0
$\begingroup$

I am writting the following proof and would like to know if you guys think it is good enough, or should I maybe explain it further. My concern is that I maybe should prove why the last claim I make (about the integral being different) is true.

Consider the generalization of the Poisson process with a time-dependant intensity of arrivals $\lambda_t$ such that $\lim_{T \to \infty}\int_{0}^T\lambda_tdt=\infty$. The time $t\geq 0$ at which the event occurs has c.d.f: \begin{equation*} \mathbb{P}(t\leq\tau|t>s)=1-e^{-\int_0^{\tau}\lambda_tdt} \end{equation*}

Claim: in general, $\mathbb{P}(t\leq\tau|t>s)\neq\mathbb{P}(t\leq\tau-s)$

Proof. Using the definition of conditional probability: \begin{align*} \mathbb{P}(t\leq\tau|t>s)&=\frac{\mathbb{P}(s

if $\lambda_t$ is not a constant for all $t$.

  • 0
    What is $\tau$ here? Some kind of stopping time? Also by your notation, it appears as if $t$ is a random variable...2017-02-19
  • 0
    Yes! That's the CDF of $t$, I believe @GautamShenoy2017-02-19

0 Answers 0