# Inverse Laplace transform to get CDF

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?

• What is $T_A$? Is it the hitting time of $0$ when the process starts at $A$, or vice versa? Commented Oct 4, 2017 at 8:15
• T_A is the hitting time to 0 when the process starts at A Commented Oct 4, 2017 at 15:28
• Ok think of T_A and T_B as first passage times Commented Oct 4, 2017 at 15:58
• How are processes starting at $A$ and $B$ coupled then? In order to compare $T_A$ and $T_B$, both random variables need to be defined on a common probability space. Commented Oct 4, 2017 at 21:25
• They are independent birth death processes. $T_A$ and $T_B$ are both positive r.vs Commented Oct 5, 2017 at 4:50

## 1 Answer

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.

• Hey Mateusz, thanks for you proof. So my main question is, am i supposed to evaluate that integral from $-\infty$ to positive $\infty$? I.e would i be taking inverse of 1/2 - (integral from negative infinity to infinity?) Commented Oct 13, 2017 at 17:38
• @rajn: I am sorry, but I do not understand your question. What kind of "inverse" are you asking about? Commented Oct 13, 2017 at 18:05
• I'm trying to find the inverse of $\phi_A(s)\phi_B(-s) / s$ evaluated at t = 0. Is that possible? Commented Oct 14, 2017 at 3:36
• @rajn: Yes, by inverting the Laplace transform. The correct expression is given in the last display in my answer. Commented Oct 14, 2017 at 6:52
• Oh ok. Thanks for the answer. Question: If I implement the last expression in display in a computer, how would I go about doing this? I’m using python. Would I be using Euler algorithm? Commented Oct 15, 2017 at 17:08