mathbb{E} (\mathbf{1}_A M_t). \end{align*}$$

Since ${(M_t)_{t \ge 0}}$ is always a non-negative ${\mathbb{P}}$ -martingale, it must converge ${\mathbb{P}}$ -almost surely to some non-negative limit ${M_\infty}$ . Particularly interesting are those martingales whose limit is not almost surely 0.

When investigating whether the limit of M is ${\mathbb{P}}$ -almost surely 0 or not, we shall make use of the following result, commonly known as the spine decomposition.

Lemma 2. (Spine decomposition.) Let ${(M_t)_{t \ge 0}}$ be the ${\mathbb{P}}$ -martingale of the form (24) constructed from some given ${\tilde{\mathbb{P}}}$ -martingale ${(\skew1\tilde{M}^{(1)}_t)_{t \ge 0}}$ with the property (23). Then

$$\tilde{\mathbb{Q}} ( M_t \!{\mid}\! \tilde{\mathcal{G}}_\infty ) = spine(t) + \sum_{n=1}^{n_t} (A_n - 1) spine(S_n),$$

where

$${\operatorname{spine}(t) = \skew1\tilde{M}^{(1)}_t \,\mathrm{e}^{- \hat{\beta}t - \hat{\beta}_0 \tilde{L}_t}}$$

and, as defined earlier, ${S_n}$ is the time of the nth branching event along the path of the spine process and ${A_n}$ is the corresponding number of offspring produced.

Note that we have used ${\tilde{\mathbb{Q}}}$ to denote the expectation corresponding to probability measure ${\tilde{\mathbb{Q}}}$ , which is a common practice.

We have already used the fact that if ${A \in \mathcal{F}_t}$ for some ${t \ge 0}$ , then

$${\mathbb{Q}(A) = \int_A M_t \,\mathrm{d}\mathbb{P}.}$$

Let us also recall that if ${A \in \mathcal{F}_\infty}$ , then

(25) $$\begin{equation} \mathbb{Q}(A) = \int_A M_\infty \,\mathrm{d}\mathbb{P} + \mathbb{Q} \Big( A \cap \Big\{ \limsup_{t \to \infty} M_t = \infty\Big\} \Big). \end{equation}$$

The latter identity can be found in [7, p. 241].

Proposition 5. Let

(26) $$\begin{equation} M_t^\pm \,{:\!=}\, \sum_{u \in N_t} \,\mathrm{e}^{- \hat{\beta}_0 |X^u_t| - \frac{1}{2} \hat{\beta}_0^2 t - \hat{\beta} t} , \quad t \ge 0{,} \end{equation}$$

be the ${\mathbb{P}}$ -martingale of the form (24) constructed by taking

(27) $$\begin{align} \skew1\tilde{M}^{(1)}_t = \mathrm{e}^{- \hat{\beta}_0 |\xi_t| + \hat{\beta}_0 \tilde{L}_t - \frac{1}{2} \hat{\beta}_0^2 t} , \quad t \ge 0.\label{eqn27}\end{align}$$

$$\begin{align}\nonumber \end{align}$$

(a) If condition (10) (the ${X \log X}$ condition on the offspring distribution) is satisfied, then

$$M^\pm_\infty > 0 \quad \mathbb{P} \text{-a.s.}$$

(b) If condition (10) is not satisfied, then

$$M^\pm_\infty = 0 \quad \mathbb{P} \text{-a.s.}$$

Proof. Let ${\mathbb{Q}^\pm}$ and ${\tilde{\mathbb{Q}}^\pm}$ be the probability measures associated with martingales (26) and (27) as previously described in this subsection.

A standard argument which relies only on point-recurrence of ${(\xi_t)_{t \ge 0}}$ under ${\tilde{\mathbb{P}}}$ (see e.g. [Reference Harris and Harris10]) tells us that ${\mathbb{P}(M^\pm_\infty > 0) \in \{0, 1\}}$ .

So if we could show that ${\mathbb{Q}^\pm(\limsup_{t \to \infty} M^\pm_t = \infty) = 1}$ , then by taking ${A = \Omega}$ in (25) we would get ${1 = \mathbb{E} M^\pm_\infty + 1}$ and hence ${M^\pm_\infty = 0\ \mathbb{P}}$ -almost surely.

On the other hand, if we could show that ${\mathbb{Q}^\pm(\limsup_{t \to \infty} M^\pm_t = \infty) = 0}$ then, again by taking ${A = \Omega}$ in (25), we would get ${1 = \mathbb{E} M^\pm_\infty + 0}$ and hence ${\mathbb{P}(M^\pm_\infty > 0) > 0}$ , which from the 0–1 law above is the same as ${\mathbb{P}(M^\pm_\infty > 0) = 1}$ .

Thus, to prove the proposition, it is sufficient to show that, if condition (10) is satisfied, then ${\mathbb{Q}^\pm(\limsup_{t \to \infty} M^\pm_t = \infty) = 0}$ , and if condition (10) is not satisfied, then ${\mathbb{Q}^\pm(\limsup_{t \to \infty} M^\pm_t = \infty) =1}$ .

Let us observe that because of the effect of martingale ${\skew1\tilde{M}^{(3)}}$ , for any choice of ${c > 0}$ we have

$$\begin{align*} \dfrac{1}{c} \tilde{\mathbb{E}} (A_1' \log A_1' ) &= \dfrac{1}{c} m \tilde{\mathbb{Q}}^\pm (\log A_1' ) \\ &= m \int_0^\infty \tilde{\mathbb{Q}}^\pm\bigg( \dfrac{1}{c} \log A_1' \geq t\bigg) \,\mathrm{d}t \\ &= m \sum_{n=0}^\infty \int_n^{n+1} \tilde{\mathbb{Q}}^\pm (\log A_1' \geq ct ) \,\mathrm{d}t. \end{align*}$$

Hence, by monotonicity of ${\tilde{\mathbb{Q}}^\pm (\log A_1' \geq ct)}$ , we have

(28) $$\begin{equation} m c \sum_{n=1}^\infty \tilde{\mathbb{Q}}^\pm (\log A_1' \geq c n ) \leq \tilde{\mathbb{E}} (A_1' \log A_1' ) \leq m c \sum_{n=0}^\infty \tilde{\mathbb{Q}}^\pm (\log A_1' \geq c n ). \end{equation}$$

Then, since ${(A_n')_{n \geq 1}}$ are i.i.d. random variables, it follows from (28) that

$$\begin{align*} \sum_{n = 1}^\infty \tilde{\mathbb{Q}}^\pm (A_n' \geq \mathrm{e}^{cn} ) = \sum_{n = 1}^\infty \tilde{\mathbb{Q}}^\pm (\log A_1' \geq c n) < \infty \quad \iff \quad \tilde{\mathbb{E}} (A_1' \log A_1' ) = \sum_{n=1}^\infty p_n n \log n < \infty{,} \end{align*}$$

and then, by the first and second Borel–Cantelli lemmas,

(29) $$\begin{equation} \tilde{\mathbb{Q}}^\pm ( \{ A_n' \geq \mathrm{e}^{cn}\} \text{ i.o.}) = \begin{cases} 0& \displaystyle \text{if } \sum_{n=1}^\infty p_n n \log n < \infty, \\\\ 1& \displaystyle \text{if } \sum_{n=1}^\infty p_n n \log n = \infty. \end{cases} \end{equation}$$

An identical argument gives

(30) $$\begin{equation} \tilde{\mathbb{Q}}^\pm ( \{ A_n^0 \geq \mathrm{e}^{cn}\} \text{ i.o.}) = \begin{cases} 0& \displaystyle \text{if } \sum_{n=1}^\infty q_n n \log n < \infty, \\\\ 1& \displaystyle \text{if } \sum_{n=1}^\infty q_n n \log n = \infty. \end{cases} \end{equation}$$

Let us emphasize that dichotomies (29) and (30) hold for any choice of ${c>0}$ .

