In a Poisson process measured with some efficiency, is the measured count still Poisson? Situation:
Say I have a Poisson process, like radioactive decay, producing R particles per second. I measure with a detector. There is a probability P that a particle will be detected by the detector.
Things I think I know:


*

*The inter-arrival time of the particle emission is exponentially
distributed with parameters based on R. 

*The number of particles emitted before detection is given by a
negative binomial based on P.

*If a number N is sampled from (2), a single sample of inter-arrival time for detected particles can be given by the sum of N samples from (1). This sum can be obtained by sampling from a gamma distribution with parameters based on N and R.


My question:
If a single inter-arrival time can be calculated by sampling from a gamma based on N and R, how does the number of detector counts in an interval end up being Poisson again? (To be Poisson, the inter-arrival time for the detector must be exponential, not distributed according to some weird gamma thing.) Of course N is fluctuating, but I can't see how this works out.
However, I am almost completely sure detector counts are in fact Poisson distributed. Could somebody show me the math?
Thanks for the help!
EDIT:
I found this paper:
Fried, D. L. "Noise in photoemission current." Applied Optics 4.1 (1965): 79-80.
Which shows the result that a binomially selected poisson random variable is also Poisson with a rate given by PR. This confirms the comment by jbowman. Still, I would be interested in seeing the explanation of how my process of generating the inter-arrival time at the detector using the negative binomial and gamma distribution is incorrect. This is my major mental hiccup. Thank you.
EDIT 2: 
I wrote this matlab script to test whether what I was trying with the gamma distribution worked. Turns out that somehow the gamma inter-arrival times generated with a geometrically distributed N are exponential and agree with the inter-arrival times suggested by Poisson(PR). (ia2 and ia3 are identically distributed). Any idea how this works out analytically? It was not intuitively obvious to me!
close all
n = 100000;
ia1 = exprnd(1,n,1); % create exponentially distributed inter-arrival times
t1 = cumsum(ia1); % running sum (the real experiment time)

mask = (rand(n,1) > 0.5); % flip a coin
t2 = t1(mask); % get only the events for which "the coin landed on heads"
ia2 = diff(t2); % calculate the inter-arrival times at the detector.

% plot the distributions
figure; hist(ia1,100); title('exponential inter-arrival times');
figure; hist(ia2,100); title('binomial sampled inter-arrival times');

%%
spacing = geornd(0.5,n,1) + 1; % how many events before we get heads
ia3 = gamrnd(spacing,ones(n,1)); % generate the interarrival times with gamma
figure; hist(ia3,100); title('geom/gamma inter-arrival times');

 A: A quick non-technical argument might use Jackson networks. In your case total external arrivals is rate $R$, and there are no internal transitions (observed particles don't switch to the unobserved queue). The splitting proportion between the observed and unobserved nodes $p_{0i}$ is $P$, so the 
$\lambda_{obs}=RP$
If you're looking for first principles, call $O(t)$ the observed counting process, and $N(t)\sim PP(r)$ the total counting process. Where each arrival in $N(t)$ gets logged in $O(t)$ with probability $p$. So that if for some $s$ we have $N(s)=n$ then $O(s)$ has a binomial($n,p$) distribution.
This approach uses probability generating functions:
$E[z^{O(t)}|N(t)=n]=\sum_{j=0}^{n}z^{j} {n \choose j}p^j(1-p)^{n-j}=(1-p+pz)^{n}$
Last equality by the binomial theorem. Then, unconditionally, since $N(t)\sim Poisson(rt)$:
$E[z^{O(t)}]=E[E[z^{O(t)}|N(t)=n]]=\sum_{n=0}^{\infty}(1-p+pz)^{n}\frac{rt^n}{n!}e^{-rt}=e^{-rt}e^{rt(1-p+pz)}=e^{rpt(z-1)}$
Which is the probability generating function of a Poisson($rpt$) random variable.
