The queue model is such that:
$V \rightarrow V + 1$ Poisson with rate $\lambda dt$ deposition rate
$V \rightarrow V - 1$ Poisson with rate $(\mu + \nu) dt$ execution plus cancellation rates
The goal is to calculate the first hitting time ($T_1$), the time at which the queue volume hits zero, for the first starting at volume of V ($E[T_1|V]$).
$\Phi(\tau, V)$ is the probability that the first hitting time is $\tau$ given you started at V.
$\Phi(\tau, V) = \int_0^{\infty}d \tau_1 \lambda e^{-(\nu+\mu+\lambda)tau_1}\Phi(\tau-\tau_1, V+1) + \int_0^{\infty}d \tau_1 (\nu + \mu) e^{-(\nu+\mu+\lambda)tau_1}\Phi(\tau-\tau_1, V-1)$
The Laplace transform of $\Phi(\tau, V)$ defined to be is:
$\hat{\Phi}(\tau,V) = : \int_0^{\infty} d\tau \Phi(\tau, V) e^{-z\tau}$
Applying that to the first equation eventually gives:
$\hat{\Phi}(z, V) = \frac{\lambda}{\lambda + \mu+\nu+z} \hat{\Phi}(z, V+1) + \frac{\mu + \nu}{\lambda + \mu+\nu+z} \hat{\Phi}(z, V-1)$
By expanding for small z:
$\hat{\Phi}(z, V) = 1 - E[T_1|V] + \frac{z^2}{2}E[T_2^2|V] + O(z^3)$
Applying the expansion to the equation above the expansion gives:
$ -z E[T_1|V] = -z\frac{\lambda}{\lambda+\mu+\nu} E[T_1|V+1] -z\frac{\mu + \nu}{\lambda+\mu+\nu} E[T_1|V-1] -z\frac{1}{\lambda+\mu+\nu} $
I am confused here by two parts:
1.) Why has the denominator changed from $(\lambda +\mu+\nu+z)$ to $(\lambda +\mu+\nu)$?
2.) Where has the last term $-z\frac{1}{\lambda+\mu+\nu}$ come from?
I realise that the ones from the original expansion cancel so I know it's nothing to do with them.
Can someone please explain.