(a) Assume now that condition (10) holds, and recall from Lemma 2.3 that

$$\begin{align*} \tilde{\mathbb{Q}}^\pm ( M^\pm_t \!{\mid}\! \tilde{\mathcal{G}}_\infty ) &= \operatorname{spine}(t) + \sum_{n=1}^{n_t'} (A_n' - 1) \operatorname{spine}(S_n') + \sum_{n=1}^{n^0_t} (A_n^0 - 1) \operatorname{spine}(S^0_n)\\&\leq 1 + \sum_{n=1}^\infty (A_n' - 1) \operatorname{spine}(S_n') + \sum_{n=1}^\infty (A_n^0 - 1) \operatorname{spine}(S^0_n), \end{align*}$$

where

$${\operatorname{spine}(t) = \mathrm{e}^{- \hat{\beta}_0 |\xi_t| - \frac{1}{2} \hat{\beta}_0^2t - \hat{\beta}t}.}$$

Recall that martingale (27), when used as the Radon–Nikodym derivative, has the effect of putting constant drift of magnitude ${\hat{\beta}_0}$ towards the origin onto ${(\xi_t)_{t \ge 0}}$ or, equivalently, the effect of putting constant drift ${\hat{\beta}_0}$ onto ${(\tilde{L}_t - |\xi_t|)_{t \ge 0}}$ . Thus we can see that, ${\tilde{\mathbb{Q}}^\pm}$ -almost surely,

$$\dfrac{\tilde{L}_t - |\xi_t|}{t} \to \hat{\beta}_0,\quad \dfrac{\xi_t}{t} \to 0,\quad \dfrac{\tilde{L}_t}{t} \to \hat{\beta}_0$$

and consequently

$$\frac{n_t'}{t} \to m \beta\quad\text{and}\quad \frac{n^0_t}{\tilde{L}_t} \to m_0 \beta_0\quad \text{as $t \to \infty$.}$$

It follows that ${S_n' \sim \frac{1}{(m \beta)}n}$ as ${n \to \infty\ \tilde{\mathbb{Q}}^\pm}$ -a.s., so for any ${\delta > 0}$

(31) $$\begin{equation} \mathrm{e}^{-(K+\delta)n} \leq \operatorname{spine}(S_n') \leq \mathrm{e}^{-(K-\delta)n} \quad \text{eventually } \tilde{\mathbb{Q}}^\pm \text{-a.s.}{,} \end{equation}$$

where

$$K = \bigg(\frac{1}{2} \hat{\beta}_0^2 + \hat{\beta}\bigg)\frac{1}{m \beta} > 0{,}$$

which together with the first line of (29) yields

$${\sum_{n=1}^\infty (A_n' - 1) \operatorname{spine}(S_n') < \infty \quad \tilde{\mathbb{Q}}^\pm \text{-a.s.}}$$

Similarly,

$$S^0_n \sim \frac{1}{m_0 \beta_0 \hat{\beta}_0}n\quad \text{as $n \to \infty$,}$$

which together with the first line of (30) yields

$${\sum_{n=1}^\infty (A^0_n - 1) \operatorname{spine}(S^0_n) < \infty \quad \tilde{\mathbb{Q}}^\pm \text{-a.s.}}$$

We have thus shown that

$${\limsup_{t \to \infty} \tilde{\mathbb{Q}}^\pm ( M^\pm_t \!{\mid}\! \tilde{\mathcal{G}}_\infty ) < \infty.}$$

Applying conditional Fatou’s lemma we get

$$\tilde{\mathbb{Q}}^\pm ( \liminf_{t \to \infty} M^\pm_t \!{\mid}\! \tilde{\mathcal{G}}_\infty) \leq \liminf_{t \to \infty} \tilde{\mathbb{Q}}^\pm (M^\pm_t \!{\mid}\! \tilde{\mathcal{G}}_\infty) \leq \limsup_{t \to \infty} \tilde{\mathbb{Q}}^\pm (M^\pm_t \!{\mid}\! \tilde{\mathcal{G}}_\infty) < \infty{,}$$

which implies that ${\liminf_{t \to \infty} M^\pm_t < \infty}$ ${\tilde{\mathbb{Q}}^\pm}$ -a.s. and hence also ${\mathbb{Q}^\pm}$ -a.s. (as ${\{ \liminf_{t \to \infty} M^\pm_t < \infty \} \in \mathcal{F}_\infty}$ ). Then, since ${{1}/{M^\pm}}$ is a positive supermartingale under ${\mathbb{Q}^\pm}$ (in fact a true martingale as there is no extinction), it must converge, so

$${\limsup_{t \to \infty} M^\pm_t = \liminf_{t \to \infty} M^\pm_t < \infty \quad \tilde{\mathbb{Q}}^\pm\text{-a.s.}{,}}$$

which is sufficient to prove part (a) of the proposition.

(b) Assume now that ${\sum_{n \geq 1} p_n n \log n = \infty}$ . Then, counting only particles born from the spine, we obtain

$$M^\pm_{S_n'} \geq A_n' \operatorname{spine}(S_n'){,}$$

so that the first inequality in (31) and the second line in (29) give us that ${\tilde{\mathbb{Q}}^\pm}$ , and hence also ${\mathbb{Q}^\pm}$ -almost surely

$${\limsup_{n \to \infty} M^\pm_{S_n'} = \infty.}$$

Therefore we also get

$${\limsup_{t \to \infty} M^\pm_t = \infty \quad \mathbb{Q}^\pm {\text{-a.s.,}}}$$

which proves the sought result.

If ${\sum_{n \geq 1} q_n n \log n = \infty}$ , then we arrive at the same conclusion by replacing ${(S_n')_{n \geq 1}}$ with ${(S^0_n)_{n \geq 1}}$ and ${(A_n')_{n \geq 1}}$ with ${(A^0_n)_{n \geq 1}}$ in the above argument.

From Proposition 5 we can now easily derive the required lower bound for Theorem 1.

Proposition 6. (Lower bound for Theorem 1.) Suppose that condition (10) on the offspring distribution is satisfied. Then

$$\mathop {\lim \inf }\limits_{t \to \infty } 1/t\log |{N_t}| \ge 1/2\hat \beta _0^2 + \hat \beta \quad n{\rm{ - a}}{\rm{.s}}{\rm{.}},$$

Proof.

$$|N_t| \,\mathrm{e}^{- \frac{1}{2} \hat{\beta}_0^2 t - \hat{\beta}t} \geq \sum_{u \in N_t} \,\mathrm{e}^{- \hat{\beta}_0|X^u_t| - \frac{1}{2} \hat{\beta}_0^2t - \hat{\beta}t} = M^\pm_t.$$

Then

$${\frac{\log |N_t|}{t} \geq {1}/{2} \hat{\beta}_0^2 + \hat{\beta} + \frac{\log M^\pm_t}{t}{,}}$$

and since, under condition (10), ${M^\pm_\infty > 0}$ ${\mathbb{P}}$ -almost surely, it follows that

$${\liminf_{t \to \infty} {1}/{t} \log |N_t| \geq {1}/{2} \hat{\beta}_0^2 + \hat{\beta} \quad \mathbb{P} \text{-a.s.}}$$

In fact we have an even stronger inequality:

$$\begin{align*} \liminf_{t \to \infty} \,\mathrm{e}^{- \frac{1}{2} \hat{\beta}_0^2 t - \hat{\beta}t} |N_t| \geq M^\pm_\infty > 0 \quad \mathbb{P} \text{-a.s.}\\ \end{align*}$$

Propositions 6 and 2 together prove Theorem 1.

In the rest of this subsection we would like to present some results for a purely homogeneous BBM ( ${\beta_0 = 0}$ ), which we shall make use of in the next section. We begin by stating the following result from [14, Theorem 1].

