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Given a random variable $$L_n = \sum_{i=1}^n X_i Z_i $$ where $X_i \sim Bernoulli(p)$ and $Z_i \sim \exp(\theta)$. I assume $\{Z_i \}$ is i.i.d. and the sequences $\{X_i \}$ and $\{Z_i \}$ are independent.
According to Cramér’s large deviation theorem, $$\lim_{n \rightarrow \infty} \frac{1}{n} \log \left( \textbf{P}\{ L_n > n \cdot a \}\right) = - \Lambda^*(a) \\ \Lambda^*(a) = \sup_{\xi \in \mathbb{R}} \left( \xi a - \Lambda (\xi) \right) \\ \Lambda (\xi) = \log \left(\kappa(\xi) \right) $$ where $\kappa$ denotes the moment generating function for $X_1 Z_1$ and $a>0$. Note that $\kappa(\xi) = \frac{p \theta}{\theta - \xi}+1-p$. I have calculated the $\xi^*$ that solves $\Lambda^*(a)$ $$\xi(a,p,\theta)^*= \frac{\theta (2-p) - \sqrt{p \cdot (\theta^2p + \frac{4(p+\theta)}{a}) } }{2(1-p)}.$$ Now, by using the approximation $ \textbf{P}\{ L_n > n \cdot a \} \approx e^{-n \Lambda^*(a)} $, I want to calculate the $\alpha$-quantile $(VaR_{\alpha})$, where I think of $n$ as "large" and $\alpha$, $\theta$ and $p$ as given. Formally, I can get this far: \begin{align*} \text{VaR}_\alpha(L) = & \text{inf} \{ x \in \mathbb{R} : P(L>x) \leq 1-\alpha \} \\ = & \text{inf} \{ a \in \mathbb{R} : P \left( L>n \cdot a \right ) \leq 1-\alpha ) \\ \approx& \inf \{ a \in \mathbb{R} : e^{-n \Lambda^*(a ) } = 1-\alpha \} \end{align*} where the equality comes from $\Lambda^*(a)$ being smooth. Thus I have to find the smallest $a$ that solves $e^{-n\Lambda^*(a )} = \alpha$ for a "large" n. \begin{align*} e^{-n\Lambda^*(a )} = 1- \alpha \\ e^{-n(\xi^* a - \log (\kappa (\xi^*)) )} =1- \alpha \\ \kappa(\xi^*)^n e^{-n \xi^* a} = 1- \alpha \end{align*} However, I cannot directly solve this equation since $a$ is in $\xi^*$.
How do I proceed?

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