Superposition of renewal processes: Variance of lifetimes 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!
 A: 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
$$\int_0^\infty x^2 g(x)\ \mathsf dx -\left(\frac\mu n\right)^2. $$