Proposition 7. Consider a BBM with ${\beta_0 = 0}$ (only homogeneous branching present). For ${\lambda \,{\in}\, \mathbb{R}}$ , let

(32) $$\begin{equation} M_t^\lambda \,{:\!=}\, \sum_{u \in N_t} \,\mathrm{e}^{ \lambda X^u_t - \frac{1}{2} \lambda^2 t - \hat{\beta} t} , \quad t \ge 0{,} \end{equation}$$

be the ${\mathbb{P}}$ -martingale of the form (24) derived through the procedure described at the beginning of this subsection by taking

(33) $$\begin{equation} \skew1\tilde{M}^{(1)}_t = \mathrm{e}^{\lambda \xi_t - \frac{1}{2} \lambda^2 t} , \quad t \ge 0. \end{equation}$$

(a) If ${\sum_{n \geq 1} p_n n \log n < \infty}$ and $|\lambda | < {\rm{ (}}2\hat \beta {)^{1/2}}$ , then

$$M^\lambda_\infty > 0 \quad \mathbb{P} \text{-a.s.}$$

(b) If ${\sum_{n \geq 1} p_n n \log n < \infty}$ and $|\lambda | < {\rm{ (}}2\hat \beta {)^{1/2}}$ , then

$$M^\lambda_\infty = 0 \quad \mathbb{P} \text{-a.s.}$$

(c) If ${\sum_{n \geq 1} p_n n \log n = \infty}$ , then

$$M^\lambda_\infty = 0 \quad \mathbb{P} \text{-a.s.}$$

The proof is essentially the same as that of Proposition 5. If we define ${\mathbb{Q}^\lambda}$ and ${\tilde{\mathbb{Q}}^\lambda}$ as probability measures associated with martingales (32) and (33), then we would see that under ${\tilde{\mathbb{Q}}^\lambda}$ the ${\operatorname{spine}(t)}$ term would grow exponentially if $|\lambda | > {\rm{ (}}2\hat \beta {)^{1/2}}$ and decay exponentially if $|\lambda | < {\rm{ (}}2\hat \beta {)^{1/2}}$ , which together with dichotomy (29) would lead to the required result.

We shall now make use of Proposition 7 to get lower bounds on ${|N_t^{\lambda t}|}$ in purely homogeneous branching systems.

Proposition 8. Consider a BBM with ${\beta_0 = 0}$ (only homogeneous branching present). If $\lambda \in (0,{\rm{ (}}2\hat \beta {)^{1/2}})$ and ${\sum_{n \geq 1} p_n n \log n < \infty}$ , then

$$\mathop {\lim \inf }\limits_{t \to \infty } 1/t\log |N_t^{\lambda t}| \ge \hat \beta - \frac{{{\lambda ^2}}}{2}\quad n{\rm{ - a}}{\rm{.s}}{\rm{.}}$$

Proof. For any choice of ${\delta > 0}$ such that $\lambda + \delta < {\rm{ (}}2\hat \beta {)^{1/2}}$ , we have the following lower bound on ${|N_t^{\lambda t}|}$ :

(34) $$\begin{align} |N_t^{\lambda t}| &\geq \sum_{u \in N_t} \mathbf{1}_{\{ \lambda t \leq X^u_t \leq (\lambda + 2 \delta)t\}}\nonumber\\&\geq \sum_{u \in N_t} \,\mathrm{e}^{(\lambda + \delta)X^u_t - (\lambda + \delta)(\lambda + 2 \delta)t} \mathbf{1}_{\{ \lambda t \leq X^u_t \leq (\lambda + 2 \delta)t\}}\nonumber\\&= \mathrm{e}^{\hat{\beta}t - \frac{1}{2} (\lambda + \delta)^2t - \delta(\lambda + \delta)t} \sum_{u \in N_t} \,\mathrm{e}^{(\lambda + \delta)X^u_t - \frac{1}{2} (\lambda + \delta)^2t - \hat{\beta}t} \mathbf{1}_{\{ \lambda t \leq X^u_t \leq (\lambda + 2 \delta)t\}}. \label{eqn34}\end{align}$$

We now claim that as ${t \to \infty}$

(35) $$\begin{equation} \sum_{u \in N_t} \,\mathrm{e}^{(\lambda + \delta)X^u_t - \frac{1}{2} (\lambda + \delta)^2t - \hat{\beta}t} \mathbf{1}_{\{ \lambda t \leq X^u_t \leq (\lambda + 2 \delta)t\}} \to M^{\lambda + \delta}_\infty \quad \mathbb{P} \text{-a.s.,} \end{equation}$$

where ${M^{\lambda + \delta}}$ is the same martingale as in Proposition 7. Indeed,

(36) $$\begin{align} &\sum_{u \in N_t} \,\mathrm{e}^{(\lambda + \delta)X^u_t - \frac{1}{2} (\lambda + \delta)^2t - \hat{\beta}t} \mathbf{1}_{\{ X^u_t > (\lambda + 2 \delta)t\}} \nonumber\\& \qquad \leq \sum_{u \in N_t} \,\mathrm{e}^{(\lambda + \delta)X^u_t - \frac{1}{2} (\lambda + \delta)^2t - \hat{\beta}t} \mathbf{1}_{\{ X^u_t > (\lambda + 2 \delta)t\}} \,\mathrm{e}^{\delta X^u_t - \delta(\lambda + 2 \delta)t} \nonumber\\ &\qquad = \mathrm{e}^{- \frac{1}{2} \delta^2 t} \sum_{u \in N_t} \,\mathrm{e}^{(\lambda + 2\delta)X^u_t - \frac{1}{2} (\lambda + 2\delta)^2t - \hat{\beta}t} \mathbf{1}_{\{ X^u_t > (\lambda + 2 \delta)t\}} \nonumber\\&\qquad \leq \mathrm{e}^{- \frac{1}{2} \delta^2 t} M^{\lambda + 2 \delta}_t \to 0 \quad \mathbb{P} \text{-a.s.}{,} \label{eqn36}\end{align}$$

using the fact that ${M^{\lambda + 2 \delta}}$ converges ${\mathbb{P}}$ -almost surely to a finite limit. Similarly, we have

(37) $$\begin{align} &\sum_{u \in N_t} \,\mathrm{e}^{(\lambda + \delta)X^u_t - \frac{1}{2} (\lambda + \delta)^2t - \hat{\beta}t} \mathbf{1}_{\{ X^u_t < \lambda t\}} \nonumber\\&\qquad \leq \sum_{u \in N_t} \,\mathrm{e}^{(\lambda + \delta)X^u_t - \frac{1}{2} (\lambda + \delta)^2t - \hat{\beta}t} \mathbf{1}_{\{ X^u_t < \lambda t\}} \,\mathrm{e}^{-\delta X^u_t + \delta \lambda t} \nonumber\\ &\qquad = \mathrm{e}^{- \frac{1}{2} \delta^2 t} \sum_{u \in N_t} \,\mathrm{e}^{\lambda X^u_t - \frac{1}{2} \lambda^2t - \hat{\beta}t} \mathbf{1}_{\{ X^u_t < \lambda t\}} \nonumber\\&\qquad \leq \mathrm{e}^{- \frac{1}{2} \delta^2 t} M^{\lambda }_t \to 0 \quad \mathbb{P} \text{-a.s.} \label{eqn37}\end{align}$$

Thus from (36) and (37) it follows that

$$\begin{align*} &\sum_{u \in N_t} \,\mathrm{e}^{(\lambda + \delta)X^u_t - \frac{1}{2} (\lambda + \delta)^2t - \hat{\beta}t} \mathbf{1}_{\{ \lambda t \leq X^u_t \leq (\lambda + 2 \delta)t\}}\\&\qquad = M^{\lambda + \delta}_t - \sum_{u \in N_t} \,\mathrm{e}^{(\lambda + \delta)X^u_t - \frac{1}{2} (\lambda + \delta)^2t - \hat{\beta}t} \mathbf{1}_{\{ X^u_t > (\lambda + 2 \delta)t\}}\\&\qquad \quad\, - \sum_{u \in N_t} \,\mathrm{e}^{(\lambda + \delta)X^u_t - \frac{1}{2} (\lambda + \delta)^2t - \hat{\beta}t} \mathbf{1}_{\{ X^u_t < \lambda t\}} \to M^{\lambda + \delta}_\infty \quad \mathbb{P} \text{-a.s.}{,} \end{align*}$$

proving (35). Moreover, from part (a) of Proposition 7 we know that ${M^{\lambda + \delta}_\infty > 0\ \mathbb{P}}$ -almost surely. Hence from (34) and (35) we get

$${\liminf_{t \to \infty} {1}/{t} \log |N_t^{\lambda t}| \geq \hat{\beta} - {1}/{2} (\lambda + \delta)^2 - \delta(\lambda + \delta) \quad \mathbb{P} \text{-a.s.}{,}}$$

which proves the proposition after letting ${\delta \to 0}$ .

