Let's say we have a 3-state CTMC represented by $$ \require{enclose} \begin{array}{ccc} & & \Large{\enclose{circle}{1}} \\\ & \lambda_1 \Large{\nearrow} & \\\ \Large{\enclose{circle}{0}} & & \Large{\downarrow} \lambda_2 \\\ & \lambda_2 \large{\searrow} & & \\\ & & \Large{\enclose{circle}{2}} \end{array} $$
where the time to go from $0$ to $1$ is distributed as $T_1 \sim Exp(\lambda_1)$, and the time to go from $0$ to $2$ and from $1$ to $2$ is $T_2 \sim Exp(\lambda_2)$.
The system starts at $t_0=0$ in state $0$, i.e. $P(X_0=0) = 1$. I am looking to find the probability that the state $2$ is reached before state $1$ within some time interval $\Delta t$ given that $X_t = 0$, i.e.
$$ P(t < T_2 \le t+\Delta t \ \cap \ T_2 < T_1 |X_t = 0) \equiv p_{0,2}(t,t+\Delta t) $$
I think this can be evaluated as
$$ p_{0,2}(t,t+\Delta t) \ = \int_t^{t + \Delta t}\lambda_2\exp(-\lambda_2 t_2) \int_{t_2}^{\infty}\lambda_1\exp(-\lambda_1 t_1) dt_1 dt_2 \\ = \int_t^{t + \Delta t}\lambda_2\exp(-\lambda_2 t_2) \exp(-\lambda_1 t_2)dt_2 \\ = \frac{\lambda_2}{\lambda_1 + \lambda_2} \left[\exp(-(\lambda_1 + \lambda_2)t) - \exp(-(\lambda_1 + \lambda_2)(t + \Delta t)) \right] $$
Is this correct or do I need to normalize by something since we are NOT at $t=0$ but $X(t) = 0$? Since we have two processes, should the normalizing factor be $P(T_1 > t \cap T_2 > t)$ to give
$$ P(T_2 \le t+\Delta t \ \cap \ T_2 < T_1 | X_t = 0) = \frac{P(t < T_2 \le t+\Delta t \ \cap \ T_2 < T_1)}{P(T_1 > t \cap T_2 > t)} ? $$
If this is the correct normalization to be applied, then I think we have
$$ P(T_1 > t \cap T_2 > t) = \exp(-\lambda_1 t)\exp(-\lambda_2 t) $$
which gives
$$ P(T_2 \le t+\Delta t \ \cap \ T_2 < T_1 | X_t = 0) = \frac{\lambda_2}{\lambda_1 + \lambda_2} \left[1 - \exp(-(\lambda_1 + \lambda_2)\Delta t) \right] . $$
On the other hand, this process, I believe, can be expressed using an infinitesimal generator $\mathbf{Q}$ as
$$ \mathbf{Q} = \pmatrix{-(\lambda_1 + \lambda_2) &\lambda_1 &\lambda_2\\ 0 &-\lambda_2 &\lambda_2\\ 0 &0 &0} $$
The eigenvalues of this matrix are $\{-(\lambda_1 + \lambda_2), -\lambda_2, 0\}$ and the corresponding eigenvectors are
$$ \mathbf{V} = \pmatrix{1 &1 &1\\ 0 &1 &1\\ 0 &0 &1} $$
so the solution of the forward equation, if I am not mistaken, is
$$ \mathbf{P}(t) = \exp(\mathbf{Q}t) \ = \mathbf{V} \exp\pmatrix{-(\lambda_1 + \lambda_2)t &0 &0\\ 0 &-\lambda_2 t &0\\ 0 &0 &0} \mathbf{V}^{-1} \\ = \mathbf{V} \pmatrix{\exp(-(\lambda_1 + \lambda_2)t) &0 &0\\ 0 &\exp(-\lambda_2 t) &0\\ 0 &0 &1} \mathbf{V}^{-1} \\ = \pmatrix{1 &1 &1\\ 0 &1 &1\\ 0 &0 &1} \pmatrix{\exp(-(\lambda_1 + \lambda_2)t) &0 &0\\ 0 &\exp(-\lambda_2 t) &0\\ 0 &0 &1} \pmatrix{1 &-1 &0\\ 0 &1 &-1\\ 0 &0 &1} \\ =\pmatrix{\exp(-(\lambda_1 + \lambda_2)t) &\exp(-\lambda_2 t) - \exp(-(\lambda_1 + \lambda_2)t) &1-\exp(-\lambda_2 t)\\ 0 &\exp(-\lambda_2 t) &1-\exp(-\lambda_2 t)\\ 0 &0 &1} $$
However it's not clear to me how to extract $p_{0,2}(t,t+\Delta t)$ from this $\mathbf{P}(t)$?
The results for $(\mathbf{P}(t))_{0,2} = 1-\exp(-\lambda_2 t)$ also feels strange to me since it seems to ignore the possibility of getting from $0$ to $2$ via $1$ and so I would have expected there to be $\lambda_1$ terms in it - what is going on here?
I suspect that the issue might be with the way I expressed $\mathbf{Q}$ since I only have two competing exponentials, not 3. Or do I? If the mechanism governing transition from $0$ to $2$ and from $1$ to $2$ is the same, do we consider it as one exponential RV or two for the purpose of representing an CTMC?