Event order from independent Poisson processes equal to selecting uniformly at random? Event order from independent Poisson processes equal to selecting uniformly at random?
Assume we have $N$ independent Poisson point processes with rate $R$. The processes are labelled with an ID and whenever an event occurs in one of the processes, we note the ID of that process. We do so until we have recorded $T$ IDs.
Now, can we say that a sequence of IDs generated that way is equally likely to occur to picking $T$ times an ID from the set of all possible IDs uniformly at random?
My guess would be that the answer is yes. I have found the following statement in [1,Chapter VII.5.2]: Denote with $\tau_i^k$ the $i$th arrival time of an event from process number $k$. The probability that the first event is caused by process $k$ is
$$ P(\min(\tau_0^1, \tau_0^2, ..., \tau_0^N) = \tau_0^k) = \frac{1}{N}$$
I am however unsure if this statement extends to all $\tau_{i>0}^k$. Since the Poisson process is memoryless it would not surprise me if it does, but I lack intuition on how to prove it and would appreciate any hint.
[1] Gut, A. (1995). An Intermediate Course in Probability. 
 A: As you said, Poisson processes are memoryless. So your assumption is true. 
Providing an elegant formal argument is not very easy and may seem useless.
You can first prove that the random variable $P(ID/h)$ is equiprobable for any h.
where h is both a point in time and the date of all events before t
and ID the ID of the process of the next event
Proof : in a Poisson process the period $P_i$ you have to wait for the next event of each process i given any context h are random exponential random variables with same parameter and independent from each other. 
So by symmetry, the variable with the smaller value is equiprobable.
Then $P(ID/s)$ is also equiprobable for any s
where s is a sequence of IDs in the order of past events as you record it (a less detailed context than h)
Proof  :  $P(ID/s) = \sum_c{P(h/s) P(ID/h,s)} = \sum_c{P(h/s) P(ID/h)}=\sum_c{P(h/s) \frac{1}{N}}=\frac{1}{N}$
You can then easily prove that  all sequences are equiprobable using the property :
$P(X_1,X_2,X_3,...) = P(X_1)P(X_2/X_1)P(X_3/X1,X_2)...$
