Assume we have a process $X$ with two states: $x_1$ and $x_0$. The process itself is a Markov Process, so its behaviour depends only on the current state, and thus, it can be described by a transition matrix containing the instantaneous probabilities of transition between states. This matrix is usually written as:
$$Q_X = \begin{pmatrix}-q_{x_0} & q_{x_0x_1}\\ q_{x_1x_0} & -q_{x_1}\end{pmatrix},$$
where the elements $q_{x_ix_j}$ stand for the probability of transition from state $x_i$ to state $x_j$ provided a transition from the $x_i$ state, and the elements in the diagonal describe the probability on remaining in a state, and are computed such that each row sums up to zero.
I know that, in theory, the probability distribution of remaining in a certain state is an exponential. This can be proved by taking the instant probability of transition from $x_i$ to $x_j$, $p_{x_ix_j}$ which is the probability of transition in $\delta t$ as $\delta t \rightarrow 0$. In an interval of time $[0,T]$ and with $X(t=0)=x_i$, the probability of remaining in the same state up to $t=T$ can be expressed as an infinite product of the probability of not transitioning in infinitely many smaller intervals:
$$P(\textrm{not transitioning from }X(0)=x_i\textrm{ to }X(T)=x_i) = \lim_{n \rightarrow \infty}\prod_{i=0}^n(1-p_{x_ix_j}\frac{T}{n})=\lim_{n \rightarrow \infty}(1-p_{x_ix_j}\frac{T}{n})^n.$$
The last limit is the definition of the exponential, so, finally:
$$P(\textrm{not transitioning from }X(0)=x_i\textrm{ to }X(T)=x_i)=e^{-p{x_ix_j}T}.$$
So far so good, but I can't figure out why the following reasoning is wrong: Instead of taking the instantaneousprobability of transitioning, let's try to get the same result from the instantaneous probability of remaining in the same state, $p_{x_i}$, which should be $p_{x_i} = 1-p_{x_ix_j}$ (as there are only two states in this scenario). Clearly we can follow the same strategy again:
$$P(\textrm{not transitioning from }X(0)=x_i\textrm{ to }X(T)=x_i) = \lim_{n \rightarrow \infty}\prod_{i=0}^n(p_{x_i}\frac{T}{n})=\lim_{n \rightarrow \infty}(p_{x_i}\frac{T}{n})^n,$$
but this time it is clear that the limit approaches $0$ as $n \rightarrow \infty$, and even if we substitute $p_{x_i} = 1-p_{x_ix_j}$, the limit vanishes as well.
I know there's something phishy about $p_{x_i} = 1-p_{x_ix_j}$, since it would be fixed in the case that this relation only held when multiplying the instantaneous probabilities by $\delta t = \frac{T}{n}$, the infinitesimally small interval of time in which they are applied, but on the other hand this would mean that the instantaneous probabilities don't sum up to one, which bugs me as well.
I will appreciate any information on the subject, even as sources to check. Thank you in advance.