# Poisson Process from independent non-identical exponential RVs

I know, I can define a Poisson Process using a sequence of i.i.d. exponential random variables, i.e. let $$\tau_1, \tau_2, \tau_3, ... \sim \mathrm{Exp}(\lambda)$$, then $$T_i = \sum_{j=1}^I \tau_j$$ are realizations of a Poisson Process with rate $$\lambda$$.

Now, I am wondering, if I let each $$\tau_i \sim \mathrm{Exp}(\lambda_i)$$, but still independent,

1. Are the $$T_i$$ still a Poisson process?
2. Possibly a non-homogenous one? With what intensity function?
3. And what transformation would turn the resultant Poisson Process into a homogenous Poisson Process?

## 1 Answer

Unfortunately, the $$T_i$$ are not a poisson process. One way of showing it is, for example, by considering the number $$N$$ of $$T_i$$ that falls into the interval $$[0,1]$$. It's supposed to be a Poisson rv if the $$T_i$$ are a poisson process. But, $$\mathbb{P}(N=0)=\mathbb{P}(\tau_1\geq 1)=e^{-\lambda_1}$$ and $$\mathbb{P}(N=1)=\mathbb{P}(\tau_1\leq 1~\mbox{and}~\tau_1+\tau_2\geq 1)=\left\{\begin{array}{l} \lambda_1 e^{-\lambda_1}~\mbox{if}~\lambda_1=\lambda_2 \\ \frac{\lambda_1 e^{-\lambda_2}}{\lambda_2-\lambda_1}\left(e^{\lambda_2-\lambda_1}-1\right)~\mbox{otherwise.}\end{array}\right.$$ So, $$\lambda_1=\lambda_2$$ and similarly we can show that all $$\lambda_i$$ have to be equal.

An instinctive way of seeing that $$T_i$$ is not a poisson process is imagining, for example, that $$\lambda_1=1$$ and $$\lambda_i=10$$ for all $$i\geq 2$$. If we know that $$T_1<1$$ then we know that the average number of $$T_i$$ that falls in $$[1,2[$$ is $$10$$ but if $$T_1>1$$ then this is not the case anymore, it will be lower. So, knowing what happens on the interval $$[0,1[$$ has an influence on what happens on the interval $$[1,2[$$. This is never the case for a Poisson point process.