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?

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    $\begingroup$ It is not correct that $(1/(2\lambda_1)^2)/(1/\lambda_1^2)=1/2$. Nor would the variance be $1/\lambda^2$; rather it is $1/\lambda$. For a Poisson distribution, the variance is equal to the expected value. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 27 '14 at 16:13
  • $\begingroup$ What @MichaelHardy said. Plus, I think I know what a Poisson process is, but I cannot make anything of the question... (Upvotes as mystifying as ever.) $\endgroup$ – Did Aug 27 '14 at 16:20
  • $\begingroup$ You don't get $2N$ samples; you get two samples, each of size $N$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 27 '14 at 16:23
  • $\begingroup$ Well I have a system, where two Poisson processes increase a variable. In between the variable smoothly decreases. The question is, whether I have to have two simulations and two variables (which I sum at the end) or if I can just somehow sum both processes and run just one simulation. I tried but I get different variance of the "number of arrivals" for separate and joint simulation. $\endgroup$ – Morty Aug 27 '14 at 16:29
  • $\begingroup$ Ok, now I see that the problem is that I computed variance of SUM of two processes, which is really twice bigger, while the second simulation actually computed variance of MEAN of two processes which is suprisingly twice lower (I do not see why but they say it here: web.williams.edu/Mathematics/sjmiller/public_html/BrownClasses/… ). $\endgroup$ – Morty Aug 27 '14 at 17:02

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