How to use Inverse Laplace transform to obtain the Cumulative distribution function (CDF)?

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Questions:

1. Given the Laplace transform of the CDF, which is $\mathcal{L}[G_T(t)]$, I want to use the inverse Laplace to obtain $G_T(t)$ evaluated at 0. But, by definition, inverse Laplace using algorithms in python are all one sided from $[0,\infty]$. My random variable $T$ is given by difference of two first passage times, $T = T_A - T_B $. Won't this be negative? When I implement this function in python, and evaluate it at 0, it gives me 0 as answer.

2. In the paper, pg. 17 section 5.2.2, it says they are shifting the random variable $X$ under study by a constant $c$ such that $\mathbb{P}(X + c > 0)$ is approximately $1$, and then inverting the corresponding one sided Laplace transform. How would I do that in this context here?

Background:

I am trying to replicate a research paper by Cont, Stoikov, & Talreja, "A stochastic model for order book dynamics", and came across the following problem.

I am trying to find probability of the mid price increase. It's stated as follows:

Suppose X is a birth death process (represents population size) that evolves by:

$X \to X+1 $ if a birth occurs with rate $\mu$

$X \to X-1 $ if a death occurs with rate $\theta$

Suppose $T_A$ is first passage time of a birth-death (BD) process from state $A$ to state $0$, and suppose $T_B$ is first passage time of another BD process from state $B$ to state $0$. They are both independent.

I need to find $\mathbb{P}(T_A < T_B)$. That is, probability that a population of size $A$ goes to $0$ before population of size $B$ goes to 0.

By definition:

$$T_A = T_{A,A-1} + T_{A-1,A-2} + ... + T_{1,0}$$

where $T_{i,i-1}$ represents first passage time from state $i$, to state $i-1$.

I read some articles online that mentioned that if $T = T_A - T_B$, then the CDF is defined by:

$$G_T(t) := \mathbb{P}(T \le t)$$

is what I need. The paper suggested taking inverse Laplace of a CDF to obtain the CDF and evaluate it at $0$. It first suggested finding Laplace transform of T, which is given by

$$\mathcal{L}[T] = \mathbb{E}(e^{-ST}) = -\frac{\mathcal{L}[T_A](s) \mathcal{L}[T_B](-s)}{s}$$

Then it suggested taking laplace of $G_T(t)$ i.e $\mathcal{L}[G_T(t)]$. However, $$\mathcal{L}[G_T(t)] = \frac{\mathcal{L}[T]}{s} = \frac{\mathcal{L}[T_A](s) \mathcal{L}[T_B](-s)}{s}$$

Then the paper suggests taking the inverse of above evaluated at 0 to get $\mathbb{P}(T_A < T_B)$.