\mathrm{e}^{\hat{\beta} t} \int_0^\infty \int_x^{\infty} \mathrm{e}^{\hat{\beta}_0 l} \frac{y + l}{\sqrt{2 \pi t^3}} 
\exp \Big\{ - \frac{(y+l)^2}{2t}\Big\} \mathrm{d}y \mathrm{d}l\\
&= \mathrm{e}^{\hat{\beta} t} \int_0^\infty \mathrm{e}^{\hat{\beta}_0 l} \frac{1}{\sqrt{2 \pi t}} 
\exp \Big\{ - \frac{(x+l)^2}{2t}\Big\} \mathrm{d}l\\
&= \mathrm{e}^{\hat{\beta} t} \int_0^\infty \frac{1}{\sqrt{2 \pi t}} \exp \Big\{ - \frac{1}{2t} \big( l - (\hat{\beta}_0t - x)\big)^2 
+ \frac{\hat{\beta}_0^2}{2}t - \hat{\beta}_0 x \Big\} \mathrm{d}l\\
&= \mathrm{e}^{\hat{\beta}t + \frac{\hat{\beta}_0^2}{2}t - \hat{\beta}_0 x} \int_{- (\hat{\beta}_0 \sqrt{t} - \frac{x}{\sqrt{t}})}^\infty 
\frac{1}{\sqrt{2 \pi}} \mathrm{e}^{- \frac{z^2}{2}} \mathrm{d}z\\
&= \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{.}\end{aligned}$$ For the expected total population we could have followed a similar calculation: $$\label{eq_en2}
E|N_t| = \tilde{E} \Big[ \mathrm{e}^{\hat{\beta_0} \tilde{L_t} + \hat{\beta} t} \Big] = \cdots = 
2 \Phi \Big( \hat{\beta}_0 \sqrt{t}\Big) \exp \Big\{ \big( \frac{1}{2} \hat{\beta}_0^2 + \hat{\beta}\big) t \Big\}$$

Let us now prove the upper bound for Lemma \[population\_as\].

\[upper\_pop\] $$\label{eq_upper_pop}
\limsup_{t \to \infty} \frac{1}{t} \log |N_t| \leq \frac{1}{2} \hat{\beta}_0^2 + \hat{\beta} \qquad P \text{-a.s.}$$

Fix $\epsilon > 0$. Then by the Markov inequality and $$\begin{aligned}
P \Big( \frac{1}{n} \log |N_n| > \frac{1}{2}\hat{\beta}_0^2 + \hat{\beta} + \epsilon \Big) &= 
P \Big( |N_n| > \mathrm{e}^{(\frac{1}{2}\hat{\beta}_0^2 + \hat{\beta} + \epsilon)n} \Big)\\ 
&\leq \mathrm{e}^{-(\frac{1}{2}\hat{\beta}_0^2 + \hat{\beta} + \epsilon)n} E|N_n|
< 2 \mathrm{e}^{- \epsilon n}.\end{aligned}$$ 