# Waiting time for the first event in an Exponential Distribution with different rate parameters

I have a random variable $t$ that follows an exponential distribution with rate $\lambda_1$ in the interval $[0,t_1]$ and an exponential distribution with rate $\lambda_2$ in the interval $[t_1,\infty]$. I want to estimate the probability that there is no event in the time interval $[0,t_2]$, where $t_2>t_1$.

Here is how I have approached this problem.

$$$P(t>t_1+t_2)\\ =\{1-\int_{-\infty}^{t_1}f_{\lambda_1}(t)\} \{1-\int_{t_1}^{t_2}f_{\lambda_2}(t)\}$$$

Is the correct way to approach this problem? Should the second integral be $\{1-\int_{0}^{t_2}f_{\lambda_1}(t)\}$?

In general, what is the procedure to find the waiting time to the 1st event in a case where the inter-arrival process is non-homogenous?

• Suppose $\lambda_1$ happened to equal $\lambda_2$. You would likely use a simpler expression for the waiting time involving just one integral. Is your formula equivalent in that case to the simple expression? – whuber Aug 17 '16 at 18:53
• That is a good point. It is not. – statBeginner Aug 17 '16 at 18:58
• OK. The simple expression can be written as an integral. There is a natural way to break it into an integral over two regions separated at the value $t_1$, which re-expresses it as a sum of two integrals. Consider why that is so and apply that insight to the case $\lambda_1 \ne \lambda_2$. – whuber Aug 17 '16 at 19:02
• Unless $\lambda_1 = \lambda_2$, the density of the "random variable" does not integrate to 1. If $\lambda_1 \ge \lambda_2$, it would integrate out to 1 if the $\lambda_1$ rate section ended at $t_1$, then there was a gap of 0 density, and the $\lambda_2$ rate section started at $t_3 = (\lambda_1/\lambda_2) t_1$. Barring that, a time-inhomogeneos Possion Process, yes, a proper random variable, no. – Mark L. Stone Aug 17 '16 at 19:49
• I think Math1000 has given a nice demonstration. It differs from your approach by recognizing you have to compute the probability by multiplying a marginal by a conditional probability rather than adding those two probabilities. – whuber Aug 18 '16 at 19:44

Let $\{N(t):t\in\mathbb R_+\}$ be a (non-homogeneous) Poisson process with intensity $$\lambda(t) = \lambda_1\mathsf 1_{[0,t_1)}(t) + \lambda_2\mathsf 1_{[t_1,\infty)}(t).$$ Then for $t_2>t_1$, we have \begin{align} \mathbb P(N(t_2)=0) &= \mathbb P(N(t_1)=0,N(t_2)-N(t_1)=0)\\ &= \mathbb P(N(t_1)=0)\mathbb P(N(t_2)-N(t_1)=0)\\ &= e^{-\lambda_1 t_1}e^{-\lambda_2(t_2-t_1)}. \end{align}