Say I have a birth-death process, with birth having Poisson distribution with parameter $\lambda$ and death having poisson distribution $\mu$. Assuming that both stochastic processes, birth and death, have time homogeneity, then my textbook is telling me that the expected wait time is exponentially distributed: $\tau ~ \mathcal{E}(\lambda + \mu)$. Now, I need to find the total expected time when in its invariant distribution.

I have two questions about this:

1) since the sum of two poisson processes has a poisson distribution, with parameter given by the sum of the parameters, (see Poisson Distribution of sum of two random independent variables $X$, $Y$) Why is it $\lambda + \mu$ and not $\lambda - \mu$?

2) Obviously, the total epected time is $\frac{1}{\lambda + \mu}$, since it is exponentially distributed...but iss this distribution, and hence the expected value going to be different when in its stationary distribution? It seems like not to me, but then I wonder why it is used in the question.



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