Poisson mixture process independence used to devastating effect on the Coupon collectors problem Proposition 5.2 of the book, Introduction to probability models by Sheldon Ross says that if we have a Poisson process and each event in the process is of type-1 with probability $p$ and type-2 with probability $1-p$, then the number of type-1 and type-2 events are independent Poisson processes with rates $\lambda p$ and $\lambda (1-p)$ respectively. The independence is key here. It is then used as a powerful tool in example 5.17, where Ross addresses the coupon collectors problem. Quoting:
There are $m$ different types of coupons. Each time a person collects a coupon it is, independently of ones previously obtained, a type $j$ coupon with probability $p_j$, 
$\sum\limits_{j} p_j = 1$. Let $N$
denote the number of coupons one needs to collect in order to have a complete collection of at least one of each type. Find $E[N]$.
In the solution, he starts with the straightforward approach, denoting by $N_j$ the number of coupons that must be collected to obtain a type $j$ coupon. We can then express $N$ as:
$$N = \max_{1\leq j \leq m} N_j \tag{1}$$
He notes that the $N_j$ are geometric, but this method runs into a wall when we realize that the $N_j$'s aren't independent. And this makes sense. If there were only two types of coupons, they would be competing each time we collected a coupon. So, if we need very few coupons to collect one for the first kind, it tells us it's a common coupon and so, we now know that we'll have to wait a long time to see the second coupon (meaning $N_1$ and $N_2$ are negatively correlated).
Now, Ross considers the coupons arriving according to a Poisson process with rate $1$. By proposition 5.2, the counting processes defining the arrivals of each of the coupon types (say $j$) are independent Poisson process with rates $1 . p_j$. Now, define $X$ the time at which all coupons are collected and $X_j$ the time at which the first type $j$ coupon is collected. We get an equation very similar to (1):
$$X = \max_{1\leq j \leq m} X_j \tag{2}$$
Now, we don't run into the wall since by proposition 5.2, the $X_j$'s are independent. However, I haven't been convinced by the arguments presented for this. Why does the reasoning we used to conclude that the $N_j$'s are negatively correlated not apply to the $X_j$'s as well?
 A: The fact that $X_j$'s are independent follows directly from the fact that a Poisson process can split into one with rate $\lambda p$ and one with rate $\lambda (1-p)$ (but of course here it splits into $N$ such processes, not just $2$ such processes).  So that's the "math" explanation.
If you would like a more "intuitive" explanation, esp. on why the $X_j$'s behave differently from the $N_j$'s, try this hand-wavy one.  Imagine $N=2$, and you get $1$ coupon, then it is either type $1$ or $2$, and they are mutually exclusive (or "negatively correlate").  But if you wait $1$ unit of time in the Poisson formulation, you can get any number of coupons of either type.  Crucially, the fact that you get one (or more) coupon of type $1$ does not affect the prob of you getting one (or more) coupon of type $2$ in that same unit of time - that is the magic of splitting Poisson processes. 
 E.g. imagine you get a type-$1$ coupon in at time $t=0.6$, that does not change the prob that you get a type-$2$ coupon in the time interval $(0.6,0.6+\epsilon]$ for any $\epsilon$.  
Allow me to vaguely define $A_i$ as the event "getting a coupon of type $i$" (under some to-be-specified circumstances), then:


*

*Conditioned on you getting $1$ coupon (total), then $A_1, A_2$ are mutually exclusive.

*In fact, for any $n \in \mathbb{N}, T \in \mathbb{R}$, conditioned on you waiting $T$ time and getting $n$ coupons (total), then $A_1, A_2$ are dependent ("negatively correlated").