Proposition 9. Consider a BBM with ${\beta_0 = 0}$ (only homogeneous branching present). Let

(38) $$\begin{equation} \tilde{N}_t^{\lambda} \,{:\!=}\, \{ u \in N_{t+1} \colon X_s^u > \lambda s\ \text{for all } s \in [t, t+1] \}. \end{equation}$$

If $\lambda > {(2\hat \beta )^{1/2}}$ and ${\sum_{n \geq 1} p_n n \log n < \infty}$ , then

$$\mathop {\lim \inf }\limits_{t \to \infty } 1/t\log {\mathbb{P}}(|\tilde N_t^\lambda | > 0) \ge \hat \beta - \frac{{{\lambda ^2}}}{2}.$$

In particular, it is also true that

(39) $$\begin{equation} \liminf_{t \to \infty} {1}/{t} \log \mathbb{P} (|N_t^{\lambda t}| > 0 ) \geq \hat{\beta} - \frac{\lambda^2}{2}. \end{equation}$$

Proof. For any choice of ${\delta > 0}$ and ${K > 0}$ , consider the following events:

$$S^{\lambda, t} \,{:\!=}\, \bigg\{ \begin{aligned}\text{{there exists} } u \in N_{t+1} \colon & X^u_s > \lambda s\ \text{{for all} } s \in [t, t+1] \\ & X^u_s \leq K + (\lambda + 2 \delta)s\ \text{{for all} } s \in [0, t+1] \end{aligned} \bigg\}$$

and

$${\tilde{S}^{\lambda, t} \,{:\!=}\, \{ \xi_s > \lambda s\ \text{{for all} } s \in [t, t+1],\ \xi_s \leq K + (\lambda + 2 \delta)s\ \text{{for all} } s \in [0, t+1] \}.}$$

One can then see that ${S^{\lambda, t} \in \mathcal{F}_{t+1} \subseteq \tilde{\mathcal{F}}_{t+1}}$ , ${\tilde{S}^{\lambda, t} \in \mathcal{G}_{t+1} \subseteq \tilde{F}_{t+1}}$ and that

$${\tilde{S}^{\lambda, t} \subseteq S^{\lambda, t} \subseteq \{|\tilde{N}_t^{\lambda}| > 0 \} \subseteq \{|N_t^{\lambda t}| > 0 \}.}$$

We then have the following lower bound on ${\mathbb{P}(|\tilde{N}_t^{\lambda}| > 0)}$ :

$$ \mathbb{P} ( |\tilde{N}_t^{\lambda}| > 0 ) \geq \mathbb{P} ( S^{\lambda, t} ) = \mathbb{E} \bigg( \mathbf{1}_{S^{\lambda, t}} \frac{M^{\lambda + \delta}_{t+1}}{M^{\lambda + \delta}_{t+1}} \bigg) = \mathbb{Q}^{\lambda + \delta} \bigg( \mathbf{1}_{S^{\lambda, t}} \frac{1}{M^{\lambda + \delta}_{t+1}}\bigg) = \tilde{\mathbb{Q}}^{\lambda + \delta} \bigg( \mathbf{1}_{S^{\lambda, t}} \frac{1}{M^{\lambda + \delta}_{t+1}}\bigg),$$

where ${M^{\lambda + \delta}}$ , ${\mathbb{Q}^{\lambda + \delta}}$ , and ${\tilde{\mathbb{Q}}^{\lambda + \delta}}$ are the same as in Proposition 7. Then

$$\tilde{\mathbb{Q}}^{\lambda + \delta} \bigg( \mathbf{1}_{S^{\lambda, t}} \frac{1}{M^{\lambda + \delta}_{t+1}}\bigg) \geq \tilde{\mathbb{Q}}^{\lambda + \delta} \bigg( \mathbf{1}_{\tilde{S}^{\lambda, t}} \frac{1}{M^{\lambda + \delta}_{t+1}}\bigg) \geq \tilde{\mathbb{Q}}^{\lambda + \delta} \bigg( \mathbf{1}_{\tilde{S}^{\lambda, t}} \frac{1}{\tilde{\mathbb{Q}}^{\lambda + \delta} ( M^{\lambda + \delta}_{t+1} \vert \tilde{\mathcal{G}}_\infty)}\bigg){,}$$

using the conditional Jensen inequality and the pull-through property of conditional expectation in the last inequality. We now recall that

$$\tilde{\mathbb{Q}}^{\lambda + \delta} ( M^{\lambda + \delta}_{t+1} \!{\mid}\! \tilde{\mathcal{G}}_\infty) = \operatorname{spine}(t+1) + \sum_{n=1}^{n_{t+1}} (A_n - 1)\operatorname{spine}(S_n),$$

where

$${\operatorname{spine}(t) = \mathrm{e}^{(\lambda + \delta)\xi_t - \frac{1}{2} (\lambda + \delta)^2t - \hat{\beta}t}.}$$

On the event ${\tilde{S}^{\lambda, t}}$ we have that for all ${s \in [0,t+1]}$

$$\begin{align*} \operatorname{spine}(s) &\leq \exp \bigg\{(\lambda + \delta) ( K +(\lambda + 2\delta)s ) - \frac{1}{2}(\lambda + \delta)^2s - \hat{\beta}s \bigg\}\\ &= \mathrm{e}^{K(\lambda + \delta)} \exp \bigg\{\bigg( \frac{1}{2}(\lambda + \delta)^2 + \delta(\lambda + \delta) - \hat{\beta}\bigg)s \bigg\}\\&\leq C_\delta \exp \bigg\{\bigg( \frac{1}{2}(\lambda + \delta)^2 + \delta(\lambda + \delta) - \hat{\beta}\bigg)t \bigg\}, \end{align*}$$

where ${C_\delta}$ is some positive constant. Also from dichotomy (29) we know that ${A_n < \mathrm{e}^{\delta n}}$ eventually ${\tilde{\mathbb{Q}}^{\lambda + \delta}}$ -almost surely. Thus

$${\sum_{n=1}^{n_{t+1}} A_n \leq \sum_{n =1}^{n_{t+1}} \,\mathrm{e}^{\delta n} + Y \leq C_\delta' \,\mathrm{e}^{\delta n_{t+1}} + Y,}$$

where ${Y = \sum_{n=1}^\infty A_n \mathbf{1}_{\{A_n > \mathrm{e}^{\delta n}\}}}$ is a ${\tilde{\mathbb{Q}}^{\lambda + \delta}}$ -almost surely finite random variable independent of ${n_{t+1}}$ and ${(\xi_s)_{0 \leq s \leq t+1}}$ and ${C_\delta'}$ is some positive constant. Thus

$$\tilde{\mathbb{Q}}^{\lambda + \delta} ( M^{\lambda + \delta}_{t+1} \!{\mid}\! \tilde{\mathcal{G}}_\infty) \leq C_\delta \,\mathrm{e}^{( \frac{1}{2} (\lambda + \delta)^2 + \delta(\lambda + \delta) - \hat{\beta})t} ( 1 + Y + C_\delta' \,\mathrm{e}^{\delta n_{t+1}} ).$$

