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I've a question concerning the superposition of renewal processes. Assume we have $n$ independent renewal processes with the same lifetime distribution (especially mean $\mu$ and variance $\sigma^2$). Now regard the superposition of these processes. This is not necessarly a renewal process. I know that for the mean $\mu_s$ of the lifetimes of this process it holds $\mu_s=\mu/n$. Such an easy relationship does not hold for the variance $\sigma_s^2$ of the lifetimes. Does anybody know how I can compute this variance?

Thanks a lot!

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Let $g$ be the density of the interarrival times in the superimposed process, and $F$ the interrenewal distribution of a component renewal process. Let $U$ be the limiting backward recurrence time of the superimposed process, and $U_i$ the same for the $i^{\mathrm{th}}$ component renewal process. Then for $x>0$, $$\mathbb P(U>x)=\prod_{i=1}^n \mathbb P(U_i>x) = \left(\frac1\mu\int_x^\infty \bar F(u)\ \mathsf d u\right)^n, $$ where $\bar F=1-F$ is the survival function of the component interrenewal time. Differentiating we obtain the density of $U$, $$ n\frac{\bar F(x)}\mu\left(\frac1\mu\int_x^\infty \bar F(u)\ \mathsf d u\right)^{n-1}.$$ Now since the mean interarrival time is $\frac\mu n$, we have $$\frac{\bar G(x)}{\mu/n} = p\frac{\bar F(x)}\mu\left(\frac1\mu\int_x^\infty\bar F(u)\ \mathsf d u\right)^{n-1}, $$ where $\bar G$ is the survival function corresponding to $g$, that is $$\bar G(x) = \int_x^\infty g(u)\ \mathsf du.$$ It follows that $$g(x) = -\frac{\mathsf d}{\mathsf d x}\left[\bar F(x)\left(\frac1\mu\int_x^\infty\bar F(u)\ \mathsf d u\right)^{n-1}\right]. $$ Therefore the variance of the lifetimes would be $$\int_0^\infty x^2 g(x)\ \mathsf dx -\left(\frac\mu n\right)^2. $$

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  • $\begingroup$ Can we then conclude that the superimposed process is renewal or stationary? $\endgroup$ – user52227 Dec 9 '19 at 14:11
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    $\begingroup$ As for the superimposed process being renewal, see here: math.stackexchange.com/questions/1536488/… As for the superimposed process being stationary, I do not know. But I would not expect this to be the case. $\endgroup$ – Math1000 Dec 9 '19 at 18:03

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