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, 


*

*Are the $T_i$ still a Poisson process? 

*Possibly a non-homogenous one? With what intensity function?

*And what transformation would turn the resultant Poisson Process into a homogenous Poisson Process?

 A: 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.