Then, using the fact that ${p{1}/{a + b} \geq p{1}/{2ab}}$ whenever a, ${b \geq 1}$ , we get

$$\begin{align*} &\tilde{\mathbb{Q}}^{\lambda + \delta} \bigg( \mathbf{1}_{\tilde{S}^{\lambda, t}} \frac{1}{\tilde{\mathbb{Q}}^{\lambda + \delta} ( M^{\lambda + \delta}_{t+1} \vert \tilde{\mathcal{G}}_\infty)}\bigg)\\ &\qquad \geq \frac{1}{C_\delta}\,\mathrm{e}^{(\hat{\beta} - \frac{1}{2} (\lambda + \delta)^2 - \delta(\lambda + \delta))t}\ \tilde{\mathbb{Q}}^{\lambda + \delta} \bigg( \mathbf{1}_{\tilde{S}^{\lambda, t}} \frac{1}{1 + Y + C_\delta' \,\mathrm{e}^{\delta n_{t+1}}}\bigg)\\&\qquad \geq \frac{1}{2 C_\delta C_\delta'} \,\mathrm{e}^{(\hat{\beta} - \frac{1}{2} (\lambda + \delta)^2 - \delta(\lambda + \delta))t} \tilde{\mathbb{Q}}^{\lambda + \delta} ( \tilde{S}^{\lambda, t} ) \tilde{\mathbb{Q}}^{\lambda + \delta} ( \mathrm{e}^{- \delta n_{t+1}} ) \tilde{\mathbb{Q}}^{\lambda + \delta} \bigg( \frac{1}{1 + Y} \bigg). \end{align*}$$

We note that

$${\tilde{\mathbb{Q}}^{\lambda + \delta}\bigg(\frac{1}{1 + Y}\bigg) > 0}$$

since Y is ${\tilde{\mathbb{Q}}^{\lambda + \delta}}$ -almost surely finite, ${\tilde{\mathbb{Q}}^{\lambda + \delta}(\mathrm{e}^{- \delta n_{t+1}}) = \mathrm{e}^{m \beta (t+1) (\mathrm{e}^{- \delta} - 1)}}$ since ${(n_t)_{t \ge 0}}$ is a ${\tilde{\mathbb{Q}}^{\lambda + \delta}}$ -Poisson process with rate ${m \beta}$ , and ${\tilde{\mathbb{Q}}^{\lambda + \delta} ( \tilde{S}^{\lambda, t}) \to C_{K, \delta}}$ for some positive constant ${C_{K, \delta}}$ since ${(\xi_t)_{t \ge 0}}$ is a Brownian motion with drift ${\lambda + \delta}$ under ${\tilde{\mathbb{Q}}^{\lambda + \delta}}$ . Therefore

$$\begin{align*} \liminf_{t \to \infty} {1}/{t} \log \mathbb{P} ( |\tilde{N}_t^{\lambda}| > 0 ) &\geq \liminf_{t \to \infty} {1}/{t} \log \tilde{\mathbb{Q}}^{\lambda + \delta} \bigg( \mathbf{1}_{\tilde{S}^{\lambda, t}} \frac{1}{\tilde{\mathbb{Q}}^{\lambda + \delta} ( M^{\lambda + \delta}_{t+1} \vert \tilde{\mathcal{G}}_\infty)}\bigg)\\& \geq \hat{\beta} - \frac{1}{2}(\lambda + \delta)^2 - \delta(\lambda + \delta) - m \beta(1 - \mathrm{e}^{-\delta}){,} \end{align*}$$

which proves the required result after letting ${\delta \to 0}$ .

3.1. Heuristic argument

Here we discuss the idea behind the proof in a non-rigorous way in order to help the reader understand the formal argument given in the next subsection.

Our task is to find the optimal way for a particle to reach level ${\lambda t}$ at some large time t.

In the case of spatially homogeneous branching ( ${\beta_0 = 0}$ ), the birth rate along the path of a particle is independent of the path, and so the optimal way would simply be to travel at speed ${\lambda}$ all the time (there are of course finer results available, but they are irrelevant to this discussion).

However, in the presence of the catalyst at the origin, travelling at speed ${\lambda}$ all the time might be disadvantageous as it will discard any contribution from the catalyst. Thus one might think that a better strategy for a particle would first be to stay near the origin for some positive proportion of time in order to give birth to more particles at an accelerated rate (due to both homogeneous and catalytic branching potential), and then for the remaining time let its children travel at whatever speed is necessary in order to reach the required level.

The argument goes as follows. For a large time t we let

$$\begin{equation*} q \,{:\!=}\, \begin{cases} |N_t^{\lambda t}|& \text{if } \lambda < \lambda_{\mathrm{crit}}, \\ \mathbb{P}(|N_t^{\lambda t}| > 0)& \text{if } \lambda > \lambda_{\mathrm{crit}}{,} \end{cases} \end{equation*}$$

and we want a lower bound on q.

We fix a number ${p \in [0,1]}$ . As we know from Theorem 1, at time pt there are

|N_{pt}| \approx \exp \bigg\{ \bigg(\frac{\hat{\beta}_0^2}{2} + \hat{\beta} \bigg)pt \bigg\}$$

particles in the system and about half of them lie in the upper half-plane. Next we ignore any catalytic branching that takes place between times pt and t by assuming that every particle ${u \in N_{pt}^0}$ starts an independent spatially homogeneous branching process from the position ${X^u_{pt} > 0}$ .

We let

$$\begin{equation*} q(u) \,{:\!=}\, \begin{cases} |N_T^{\frac{\lambda}{1-p}T}(u)|& \text{if } \frac{\lambda}{1-p} < \sqrt{2 \hat{\beta}},\\\\ \mathbb{P}(|N_T^{\frac{\lambda}{1-p}T}(u)| > 0)& \text{if } \frac{\lambda}{1-p} > \sqrt{2 \hat{\beta}}, \end{cases} \end{equation*}$$

where ${N_T^{(\frac{\lambda}({1-p}))T}(u)}$ is the set of particles which lie to the right of ${(\frac{\lambda}({1-p}))T}$ at time T of the spatially homogeneous process initiated by u in the time–space frame of this process and where ${T = (1-p)t}$ . Then, from Propositions 8 and 9, we know that

$$q(u) \ge \exp \{ \hat \beta T - 1/2{(\frac{\lambda }{{1 - p}})^2}T\} = \exp \{ \hat \beta (1 - p)t - \frac{{{\lambda ^2}}}{{2(1 - p)}}t\} .$$

Then, since every particle in ${N_T^{\frac{\lambda}{1-p}T}(u)}$ for every ${u \in N_{pt}^0}$ also belongs to ${N_t^{\lambda t}}$ , we can estimate

(40) $$\begin{equation} q \ge \sum_{u \in N_{pt}^0} q(u) \approx |N_{pt}^0| \exp \bigg\{ \hat{\beta}(1-p)t - \frac{\lambda^2}{2(1-p)} t \bigg\} \approx \exp \bigg\{ \bigg( \hat{\beta} + \frac{\hat{\beta}_0^2}{2}p - \frac{\lambda^2}{2(1-p)} \bigg) t \bigg\}. \end{equation}$$

The value of p which maximizes this expression is

(41) $$\begin{equation} p^{\ast} \,{:\!=}\, \begin{cases} 1 - \frac{\lambda}{\hat{\beta}_0}& \text{if } \lambda \leq \hat{\beta}_0,\\ 0& \text{if } \lambda \geq \hat{\beta}_0. \end{cases} \end{equation}$$

Substituting this value of p into (40), we get

$$\begin{array}{l} q \ge \left\{ {\begin{array}{*{20}{l}} {\exp \{ (\hat \beta + \frac{{\hat \beta _0^2}}{2} - {{\hat \beta }_0}\lambda )t\} }&{{\rm{if }}\lambda \le {{\hat \beta }_0}}&{}\\ {\exp \{ (\hat \beta - \frac{{{\lambda ^2}}}{2})t\} }&{{\rm{if }}\lambda \ge {{\hat \beta }_0}} \end{array}} \right.\\ \,\,\,\,\, = \exp \{ {\Delta _\lambda }t\} , \end{array}$$

which gives the lower bound on q that we want.

