present a few simple calculations which should motivate our main results in the next subsection.

Let us define for any $x \in \mathbb{R}$ and $t \geq 0$ $$\label{set_ntx}
N_t^x := \{ u \in N_t \ : \ X^u_t >x\}$$ to be the set of particles in the system at time $t$ whose spatial position is bigger than $x$. A simple application of the widely used “many-to-one” formula (see Subsection 2.2) gives an exact expression for $E|N_t^x|$:

\[prop\_enx\] For any $x \geq 0$ and $t \geq 0$ $$\label{eq_enx}
E |N_t^x| = \Phi \Big( \hat{\beta}_0 \sqrt{t} - \frac{x}{\sqrt{t}}\Big) \exp \Big\{ \frac{1}{2} \hat{\beta}_0^2t -\hat{\beta}_0x + \hat{\beta}t \Big\} \text{,}$$ where $\Phi(x) = \mathbb{P} (\mathcal{N}(0, 1) \leq x) = \int_{- \infty}^x (2 \pi)^{-1/2} \mathrm{e}^{- y^2/2} \mathrm{d}y$ is the cumulative distribution function of a standard normal random variable.

In particular, for any $\lambda \geq 0$ we get $$\label{enlt}
E|N_t^{\lambda t}|  = \Phi \Big( \big(\hat{\beta}_0 - \lambda\big) \sqrt{t}\Big) \exp \Big\{ \big(\frac{1}{2}\hat{\beta}_0^2 - \hat{\beta}_0\lambda + \hat{\beta}\big)t\Big\}.$$ Using the fact that $\Phi(x) \sim (2 \pi)^{-1/2}|x|^{-1} \mathrm{e}^{-x^2/2}$ as $x \to - \infty$ and $\Phi(x) \to 1$ as $x \to \infty$ we can then see that $$\label{log_e_lambda}
\frac{1}{t} \log E|N_t^{\lambda t}| \to \Delta_\lambda \text{ as } t \to \infty \text{,}$$ where 