# Markov chains derivation for absorbing states

I am given that the probability to reach a specific absorbing state $$s$$, from states 1, ..., M as $$a_1, \cdots, a_M$$, which are unique solutions to equations $$a_s = 1$$, $$a_0 = 0$$ for all absorbing states such that $$i \neq s$$, and $$a_i = \sum_i^M a_j p_{ij}$$ for all transient states $$i$$.

Can someone show me how this summation is derived. It looks like the law of total probability, by conditioning on the next states reachable from $$i$$. But it looks like a different kind of law of total probability (LOTP) that what I'm used to seeing.

Usually when I see LOTP, it's something like this: $$P(A) = \sum_k P(A, B_k) = \sum_k P(A | B_k) P(B_k).$$

For the Markov chain probability $$a_i$$, say $$i=1, M=2$$, we have

$$a_1 = p_{11}a_1 + p_{12}a2 \\$$

Note that $$p_{11} = P(X_{n+1} = 1 | X_{n} = 1) \\ p_{12} = P(X_{n+1} = 2 | X_{n} = 1) \\$$

Is this an application of LOTP?

A minor issue first, but one which may have caused you confusion: the lower index in your first sum should be 1, not $$i$$; and in the "LOTP" equation, the notation $$P(A\cap B_k)$$ would also be clearer than $$P(A, B_k)$$.
Yes, the result does use the LOTP. The key observation that allows to connect the generic LOTP equation and your problem at hand is that the index $$i$$ is "built into" the probability $$P$$ from the LOTP (so that $$P$$ may itself be regarded as a conditional probability). That is, if $$B_k$$ is the probability of going to state $$k$$ in one step, then $$P(B_k)$$ is actually the probability of going from $$i$$ to $$k$$ in one step, thus corresponding to $$p_{ik}$$. That the second factor in the equation that you wish to prove is just $$a_j$$ is due to the Markov property.