given $ X_i = A_i – B_i $ from 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

$$ M_n = max_ {1 leq k leq n} S_k. $$

Can you calculate the amount? $ mathbb {P} (M_n leq x) $ expressly? I tried it and the result is down, but it is not clear …

In Feller's Introduction to Probability Theory and its application, he has repeatedly remarked that this type of two-exponential divergence distribution is a rare but important case in which almost all erroneous calculations can be made explicit. (V1.8 example (b) page 193; XII.2 example (b) page 395; XII.3 example (b) page 401) Unfortunately, I could not find any detailed calculations in the book.

A second reference that I have looked at is the paper "*On the distribution of the maximum of sums of independent and equally distributed random variables*"by Lajos Takacs (Adv. Appl. Prob 1970) Takacs mentioned that we can calculate in some special cases $ mathbb {P} (M_n leq x) $ light. After his example on page 346 (where he only suspected) $ X_i = A_i – B_i $ from where $ B_i $ is exponential and $ A_i $ is not negative) that I could count on $ A_i sim text {Exp} ( alpha) $, $ B_i sim text {Exp} ( lambda) $, we have

$$ U (s, p) = sum_ {n = 0} ^ infty mathbb {E} left[e^{-sM_n}right]p ^ n = frac { lambda – frac {s lambda} { gamma (p)}} { lambda – s – frac { lambda alpha p} { alpha + s}} $$

from 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 get an explicit formula for $ mathbb {P} (M_n leq x) $?