$$\label{delta_lambda}
\Delta_{\lambda} = \left\{
\begin{array}{rl}
\frac{1}{2} \hat{\beta}_0^2 - \hat{\beta}_0 \lambda + \hat{\beta}& \text{if } \lambda \leq \hat{\beta}_0, \\
- \frac{1}{2} \lambda^2 + \hat{\beta}& \text{if } \lambda \geq \hat{\beta}_0 .
\end{array} \right.$$ We can then observe that $\Delta_\lambda$ takes positive or negative values according to whether $\lambda < \lambda_{crit}$ or $\lambda > \lambda_{crit}$, where

$$\label{lambda_crit}
\lambda_{crit} = \left\{
\begin{array}{rl}
\frac{\hat{\beta}}{\hat{\beta}_0} + \frac{1}{2} \hat{\beta}_0& \text{if } \hat{\beta} \leq \frac{1}{2}\hat{\beta}_0^2, \\
\sqrt{2 \hat{\beta}}& \text{if } \hat{\beta} \geq \frac{1}{2}\hat{\beta}_0^2 .
\end{array} \right.$$

Since the expected number of particles above the line $(\lambda_{crit} + \epsilon)t$ decays exponentially with $t$ and the expected number of particles above the line $(\lambda_{crit} - \epsilon)t$ grows exponentially with $t$ we may interpret $\lambda_{crit}$ for now as the speed of the rightmost particle “in expectation”.

Also using symmetry or a direct calculation we may find the expected total population at any time $t \geq 0$: $$\label{eq_en}
E |N_t| = 2 E |N_t^0| = 2 \Phi \Big( \hat{\beta}_0 \sqrt{t}\Big) \exp \Big\{ \big( \frac{1}{2} \hat{\beta}_0^2 + \hat{\beta}\big) t \Big\} \text{.}$$

In particular, $$\label{exp_pop_assymp}
E |N_t| \sim 2  \exp \Big\{ \big( \frac{1}{2} \hat{\beta}_0^2 + \hat{\beta}\big) t \Big\} \text{ as } t \to \infty \text{.}$$

Main results
------------

Our aim is to replace convergences 