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I know that the superposition of two Poisson process with rates $\lambda_1$ and $\lambda_2$ is again a Poisson process with rate $\lambda_1+\lambda_2$. Thus this process has interarrival times distributed as an exponential with parameter $\lambda_1+\lambda_2$.

What I am actually doing is numerically sampling the exponential distribution: I collect a bunch of random times sampled from $\lambda e^{-\lambda t}$. The point is that I do it for different values of $\lambda$ and put all together in the same file. Then I plot the distribution. Let's assume the simple case in which I sample an equal number of times for each value of $\lambda$ and that I make $\lambda$ vary say from $10^{-4}$ to $10^3$.

To my intuition the resulting interevent time distribution should be \begin{equation} \int_{\lambda_{min}}^{\lambda_{max}} P(\lambda)\lambda e^{-\lambda t} d\lambda. \end{equation} Given the above hypothesis of uniformly distributed $\lambda$ values I just have $P(\lambda) = const$ and the integral is then \begin{equation} \frac{1}{t}\left(\lambda_{min} e^{-\lambda_{min} t} - \lambda_{max} e^{-\lambda_{max} t}\right) + \frac{1}{t^2}\left(e^{-\lambda_{min} t} - e^{-\lambda_{max} t}\right). \end{equation}

Looking (and plotting) the above function, I expect two different regimes, one in which the interevent time distribution decays as $t^{-1}$ (large times) and one in which decays as $t^{-2}$ (small times). I don't find the second regime in the data.

Question: is it my approach that is wrong or the integral I am doing should actually give me the observed interevent time distribution? If it's correct, why the data don't show the small time regime?

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