= Solution
Use the <large-deviation speed> $a_N=N^{1+\varepsilon}$ from part (a). For every $\eta>0$, the <Chernoff bound> for a <normal distribution> gives, for all sufficiently large $N$,
$$
\mathbb P(|B_N|>\eta a_N)
\leq2\exp\!\left[-\frac{(\eta a_N-|N\mu|)^2}{2N\sigma^2}\right].
$$
Consequently
$$
\limsup_N a_N^{-1}\log\mathbb P(|B_N|>\eta a_N)=-\infty,
$$
because the exponent has order $N^{1+2\varepsilon}$, exceeding $a_N$ by a factor $N^\varepsilon$. Thus $(A+B_N)/a_N$ and $A/a_N$ satisfy <exponential equivalence>. This conclusion does not require <independence> between $A$ and $B_N$, since their difference is exactly $B_N/a_N$.
The <exponential equivalence> theorem transfers the <large deviation principle> and its <good rate function>, giving
$$
\boxed{\text{speed }N^{1+\varepsilon},\qquad I_{A+B}(x)=
\begin{cases}\lambda x,&x\geq0,\\+\infty,&x<0.\end{cases}}
$$
Large positive excursions are cheapest through the single <exponential random variable>; a negative excursion would require the <Gaussian random variable> sum to move on its faster scale. If $\sigma=0$, the deterministic shift $N\mu/a_N\to0$ gives the same conclusion directly.
Back to article page
