I have two Poisson processes with the same rate $\lambda_1$. If I choose to model them using single superposed process, it will have rate $\lambda_2 = 2 \lambda_1$. From here it is obvious that the combined distribution will have $2\!\times$ higher variance: $(2\lambda_1)/(\lambda_1)=2$.
Now I run a simulation. If I pull random values from corresponding exponential distribution for single process until I exhaust the time interval $t$ and count how many arrivals did fit then I get some count of arrivals. I do it $N$ times and compute variance and that matches to the Poisson variance for single process $\lambda_1$.
If I want to simulate both processes I run the simulation twice, I get $2N$ samples and after I compute their variance it did not change (it is $\lambda_1$). Or I can draw arrival times from exponential distribution with parameter $\lambda_2$ and get only $N$ samples. However, then the variance is lower ($\lambda_1/2$).
Why are these two simulations of the composed process different in variance and why is the variance twice smaller when it should be twice bigger?
Is it because the composed process discount for inner variance inside the composed process and I would get a correct result if I always averaged two sample counts from each process simulation?