*But, conditioned on you waiting $1$ unit of time (and no further conditioning on how many total coupons you got during that time), then $A_1, A_2$ are independent - and this is a non-trivial fact based on splitting Poisson processes.
Am I helping or am I just being repetitive?  :)
A: I don't yet have an intuition for why your reasoning about the $N_j$s doesn't apply also to the $X_j$s, Rohit. In particular I haven't yet managed to properly grasp antkam's nice answer.
However, I have had a go at working out some details of Ross's proof of his Proposition 5.2 (via his hints, in so far as I understand them), which I'll share here in the hope that they might be useful, albeit that what I've written is rather clunky and possibly incorrect !
With all notation as in Ross's Proposition 5.2, suppose that $0<s<t$ and $k \in \{0, 1, 2, \ldots\}$. Then
\begin{align*}
P\{N_1(t)-N_1(s)=k\}
&= \sum_{j \geq k} P\{N_1(t)-N_1(s)=k, N(t)-N(s)=j\} \\
&= \sum_{j \geq k} P\{N_1(t)-N_1(s)=k \, | \, N(t)-N(s)=j\} P\{N(t)-N(s)=j\} \\
&= \sum_{j \geq k} \binom{j}{k} p^k (1-p)^{j-k} P\{N(t-s)=j\}, \tag{1}
\end{align*}
which, given $\lambda$, $p$ and $k$, depends only on $t-s$, showing that $N_1$ has stationary increments.
Question: Can a counting process have stationary increments, without having independent increments ?
In any case, let's also try to show that $N_1$ has independent increments:
Suppose $0<s<t \leq s'<t'$, and $k,k' \in \{0, 1, 2, \ldots \}$. Then
\begin{align*}
P\{N_1(t)-N_1(s)=k, N_1(t')-N_1(s')=k'\}
&= \sum_{j \geq k, j' \geq k'} P\{N_1(t)-N_1(s)=k, N_1(t')-N_1(s')=k', N(t)-N(s)=j, N(t')-N(s')=j' \} \\
&= \sum_{j \geq k, j' \geq k'} P\{(N_1(t)-N_1(s)=k, N_1(t')-N_1(s')=k') \, | \, (N(t)-N(s)=j, N(t')-N(s')=j') \} P\{N(t)-N(s)=j, N(t')-N(s')=j'\} \\
&= \sum_{j \geq k, j' \geq k'} \binom{j}{k} \binom{j'}{k'} p^{k+k'} (1-p)^{j-k+j'-k'} P\{N(t)-N(s)=j\} P\{N(t')-N(s')=j'\} \\
&= \sum_{j \geq k, j' \geq k'} \binom{j}{k} \binom{j'}{k'} p^{k+k'} (1-p)^{j-k+j'-k'} P\{N(t-s)=j\} P\{N(t'-s')=j'\} \\
&= P\{N_1(t)-N_1(s)=k\} P\{N_1(t')-N_1(s')=k'\},
\end{align*}
by (1). This shows that $N_1$ has independent increments.
Continuing to the 3rd bullet point of Ross's proof of his Proposition 5.2, in the second equation, he uses the fact that
\begin{align*}
P\{N_1(h)=1 \, | \, N(h) \geq 2 \} P\{N(h) \geq 2 \} = o(h).
\end{align*}
We know, from Definition 5.3 part (iv), that $P\{N(h) \geq 2 \} = o(h)$. I was worried about the other factor, so I tried to control it, as follows:
I believe that
\begin{align*}
P\{N_1(h)=1 \, | \, N(h) \geq 2\}
&= P\{N_1(h)=1, N(h) \geq 2\}/P\{N(h) \geq 2 \} \\
&= \left (\sum_{k \geq 2} P\{N_1(h)=1, N(h)=k\} \right ) / P\{N(h) \geq 2\} \\
&= \left (\sum_{k \geq 2} P\{N_1(h)=1 \, | \, N(h)=k\} P\{N(h)=k\} \right ) / P\{N(h) \geq 2\} \\
&= \left (\sum_{k \geq 2} \binom{k}{1} p^1 (1-p)^{k-1} e^{- \lambda h} (\lambda h)^k/k! \right ) / (1 - (P\{N(h)=0\} + P\{N(h)=1\})) \\
&= \left (p e^{-\lambda h} \sum_{k \geq 2} k (1-p)^{k-1} (\lambda h)^k/k! \right ) / (1 - (e^{-\lambda h} + \lambda h e^{-\lambda h})) \\
&= (p(1 - \lambda h + o(h))o(h) / (1 - \lambda h + o(h) + \lambda h (1 - \lambda h + o(h))) \\
&= o(h) / (1 + o(h)) \\
&= o(h),
\end{align*}
therefore
\begin{equation*}
P \{N_1(h)=1 \, | \, N(h) \geq 2 \} P \{N(h)\geq 2\} = o(h)o(h) = o(h),
\end{equation*}
as required.
Finally, regarding the claim in Proposition 5.2 that the two processes $\{N_1(t), t \geq 0 \}$ and $\{N_2(t), t \geq 0 \}$ are independent, I (rightly or wrongly) take this to mean that for all $t \geq 0$, the random variables $N_1(t)$ and $N_2(t)$ are independent.
I don't follow the first sentence of Ross's explanation

Because the probability of a type I event in the interval from $t$ to $t + h$ is independent of all that occurs in intervals that do not overlap $(t, t + h)$, it is independent of knowledge of when type II events
  occur, showing that the two Poisson processes are independent. (For
  another way of proving independence, see Example 3.23.)
  ,

but I believe I do understand the alternate one given in Example 3.23.
