Inverse Laplace transform to get CDF

Question

I have the following problem. If I can get some help, I would greatly appreciate it. I am trying to replicate a particular research paper and came across a problem:

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 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 $P(T_A < T_B)$. That is, probability that a population of size $A$ goes to $0$ before population of size $B$.

By definition:

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

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 defined by:

$$G_T(t) := P(T \leq t)$$

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

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

Then it suggested taking Laplace of $G_T(t)$ i.e. $L[G_T(t)]$. However,

$$L[G_T(t)] = \frac{L[T]}{s} = \frac{L[T_A](s) L[T_B](-s)}{s}.$$

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

Questions:

1) Given the Laplace transform of CDF, which is $L[G_T(t)]$, 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?

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

Answer 1

This seems to be a standard exercise. Anyway, here is a sketch of the solution.

Let $T$ be the hitting time of zero, and $\phi_n(s) = \mathbb{E}(e^{-s T} | X_0 = n)$ be the Laplace transform of $T$. Then $\phi_0(s) = 1$, and $$ \phi_n(s) = \frac{\mu + \theta}{\mu + \theta + s} \left( \frac{\mu}{\mu + \theta} \, \phi_{n+1}(s) + \frac{\theta}{\mu + \theta} \, \phi_{n-1}(s) \right) = \frac{\mu \phi_{n+1}(s) + \theta \phi_{n-1}(s)}{\mu + \theta + s} \, .$$ Solving this system of linear equations (given $0 \le \phi_n(s) \le 1$) leads to $$ \phi_n(s) = \left( \frac{2 \theta}{\mu + \theta + s + \sqrt{(\mu + \theta + s)^2 - 4 \mu \theta}}\right)^n . $$ The Laplace transform of $T = T_A - T_B$, with independent $T_A$ and $T_B$, is $\phi_A(s) \phi_B(-s)$. The probability that $T > 0$ can be expressed as $$ \frac{1}{2} - \frac{1}{2 \pi i} \int_{-\infty}^\infty \frac{\phi_A(i s) \phi_B(-i s)}{s} ds ,$$ with the integral understood in the principal value sense.