Let's say we have a 3-state Markov chain $X_t \in \{S_0,S_1,S_2\}$ with $X_0 = S_0$ and with $S_1$ and $S_2$ being absorbing states. Let $T_1 \sim Exp(\lambda_1)$ and $T_2 \sim Exp(\lambda_2)$ be independent exponentially distributed RVs representing times for moving from $S_0$ to $S_1$ and from $S_0$ to $S_2$ respectively. Then we have
$$
\mathbb{P}(X_t = S_0) = \exp(-(\lambda_1 + \lambda_2)t) = \exp(-\lambda_1 t)\exp(-\lambda_2 t) = \mathbb{P}(T_1 >t)\mathbb{P}(T_2 >t)
\\
\mathbb{P}(X_t = S_1) = \frac{\lambda_1}{\lambda_1 + \lambda_2} – \frac{\lambda_1}{\lambda_1 + \lambda_2}\exp(-(\lambda_1 + \lambda_2)t)
\\
\mathbb{P}(X_t = S_2) = \frac{\lambda_2}{\lambda_1 + \lambda_2} – \frac{\lambda_2}{\lambda_1 + \lambda_2}\exp(-(\lambda_1 + \lambda_2)t)
\tag{1}\label{1}
$$
But on the other hand we also have, if I understand it correctly,
$$
\mathbb{P}(X_t = S_1) = \mathbb{P}(T_1 \le t)
\\
\mathbb{P}(X_t = S_2) = \mathbb{P}(T_2 \le t)
\\
\mathbb{P}(X_t = S_1 \cap X_t = S_2) = \mathbb{P}(T_1 \le t \cap T_2 \le t) = \mathbb{P}(T_1 \le t)\mathbb{P}(T_2 \le t)
\tag{2}\label{2}
$$
but the last expression should be zero as the states are mutually exclusive.
So, we have mutually exclusive states, but they are related to independent exponential RVs. How do we reconcile the two: mutual exclusivity and independence?
Add 1
Following @Ian's answer, it looks like instead in $\eqref{2}$ we have
$$
\mathbb{P}(X_t = S_1) = \mathbb{P}(T_1 \le t \cap T_1 < T_2)
= \int_0^t \lambda_1 \exp(-\lambda_1 t_1) \int_{t_1}^{\infty} \lambda_2 \exp(-\lambda_2 t_2) \mathrm{d}t_2 \mathrm{d}t_1
= \int_0^t \lambda_1 \exp(-\lambda_1 t_1)\exp(-\lambda_2 t_1)\mathrm{d}t_1
$$
which does indeed agree with the $\eqref{1}$
Add 2
If we observe the process at time $t_0 > 0$ and find that $X(t_0) = S_0$ and we want to calculate the probability that the process will be in state $S_1$ at some time $t_1 > t_0$, do we normalize the above by the "remaining" probability of being in state $S_0$ at $t_0$? That is, does the expression become
$$
\mathbb{P}(X(t_1) = S_1 | X(t_0) = S_0) = \frac{\mathbb{P}(t_0 < T_1 \le t_1 \cap T_1 < T_2)}{\mathbb{P}(t_0 < T_1 \cap t_0 < T_2)} \; \; ?
$$
And is the denominator given by
$$
\mathbb{P}(t_0 < T_1 \cap t_0 < T_2) = \int_{t_0}^{\infty} \lambda_1 \exp(-\lambda_1 t_1) \mathrm{d}t_1 \int_{t_0}^{\infty} \lambda_2 \exp(-\lambda_2 t_2) \mathrm{d}t_2 = \mathbb{P}(t_0 < T_1)\mathbb{P}(t_0 < T_2) \; \; ?
$$
If yes, then again it seems like there is a conflict: if the times $T_1$ and $T_2$ are independent, then how can we guarantee that the states are mutually exclusive? Can't both $T_1$ and $T_2$ turn out to be the same? Or does the fact that the times are continuous means that $P(T_1 =T_2) = 0$? What if we measured time in, say, days – could we still model transitions with discrete distributions of time (say, both $T_1$ and $T_2$ follow geometric distributions) and how would we handle $P(T_1 =T_2)$ in this case?
Best Answer
Your misunderstanding is that the process stops running once $t$ reaches $\min \{ T_1,T_2 \}$, and so $P(X_t=S_1 \wedge X_t=S_2)=0$ as you would expect. The actual situation is $P(X_t=S_1)=P(T_1<T_2 \wedge \min \{ T_1,T_2 \} \leq t)$ and analogous for $S_2$.