Mixing and dividing point processes At the following figure at left side two realizations of point processes with different density (intensity) $\lambda_1$ and $\lambda_2$ is being mixed matching the center of the belonging areas to build a point process in the middle with intensity $\lambda$. Then randomly selected points as two sets extracted from it as shown at the right side.
Questions:
Is $\lambda=\lambda_1+\lambda_2$?  and Is $\lambda=\lambda_3+\lambda_4$?
If two at left side were Poisson PP, Is the middle one a Poisson PP?
How about the two at right hand side?

 A: To answer this question we need a little background and notation. In the general terminology let $N$ denote a of point process in the plane, which means that for any Borel set, $A$, in the plane, $N(A)$ is an integer valued (including $+\infty$) random variable, which counts the number of points in $A$. Moreover, $A \mapsto N(A)$ is a measure for each realization of the point process $N$. 
Associated with the point process is the expectation measure 
$$A \mapsto \mu(A) := E(N(A))$$
where the expectation is always well defined, since $N(A) \geq 0$, but may be $+\infty$. It is left as an exercise to verify that $\mu$ is again a measure. To avoid technical issues lets assume that $\mu(\mathbf{R}^2) < \infty$, which is also reasonable if the process only really lives on a bounded set such as the box in the figure that the OP posted. It implies that $N(A) < \infty$ a.s. for all $A$.
The following definitions and observations follow.  


*

*We say that $N$ has intensity $\lambda$ if $\mu$ has density $\lambda$ w.r.t. the Lebesgue measure, that is, if
$$\mu(A) = \int_A \lambda(x) \mathrm{d}x.$$

*If $N_1$ and $N_2$ are two point processes we define the superposition as the sum $N_1 + N_2$. This is equivalent to superimposing one point pattern on top of the other. 

*If $N_1$ and $N_2$ are two point processes (independent or not) with intensities $\lambda_1$ and $\lambda_2$ then the superposition has intensity $\lambda_1 + \lambda_2$. 

*If $N_1$ and $N_2$ are independent Poisson processes then the superposition is a Poisson process. To show this we first observe that $N_1(A) + N_2(A)$ is Poisson from the convolution properties of the Poisson distribution, and then that if $A_1, \ldots, A_n$ are disjoint then $N_1(A_1) + N_2(A_1), \ldots, N_1(A_n) + N_2(A_n)$ are independent because $N_1$ and $N_2$ are independent and Poisson processes themselves. These two properties characterize a Poisson process.  


Summary I: We have shown that whenever a point process is a sum, or superposition, of two point processes with intensities then the superposition has as intensity the sum of the intensities. If, moreover, the processes are independent Poisson the superposition is Poisson.
For the remaining part of the question we assume that $N(\{x\}) \leq 1$ a.s. for all singleton sets $\{x\}$. Then the point process is called simple. Poisson processes with intensities are simple. For a simple point process there is a representation of $N$ as 
$$N = \sum_i \delta_{X_i},$$
that is, as a sum of Dirac measures at the random points. If $Z_i \in \{0,1\}$ are Bernoulli random variables, a random thinning is the simple point process
$$N_1 =  \sum_i Z_i \delta_{X_i}.$$
It is quite clear that with 
$$N_2 = \sum_i (1-Z_i) \delta_{X_i}$$
it holds that $N = N_1 + N_2$. If we do i.i.d. random thinning, meaning that the $Z_i$'s are all independent and identically distributed with success probability $p$, say, then
$$N_1(A) \mid N(A) = n \sim \text{Bin}(n, p).$$
From this,
$$E(N_1(A)) = E \big(E(N_1(A) \mid N(A))\big) = E(N(A)p) = p \mu(A).$$
If $N$ is a Poisson process it should be clear that for disjoint $A_1, \ldots, A_n$ then $N_1(A_1), \ldots, N_1(A_n)$ are again independent, and 
$$
\begin{array}{rcl}
P(N_1(A) = k) & = & \sum_{n=k}^{\infty} P(N_1(A) = k \mid N(A) = n)P(N(A) = n) \\
& =&  e^{-\mu(A)} \sum_{n=k}^{\infty} {n \choose k} p^k(1-p)^{n-k} \frac{\mu(A)^n}{n!} \\
& = & \frac{(p\mu)^k}{k!}e^{-\mu(A)} \sum_{n=k}^{\infty} \frac{((1-p)\mu(A))^{n-k}}{(n-k)!} \\
& = &  \frac{(p\mu(A))^k}{k!}e^{-\mu(A) + (1-p)\mu(A)} =  e^{-p\mu(A)}\frac{(p\mu(A))^k}{k!}.
\end{array}
$$
This shows that $N_1$ is a Poisson process. Similarly, $N_2$ is a Poisson process (with mean measure $(1-p)\mu$). What is left is to show that $N_1$ and $N_2$ are, in fact, independent. We cut a corner here and say that it is actually sufficient to show that $N_1(A)$ and $N_2(A)$ are independent for arbitrary $A$, and this follows from 
$$
\begin{array}{rcl}
P(N_1(A) = k, N_2(A) = r) & = & P(N_1(A) = k, N(A) = k + r) \\
& = & P(N_1(A) = k \mid  N(A) = k + r) P(N(A) = k + r) \\
& = & e^{-\mu(A)} {k+r \choose k} p^k(1-p)^{r} \frac{\mu(A)^{k+r}}{(k+r)!} \\
& = & e^{-p\mu(A)}\frac{(p\mu(A))^k}{k!} e^{-(1-p)\mu(A)}\frac{((1-p)\mu(A))^r}{r!} \\
& = & P(N_1(A) = k)P(N_2(A) = r). 
\end{array}
$$ 
Summary II: We conclude that i.i.d. random thinning with success probability $p$ of a simple point process, $N$, with intensity $\lambda$ results in two simple point processes, $N_1$ and $N_2$, with intensities $p\lambda$ and $(1-p)\lambda$, respectively, and $N$ is the superposition of $N_1$ and $N_2$. If, moreover, $N$ is a Poisson process then $N_1$ and $N_2$ are independent Poisson processes. 
It is natural to ask if we could thin independently without assuming that the $Z_i$'s are identically distributed and obtain similar results. This is possible, but a little more complicated to formulate, because the distribution of $Z_i$ then has to be linked to the $X_i$ somehow. For instance, $P(Z_i = 1 \mid N) = p(x_i)$ for a given function $p$. It is then possible to show the same result as above but with the intensity $p\lambda$ meaning the function $p(x)\lambda(x)$. We skip the proof. The best general mathematical reference covering spatial point processes is Daley and Vere-Jones. A close second covering statistics and simulation algorithms, in particular, is Møller and Waagepetersen.
