Asymptotics of a quotient related to a simple random walk Let $\lambda_0 < \lambda_1$ and $\lambda_0 \lambda_1 > 1$ (i.e. at least $\lambda_1 > 1$). Further, let $S_n$ denote a simple random walk with increment distribution $$ P(X = 0)= P(X= 1) = 1/2.$$
Note that, using the SLLN, one can easily prove that (as $n \to \infty$) $$ \lambda_0^{S_n} \lambda_1^{n-S_n} \to \infty \ \ a.s.$$
Now, my goal is to find a deterministic sequence $(a_n)_n$ such that $$ E \left[ \frac{\lambda_0^{S_n} \lambda_1^{n-S_n}} {\left(\sum_{k=0}^{n-1} \lambda_0^{S_k} \lambda_1^{k-S_k}\right)^2}\right] \Big{/} a_n \longrightarrow c \tag{1} $$
for some constant $c > 0$ as $n \to \infty$. To put it shortly, I am interested in the asymptotic behavior of the stated expectation.
My (not very educated) guess for such a sequence is $$ a_n = E \left[ \frac{1}{\lambda_0^{S_n} \lambda_1^{n-S_n}}\right].$$ The reason for this guess is that in a deterministic case (for any $\lambda > 0$), we have as $n \to \infty$ that $$ \frac{\lambda^n}{\sum_{k=0}^{n-1} \lambda^k} \to c> 0.$$
I am looking for ideas on how to handle the expectation in (1) and deduce such a "normalizing" sequence. Handling the expression in (1) seems quite complicated... Any ideas are much appreciated!
 A: I would use a "Girsanov like" trick to modify the expectation: For a general random walk with $\mathbb{P}(X=1)=p$ that we denote the associate law $\mathbb{E}_p$.For $\alpha\in \mathbb{R}$ we have
$$\mathbb{E}_p(f(\boldsymbol{S}))=\sum_\boldsymbol{S} f(\boldsymbol{S})p^{S_n}(1-p)^{n-S_n}=\sum_{\boldsymbol{S}} f(\boldsymbol{S})\lambda_0^{\alpha S_n}\lambda_1^{\alpha(n-S_n)}(\frac{p}{\lambda_0^\alpha})^{S_n}(\frac{1-p}{\lambda_1^\alpha})^{n-S_n} = (\frac{p}{\lambda_0^\alpha}+\frac{1-p}{\lambda_1^\alpha})^n\mathbb{E}_{\tilde{p}}[f(\boldsymbol{S})\lambda_0^{\alpha S_n}\lambda_1^{\alpha(n-S_n)}]$$ with $\tilde{p}=\frac{p}{\lambda_0^\alpha}(\frac{p}{\lambda_0^\alpha}+\frac{1-p}{\lambda_1^\alpha})^{-1}$. This is to be understood as follow: one can change the law of the random walk $p\rightarrow \tilde{p}$  (adding a drift) if we add the weight $\lambda_0^{\alpha S_n}\lambda_1^{\alpha(n-S_n)}$. In our case $\tilde{p}=\frac{1}{\lambda_0^\alpha}(\frac{1}{\lambda_0^\alpha}+\frac{1}{\lambda_1^\alpha})^{-1}$.
Fist Case: If for $\alpha=1$, $$\tilde{p}\log(\lambda_0)+(1-\tilde{p})\log(\lambda_1)> 0$$
then choosing $\alpha=1$
$$\mathbb{E}_{1/2}(\frac{\lambda_0^{ S_n}\lambda_1^{(n-S_n)}}{(\sum_k \lambda_0^{ S_k}\lambda_1^{(k-S_k)})^2})= (\frac{1}{2\lambda_0}+\frac{1}{2\lambda_1})^n\mathbb{E}_{\tilde{p}}[\frac{(\lambda_0^{ S_n}\lambda_1^{(n-S_n)})^2}{(\sum_k \lambda_0^{ S_k}\lambda_1^{(k-S_k)})^2}]= (\frac{1}{2\lambda_0}+\frac{1}{2\lambda_1})^n\mathbb{E}_{\tilde{p}}[\frac{1}{(\sum_k \lambda_0^{S_k-S_n}\lambda_1^{(k-n+S_n-S_k)})^2}] $$ Because of the assumption on $\tilde{p}$, with probability that goes to 1 $\lambda_0^{S_k-S_n}\lambda_1^{(k-n+S_n-S_k)}\rightarrow 0$ exponentially fast as $n-k\rightarrow \infty$. And then we should obtain $$0<c\leq \mathbb{E}_{\tilde{p}}[\frac{1}{(\sum_k \lambda_0^{S_k-S_n}\lambda_1^{(k-n+S_n-S_k)})^2}]\leq 1 $$
Second Case :If the assumption on $\alpha=1$ fails then we choose $\alpha\leq 1$ such that $$\tilde{p}\log(\lambda_0)+(1-\tilde{p})\log(\lambda_1)=0$$ For the corresponding law with paramatter $\tilde{p}$, $S_k\log(\lambda_0)+(k-S_k)\log(\lambda_1)$ should behave for large $k$ as a brownian motion $\sigma B_k$. We could then guess
$$\mathbb{E}_{1/2}(\frac{\lambda_0^{ S_n}\lambda_1^{(n-S_n)}}{(\sum_k \lambda_0^{ S_k}\lambda_1^{(k-S_k)})^2})= (\frac{1}{2\lambda_0^\alpha}+\frac{1}{2\lambda_1^\alpha})^n\mathbb{E}_{\tilde{p}}[\frac{(\lambda_0^{ S_n}\lambda_1^{(n-S_n)})^{1+\alpha}}{(\sum_k \lambda_0^{ S_k}\lambda_1^{(k-S_k)})^2}] \approx (\frac{1}{2\lambda_0^\alpha}+\frac{1}{2\lambda_1^\alpha})^n\mathbb{E}[\frac{e^{(1+\alpha)\sigma B_n}}{(\max_k e^{\sigma B_k})^2}]\\= (\frac{1}{2\lambda_0^\alpha}+\frac{1}{2\lambda_1^\alpha})^n\mathbb{E}[e^{(1+\alpha)\sigma B_n-2\max_k \sigma B_k}] $$
where we make the assumption that the sum is dominated by the largest term. Hopefully this last espectation with the brownian motion can be estimated.