Note that if ${\lambda}$ is too large ( ${\lambda \geq \hat{\beta}_0}$ ) then ${p^\ast = 0}$ , and so the best strategy for a particle to reach level ${\lambda t}$ at time t would indeed be to travel at speed ${\lambda}$ , always being driven by a homogeneous branching potential with negligible contribution from catalytic branching. This is consistent with Remark 2 made earlier.

3.2. Lower bounds for (11) and (13)

Before we present the main body of the proof, let us give a couple of preliminary results. The first is a very crude estimate of the number of particles which lie approximately in the upper half-plane at a time t.

Proposition 10. Assume that condition (10) on the offspring distribution is satisfied. Then ${\mathbb{P}}$ -almost surely, for any ${\delta > 0}$ , there exists a finite time ${T_\delta}$ such that, for all ${t \geq T_\delta}$ ,

$$\begin{equation*} \label{eq_positive_pop} |N^{- \delta t}_t| \geq \exp \bigg\{\bigg(\hat{\beta} + \frac{\hat{\beta}_0^2}{2} - c_\delta\bigg)t \bigg\}, \end{equation*}$$

where ${c_\delta}$ is some positive constant with the property that ${c_\delta \to 0}$ as ${\delta \to 0}$ .

Proof. Let us observe that ${\Delta_\delta \to \hat{\beta} + {\hat{\beta}_0^2}/{2}}$ as ${\delta \to 0}$ . So we may write ${\Delta_\delta = \hat{\beta} + {\hat{\beta}_0^2}/{2} - c_\delta'}$ for some ${c_\delta' > 0}$ such that ${c_\delta' \to 0}$ as ${\delta \to 0}$ .

From Theorem 1 (or Proposition 6 we know that ${\mathbb{P}}$ -almost surely, for any ${\delta > 0}$ , there exists a finite time ${T_\delta'}$ such that, for all ${t \geq T_\delta'}$ ,

(42) $$\begin{equation} |N_t| \geq \exp \bigg\{\bigg(\hat{\beta} + \frac{\hat{\beta}_0^2}{2} - \frac{1}{4}c_\delta' \bigg)t \bigg\}. \end{equation}$$

We also know from Proposition 4 (equation (20)) that ${\mathbb{P}}$ -almost surely, for any ${\delta > 0}$ , there exists a finite time ${T_\delta''}$ such that, for all ${t \geq T_\delta''}$ ,

$$|N^{\delta t}_t| \leq \exp \bigg\{\bigg(\Delta_\delta + {1}/{2} c_\delta' \bigg)t \bigg\} = \exp \bigg\{ \bigg(\hat{\beta} + \frac{\hat{\beta}_0^2}{2} - \frac{1}{2}c_\delta' \bigg)t \bigg\}.$$

Thus, by symmetry it is also true that ${\mathbb{P}}$ -almost surely, for any ${\delta > 0}$ , there exists a finite time ${T_\delta'''}$ such that, for all ${t \geq T_\delta'''}$ ,

(43) $$\begin{equation} |N_t| - |N^{- \delta t}_t| \leq \exp \bigg\{ \bigg(\hat{\beta} + \frac{\hat{\beta}_0^2}{2} - \frac{1}{2}c_\delta' \bigg)t \bigg\}. \end{equation}$$

Subtracting (43) from (42) yields the result.

The next result is basically a version of Chebyshev’s inequality.

Proposition 11. Let N be a random variable supported on ${\mathbb{N}}$ and ${(S_k)_{k \geq 1}}$ a sequence of events independent of each other conditional on N. If, for some ${r \in (0, 1)}$ , it is true that ${\mathbb{P}(S_k\!{\mid}\!N) \geq r}$ ${\mathbb{P}}$ -a.s. for all ${k \geq 1}$ , then

$$\mathbb{P} \bigg( \sum_{k = 1}^N \mathbf{1}_{S_k} \leq \frac{r}{2}N \biggm\vert N \bigg) \leq \frac{4}{rN} \quad \mathbb{P} {\text{-a.s.}}$$

Sharper inequalities are of course available but are not needed here.

Proof of Proposition 11. Let us assume for simplicity that N is deterministic. Then

$$\begin{align*} \mathbb{P} \bigg( \sum_{k = 1}^N \mathbf{1}_{S_k} \leq \frac{r}{2}N \bigg) &\leq \mathbb{P} \bigg( \sum_{k = 1}^N ( \mathbf{1}_{S_k} - \mathbb{P}(S_k) ) \leq - \frac{1}{2} \sum_{k = 1}^N \mathbb{P}(S_k) \bigg) \\ &\leq \mathbb{P} \bigg( \biggm\vert \sum_{k = 1}^N ( \mathbf{1}_{S_k} - \mathbb{P}(S_k) ) \bigg\vert \geq {1}/{2} \sum_{k = 1}^N \mathbb{P}(S_k) \bigg)\\ &\leq \frac{\sum_{k=1}^N \ var(\mathbf{1}_{S_k})}{\frac{1}{4} \bigg(\sum_{k = 1}^N \mathbb{P}(S_k) \bigg)^2} \\& \leq \frac{4}{rN} \end{align*}$$

using the Markov inequality and the fact that ${var\,\,(\mathbf{1}_{S_k}) = \mathbb{P}(S_k) - \mathbb{P}(S_k)^2 \leq \mathbb{P}(S_k)}$ . The same argument will then work for N random if we replace ${\mathbb{P}({\cdot})}$ with ${\mathbb{P}({\cdot} \!{\mid}\! N)}$ and ${var({\cdot})}$ with ${var({\cdot} \!{\mid}\! N)}$ .

Proposition 12. (Lower bounds for Theorem 2.) Suppose that condition (10) on the offspring distribution is satisfied.

If ${\lambda < \lambda_{\mathrm{crit}}}$ ${(\Delta_\lambda > 0)}$ , then

(44) $$\begin{equation} \liminf_{t \to \infty} {1}/{t} \log |N_t^{\lambda t}| \geq \Delta_\lambda \quad \mathbb{P} \text{-a.s.} \end{equation}$$

If ${\lambda > \lambda_{\mathrm{crit}}}$ ${(\Delta_\lambda < 0)}$ , then

(45) $$\begin{equation} \liminf_{t \to \infty} {1}/{t} \log \mathbb{P} (|N_t^{\lambda t}| > 0 ) \geq \Delta_\lambda. \end{equation}$$

Proof. We let ${p = p^\ast}$ be the same as in (41). We note that ${p = 0}$ if and only if ${\lambda \geq \hat{\beta}_0 }$ , while if ${\lambda \geq \hat{\beta}_0}$ then ${\Delta_\lambda = \hat{\beta} - \frac{1}{2} \lambda^2}$ , so that (44) and (45) follow from Propositions 8 and 9 (equation (39)) by simply not counting any particles born due to catalytic branching. Thus, for the rest of the proof we shall assume that ${\lambda < \hat{\beta}_0}$ , so that ${p = 1 - {\lambda}/{\hat{\beta}_0} > 0}$ and ${\Delta_\lambda = \hat{\beta} + {\hat{\beta}_0^2}/{2} - \hat{\beta}_0 \lambda}$ .

We then choose some ${\delta > 0}$ and define

$$\hat \lambda {\mkern 1mu} : = {\mkern 1mu} \frac{{\lambda + \delta }}{\lambda }{{\hat \beta }_0}.$$

We also define

$$f(\delta) \,{:\!=}\, \bigg( \hat{\beta} - \frac{\hat{\lambda}^2}{2} - \delta \bigg),\quad g(\delta) \,{:\!=}\, \bigg( \hat{\beta} + \frac{\hat{\beta}_0^2}{2} - c_\delta \bigg),$$

where ${c_\delta}$ is the same as in Proposition 10. We let ${h(\delta)}$ be such that

$$\begin{align*} (1-p)f(\delta) + pg(\delta) &= \bigg( \hat{\beta} - \frac{\hat{\lambda}^2}{2} - \delta \bigg)(1-p) + \bigg( \hat{\beta} + \frac{\hat{\beta}_0^2}{2} - c_\delta \bigg)p\\ &= \hat{\beta} + \frac{\hat{\beta}_0^2}{2} \bigg( 1 - \frac{\lambda}{\hat{\beta}_0} \bigg) - c_\delta p - {1}/{2} \bigg( \frac{\lambda + \delta}{\lambda} \hat{\beta}_0 \bigg)^2 \frac{\lambda}{\hat{\beta}_0} - \delta(1-p)\\&= \Delta_\lambda - h(\delta). \end{align*}$$

Note that ${h(\delta)>0}$ and ${h(\delta) \to 0}$ as ${\delta \to 0}$ .

