How to calculate the inter arrival time of Poisson process having mean as binomial random variable?

Question

I know that to simulate normal Poisson process we have to take exponential inter arrival time, i.e., for simulating Poisson process $N2$ times we have:

for ii=1:N2
n = 0;
t = 0.0;
%lambda_me is the mean of the Poisson process%
while ( t < T )
dt = - log ( rand ( 1, 1 ) ) / lambda_me;
n = n + 1;%it is counting the number of arrivals%
t = t + dt;
end
final_n_1(ii)=n;
end

Now I have a Poisson process whose mean is a binomial random variable taking two values like $$\Lambda= \begin{cases} a & \text{with probability } p, \\ b & \text{with probability } 1-p. \end{cases}$$ What will be its inter arrival time and how should I simulate it?

Answer 1

Suppose you are 'merging' two Poisson processes one with rate $a$ and one with rate $b$ in this way. This might be a model for two radioactive sources of different intensities were both in range of a counter.

Then you have $X \sim \mathsf{Pois}(rate=ap)$ and $Y \sim \mathsf{Pois}(rate=b(1-p)),$ then $T = X + Y \sim \mathsf{Pois}(\lambda),$ where $\lambda = ap + b(1-p).$ Then the interarrival times are $\mathsf{Exp}(rate = \lambda)$ and average interarrival times are $1/\lambda.$

I confess I can't quite visualize your intended process looking at your code. Here is a brief simulation in R, that indicates $T$ based on $\lambda=16$ and $p = .4$ behaves as described.

m = 10^6; lam = 16; p = .4
x = rpois(m, lam*p);  y = rpois(m, lam*(1-p))
t = x + y;  mean(t);  sd(t)
## 15.99342    # aprx E(T) = 16
## 4.001156    # aprx SD(T) = 4.

The histogram below is the simulated distribution of $T.$ The red dots are exact probabilities from $\mathsf{Pois}(\lambda =16).$