Let me refer to the example here. Suppose $X$ is a birth death process (represents population size) that evolves by:

$X \to X+1$ if a birth occurs with rate $\mu$,

$X \to X-1$ if a death occurs with rate $\theta$.

Suppose $T_n$ is first passage time of a BD process from state $n$ to state $0$. Then: $$T_n = T + \frac{\mu}{\mu + \theta}T_{n+1} + \frac{\theta}{\mu + \theta}T_{n-1}$$

where $T$ is a random variable for the time needed to get out of state $n$, and $T ～ exp(\mu + \theta)$.

Now I would like to take the Laplace transform of both sides. The accepted answer is as follows:

$$ \phi_n(s) = \frac{\mu + \theta}{\mu + \theta + s} \left( \frac{\mu}{\mu + \theta} \, \phi_{n+1}(s) + \frac{\theta}{\mu + \theta} \, \phi_{n-1}(s) \right)$$

My question is: If $T_{n},T,T_{n-1},T_{n+1}$ are all random variables, why does the Laplace transform for the sum $\frac{\mu}{\mu + \theta}T_{n+1} + \frac{\theta}{\mu + \theta}T_{n-1}$ apply the linear rule, while it is not the case with that sum and $T$?

Let me refer to the example here. Suppose $X$ is a birth-death (BD) process (represents population size) that evolves by:

$X \to X+1$ if a birth occurs with rate $\mu$,

$X \to X-1$ if a death occurs with rate $\theta$.

Suppose $T_n$ is the first passage time of a BD process from state $n$ to state $0$. Then: $$T_n = T + \frac{\mu}{\mu + \theta}T_{n+1} + \frac{\theta}{\mu + \theta}T_{n-1}$$

where $T$ is a random variable for the time needed to get out of state $n$, and $T ～ \exp(\mu + \theta)$.

Now I would like to take the Laplace transform of both sides. The accepted answer is as follows:

$$ \phi_n(s) = \frac{\mu + \theta}{\mu + \theta + s} \left( \frac{\mu}{\mu + \theta} \, \phi_{n+1}(s) + \frac{\theta}{\mu + \theta} \, \phi_{n-1}(s) \right)$$

My question is: If $T_{n},T,T_{n-1},T_{n+1}$ are all random variables, why does the Laplace transform for the sum $\frac{\mu}{\mu + \theta}T_{n+1} + \frac{\theta}{\mu + \theta}T_{n-1}$ apply the linear rule, while it is not the case with that sum and $T$?