For ${t>0}$ we define events

$${\mathcal{A}_t}{\mkern 1mu} : = {\mkern 1mu} \{ |N_{pt}^{ - \delta pt}| \ge {{\rm{e}}^{g(\delta )pt}}\} .$$

From Proposition 10 we know that ${\mathbb{P}(\mathcal{A}_n) \text{ {eventually}} = 1}$ , so that in particular ${\mathbb{P}(\mathcal{A}_t) \to 1}$ as ${t \to \infty}$ .

Finally, for every particle ${u \in N_{pt}}$ , let us consider the subtree initiated by u at time pt. In the time–space frame of this subtree, we define ${N_s(u)}$ , ${s \geq 0}$ to be the set of particles in the subtree at time s and ${Y^v_s}$ the positions of particles ${v \in N_s(u)}$ at time s. Moreover, by analogy with (2) and (38), we define

$$N_s^x(u) \,{:\!=}\, \{ v \in N_s(u) \colon Y_s^v > x\}$$

and

$$\begin{align*} \tilde{N}_s^l(u) \,{:\!=}\, \{ v \in N_{s+1}(u) \colon Y_r^v > lr\ \text{for all } r \in [s, s+1]\}.\\ \end{align*}$$

Proof of (44), the lower bound for (11). Assume that ${\lambda < \lambda_{\mathrm{crit}}}$ so that ${\Delta_\lambda > 0}$ . There are two cases to consider, which require slightly different treatment.

Case 1: ${\hat{\beta} > \frac{1}{2} \hat{\beta}_0^2}$ (equivalently, ${{\hat \beta }_0} < {\rm{ (}}2\hat \beta {)^{1/2}}$ ). We choose ${\delta > 0}$ to be sufficiently small that $\hat \lambda < {\rm{ }}{(2\hat \beta )^{1/2}}$ , and for ${n \geq 1}$ consider events

$${\mathcal{B}_n \,{:\!=}\, \bigg\{ \sum_{u \in N_{pn}^{- \delta pn}} \mathbf{1}_{ \mathcal{B}_n(u)} < \frac{1}{4} |N_{pn}^{- \delta pn}| \bigg\},}$$

where, for every ${u \in N_{pn}^{- \delta pn}}$ ,

$${\mathcal{B}_n(u) = \{ |N_s^{\hat{\lambda}s}(u)| \geq \mathrm{e}^{ f(\delta) s}\ \text{for all } s \in [(1-p)n, (1-p)n + 1] \}.}$$

We know that, conditional on ${\mathcal{F}_{pn}}$ , events ${\mathcal{B}_n(u)}$ are independent, since all the subtrees initiated by particles ${u \in N_{pn}}$ are independent copies of the original branching process starting from positions ${X^u_{pn}}$ .

Moreover, if we ignore all the catalytic branching taking place in the subtrees initiated by particles ${u \in N_{pn}^{- \delta p n}}$ , then we can get from Proposition 8 that there exists some deterministic ${n_0}$ such that for all ${n \geq n_0}$ , ${\mathbb{P}(\mathcal{B}_n(u) \!{\mid}\! \mathcal{F}_{pn}) \geq {1}/{2} }$ . Hence, by Proposition 11,

$$\mathbb{P}(\mathcal{B}_n \!{\mid}\! \mathcal{F}_{pn}) \leq \frac{8}{|N_{pn}^{- \delta pn}|} \quad\text{$\mathbb{P}$-a.s.}$$

for all ${n \geq n_0}$ . Then, for all ${n \geq n_0}$ , we obtain

$$\begin{align*} \mathbb{P} (\mathcal{A}_n \cap \mathcal{B}_n ) = \mathbb{E} ( \mathbb{E} ( \mathbf{1}_{\mathcal{A}_n} \mathbf{1}_{\mathcal{B}_n} \!{\mid}\! \mathcal{F}_{pn} ) ) \leq \mathbb{E} \bigg( \mathbf{1}_{\mathcal{A}_n} \frac{8}{|N_{pn}^{- \delta pn}|} \bigg) \leq 8 \,\mathrm{e}^{- g(\delta) pn}{,} \end{align*}$$

which decays exponentially fast in n (for ${\delta}$ sufficiently small). Therefore ${\mathbb{P}(\mathcal{A}_n \cap \mathcal{B}_n \text{ i.o.}) = 0}$ . Then, since ${\mathbb{P}(\mathcal{A}_n \text{ {eventually}}) = 1}$ , it follows that ${\mathbb{P}(\mathcal{A}_n \cap \mathcal{B}_n^c \text{ {eventually}}) = 1}$ . So ${\mathbb{P}}$ -a.s., for all n large enough,

$$\sum_{u \in N_{pn}^{- \delta pn}} \mathbf{1}_{ \mathcal{B}_n(u)} \geq \frac{1}{4} |N_{pn}^{- \delta pn}| \geq \frac{1}{4} \mathrm{e}^{g(\delta)pn}.$$

Then, noting that, for all ${t \in [n, n+1]}$ ,

$$|N_t^{\lambda t}| \geq \sum_{u \in N_{pn}^{- \delta pn}} \mathbf{1}_{ \mathcal{B}_n(u)} \,\mathrm{e}^{f(\delta)(1-p)n}{,}$$

we get that ${\mathbb{P}}$ -a.s., for all t sufficiently large,

$$|N_t^{\lambda t}| \geq K \,\mathrm{e}^{g(\delta)pt + f(\delta)(1-p)t},$$

where K is some positive constant. Hence

$${\liminf_{t \to \infty} {1}/{t} \log |N_t^{\lambda t}| \geq \Delta_\lambda - h(\delta) \quad \mathbb{P} \text{-a.s.,}}$$

which yields the required value after letting ${\delta \to 0}$ .

Case 2: ${\hat{\beta} \leq \frac{1}{2} \hat{\beta}_0^2}$ (equivalently, ${{\hat \beta }_0} \ge {(2\hat \beta )^{1/2}}$ ). For ${n \geq 1}$ we consider events

$${\mathcal{C}_n \,{:\!=}\, \bigg\{ \sum_{u \in N_{pn}^{- \delta pn}} \mathbf{1}_{\{ |\tilde{N}_{(1-p)n}^{\hat{\lambda}}(u)| > 0\}} < {1}/{2} \mathrm{e}^{f(\delta)(1-p)n} |N_{pn}^{- \delta pn}|\bigg\}}$$

(note that $\hat \lambda > {\rm{ }}{(2\hat \beta )^{1/2}}$ ). We know that, conditional on ${\mathcal{F}_{pn}}$ , events ${\{|\tilde{N}^{\hat{\lambda}}_{(1-p)n}(u)|>0\}}$ are independent. Moreover, if we ignore all the catalytic branching taking place in the subtrees initiated by particles ${u \in N_{pn}^{- \delta p n}}$ , then we can get from Proposition 9 that there exists some deterministic ${n_0}$ such that, for all ${n \geq n_0}$ , ${\mathbb{P}(|\tilde{N}^{\hat{\lambda}}_{(1-p)n}(u)|>0 \!{\mid}\! \mathcal{F}_{pn}) \geq \mathrm{e}^{f(\delta)(1-p)n}}$ . Hence, by Proposition 11, for all ${n \geq n_0}$ ,

$${\mathbb{P} (\mathcal{C}_n \!{\mid}\! \mathcal{F}_{pn} ) \leq \frac{4}{|N_{pn}^{- \delta pn}|} \,\mathrm{e}^{-f(\delta)(1-p)n} \quad \mathbb{P} \text{-a.s.}}$$

