Given $X_i = A_i - B_i$ where $A_i\sim \text{ Exp}(\alpha)$ and $B_i \sim \text{ Exp}(\lambda)$. Define $S_k = \sum_{i=1}^k X_i$ with $S_0 = 0$, and $$M_n = \max_{1\leq k \leq n} S_k.$$ Is it possible to calculate the quantity $\mathbb{P}(M_n \leq x)$ explicitly? I have tried it and the result is written at the bottom, but it is not quite explicit...
In Feller's An Introduction to Probability theory and Its Application, he made the remark several times that this type of distribution with two exponential tails is a rare but important case where almost all calculations related to the random walk can be made explicit. (V1.8 Example (b) page 193; XII.2 Example (b) page 395; XII.3 Example (b) page 401) Unfortunately, I couldn't find any detailed calculations in the book.
A second reference that I looked at is the paper "On the distribution of the maximum of sums of mutually independent and identically distributed random variables" by Lajos Takacs (adv. Appl. Prob. 1970). Takacs mentioned that in a few particular cases we can calculated $\mathbb{P}(M_n \leq x)$ easily. Following his example on page 346 (where he only assumed $X_i = A_i - B_i$ where $B_i$ is exponential and $A_i$ is non-negative), I was able to calculate for $A_i\sim \text{ Exp}(\alpha)$, $B_i \sim \text{ Exp}(\lambda)$, we have $$U(s,p) = \sum_{n=0}^\infty \mathbb{E}\left[e^{-sM_n}\right]p^n = \frac{\lambda - \frac{s\lambda}{\gamma(p)}}{\lambda - s - \frac{\lambda \alpha p}{\alpha +s}}$$ where $\gamma(p) = \frac{\lambda - \alpha + \sqrt{(\alpha+\lambda)^2 - 4\alpha \lambda p}}{2}$, a zero of the denominator above. Is there a way to simplify this to obtain some kind of explicit formula for $\mathbb{P}(M_n \leq x)$?
Edit: I was hoping for a (simple) closed formula for doing estimates. Following the general case calculation in the solution linked below, it is seemed to be unlikely...