Then, for all ${n \geq n_0}$ , we obtain

$$\begin{align*} \mathbb{P} (\mathcal{A}_n \cap \mathcal{C}_n) &= \mathbb{E} ( \mathbb{E} ( \mathbf{1}_{\mathcal{A}_n} \mathbf{1}_{\mathcal{C}_n} \!{\mid}\! \mathcal{F}_{pn} ) ) \\ & \leq \mathbb{E} \bigg( \mathbf{1}_{\mathcal{A}_n} \frac{4}{|N_{pn}^{- \delta pn}|} \,\mathrm{e}^{-f(\delta)(1-p)n}\bigg)\\ &\leq 4\,\mathrm{e}^{-g(\delta)p -f(\delta)(1-p)}\\ &= 4 \,\mathrm{e}^{ -(\Delta_\lambda - h(\delta) ) n}, \end{align*}$$

which decays exponentially fast in n (for ${\delta}$ chosen sufficiently small). Therefore ${\mathbb{P}(\mathcal{A}_n \cap \mathcal{C}_n \text{ i.o.}) = 0}$ . Then, since ${\mathbb{P}(\mathcal{A}_n \text{ {eventually}}) = 1}$ , it follows that ${\mathbb{P}(\mathcal{A}_n \cap \mathcal{C}_n^c \text{ {eventually}}) = 1}$ . So ${\mathbb{P}}$ -a.s., for all n large enough,

$$\sum_{u \in N_{pn}^{- \delta pn}} \mathbf{1}_{\{ |\tilde{N}_{(1-p)n}^{\hat{\lambda}}(u)| > 0\}} \geq {1}/{2} \mathrm{e}^{f(\delta)(1-p)n} |N_{pn}^{- \delta pn}| \geq 4 \,\mathrm{e}^{(\Delta_\lambda - h(\delta) )n}.$$

Then, noting that, for all ${t \in [n, n+1]}$ ,

$$|N_t^{\lambda t}| \geq \sum_{u \in N_{pn}^{- \delta pn}} \mathbf{1}_{ \{ |\tilde{N}_{(1-p)n}^{\hat{\lambda}}(u)| > 0\}}{,}$$

we obtain

$$\liminf_{t \to \infty} {1}/{t} \log |N_t^{\lambda t}| \geq \Delta_\lambda - h(\delta) \quad \mathbb{P} \text{-a.s.}{,}$$

which yields the required result after letting ${\delta \to 0}$ .

Proof of (45), the lower bound for (13). Assume that ${\lambda > \lambda_{\mathrm{crit}}}$ so that ${\Delta_\lambda < 0}$ . Then necessarily $\hat \lambda > {{\hat \beta }_0} > \lambda > {\lambda _{{\rm{crit}}}} \ge {\rm{ }}{(2\hat \beta )^{1/2}}$ (the last inequality follows from the fact that, for any given ${\hat{\beta}}$ , the minimum value of ${{\hat{\beta}}/{\hat{\beta}_0} + \frac{1}{2} \hat{\beta}_0}$ over all ${\hat{\beta}_0 \in (0, \infty)}$ is ${(2\hat \beta )^{1/2}}$ ). We note that

$$\mathbb{P} ( |N_t^{\lambda t}| > 0) \geq \mathbb{P} \bigg( \bigcup_{u \in N_{pt}^{- \delta pt}} \{ |N^{\hat{\lambda}t}_{(1-p)t}(u)| > 0 \},\ |N_{pt}^{- \delta pt}| \geq \mathrm{e}^{g(\delta)pt} \bigg).$$

We know that conditional on ${\mathcal{F}_{pt}}$ events ${\{|N^{\hat{\lambda}t}_{(1-p)t}(u)|>0\}}$ are independent. Moreover, if we ignore all the catalytic branching taking place in the subtrees initiated by particles ${u \in N_{pt}^{- \delta p t}}$ , then we can get from Proposition 9 (equation (39)) that there exists some deterministic ${t_0}$ such that, for all ${t \geq t_0}$ , ${\mathbb{P} (|N^{\hat{\lambda}t}_{(1-p)t}(u)|>0 \!{\mid}\! \mathcal{F}_{pt} ) \geq \mathrm{e}^{f(\delta)(1-p)t}}$ . Hence

$$\begin{align*} &\mathbb{P} \Bigg( \bigcup_{u \in N_{pt}^{- \delta pt}} \{ |N^{\hat{\lambda}t}_{(1-p)t}(u)| > 0 \},\ |N_{pt}^{- \delta pt}| \geq \mathrm{e}^{g(\delta)pt} \Bigg)\\ &\qquad = \mathbb{E} \Bigg( \mathbf{1}_{\{|N_{pt}^{- \delta pt}| \geq \mathrm{e}^{g(\delta)pt}\}} \Bigg[ 1 - \prod_{u \in N_{pt}^{- \delta pt}} ( 1 - \mathbb{P}(|N^{\hat{\lambda}t}_{(1-p)t}(u)|>0 \!{\mid}\! \mathcal{F}_{pt}))\Bigg]\Bigg)\\ &\qquad \geq \mathbb{P} ( |N_{pt}^{- \delta pt}| \geq \mathrm{e}^{g(\delta)pt} ) [ 1 - ( 1 - \mathrm{e}^{f(\delta)(1-p)t} )^{\mathrm{e}^{g(\delta )pt}} ]\\ &\qquad \geq \mathbb{P} ( |N_{pt}^{- \delta pt}| \geq \mathrm{e}^{g(\delta)pt} ) \bigg[ \mathrm{e}^{(\Delta_\lambda - h(\delta))t} - {1}/{2} \mathrm{e}^{2(\Delta_\lambda - h(\delta))t} \bigg] \end{align*}$$

for all t large enough, and where in the last inequality we have used the fact that, for any ${a \in (0,1)}$ and ${b > 0}$ , it is true that ${(1-a)^b \leq \mathrm{e}^{-ab} \leq 1 - ab + \frac{1}{2} a^2 b^2}$ . Then, noting that as ${t \to \infty}$

$${\mathbb{P} ( |N_{pt}^{- \delta pt}| \geq \mathrm{e}^{g(\delta)pt} ) \to 1 \quad \mathbb{P} \text{-a.s.}{,}}$$

we obtain

$${\liminf_{t \to \infty} {1}/{t} \log \mathbb{P} (|N_t^{\lambda t}| > 0 ) \geq \Delta_\lambda - h(\delta){,}}$$

which yields the required result after letting ${\delta \to 0}$ .

3.3 Proof of Corollary 1.4

Proof of Corollary 1. Assume that condition (10) is satisfied. Then, for any ${\lambda < \lambda_{\mathrm{crit}}}$ , as we know from (11),

$$\mathbb{P}( R_t > \lambda t \text{ eventually} ) = \mathbb{P} ( |N_t^{\lambda t}| > 0 \text{ eventually} ) = 1.$$

Thus ${\liminf_{t \to \infty}\frac{R_t}{t} \geq \lambda}$ ${\mathbb{P}}$ -almost surely. Then, letting ${\lambda \to \lambda_{\mathrm{crit}}}$ gives

$${\liminf_{t \to \infty} {R_t}/{t} \geq \lambda_{\mathrm{crit}} \quad \mathbb{P} \text{-a.s.}}$$

Similarly, if ${\lambda > \lambda_{\mathrm{crit}}}$ , then by (12),

$${\mathbb{P} ( R_t \leq \lambda t \text{ eventually} ) = \mathbb{P} ( |N_t^{\lambda t}| = 0 \text{ eventually} ) = 1.}$$

Thus ${\limsup_{t \to \infty} \frac{R_t}{t} \leq \lambda}$ ${\mathbb{P}}$ -a.s. Then, letting ${\lambda \to \lambda_{\mathrm{crit}}}$ gives

$${\limsup_{t \to \infty} {R_t}/{t} \leq \lambda_{\mathrm{crit}} \quad \mathbb{P} \text{-a.s.}{,}}$$

which completes the proof.