I would like to understand a probabilistic result, but I would like to understand what is the fundamental theorems(Law of Total Probability, Bayes'theorem etc) underlying the procedure.
First, I have a two scenarios for each time $t:\,\,$ $S_t = 1,2$ and a history $\Omega_t = \{y_t, y_{t-1}, ....\}$.
Transition between regimes is governed by a Markov chain: \begin{equation}\label{MCH1} \mbox{Pr} (s_t = j \, | \, s_{t-1} = i, s_{t-2} = k, \cdots, \underbrace{y_{t-1}, y_{t-2}, \cdots }_{\Omega_{t-1}}) = p_{ij}, \,\,\, i, j = 1,2 \end{equation}
My goal is to understand the procedure of the two formulas. The first seems to be more easy:
1- $\operatorname{Pr}\left(s_{0}=1 \mid \Omega_{0}\right)=\frac{p_{21}}{p_{21}+p_{12}}, \,\, t = 0$
2- $\operatorname{Pr}\left(s_{t}=j \mid \Omega_{t-1}\right)=p_{1 j} \operatorname{Pr}\left(s_{t-1}=1 \mid \Omega_{t-1}\right)+p_{2 j} \operatorname{Pr}\left(s_{t-1}=2 \mid \Omega_{t-1}\right)$
Relating the first, I tryed the classic conditional probabily formula: $$ \operatorname{Pr}\left(s_{0}=1 \mid \Omega_{0}\right)=\frac{ \mbox{Pr}([s_0 = 1] \cap [\Omega_0]) }{ \mbox{Pr}(\Omega_0) } \underbrace{=}_{?} \frac{p_{21}}{p_{21}+p_{12}} $$
About the second formula, I have
\begin{align} \operatorname{Pr}\left(s_{t}=j \mid \Omega_{t-1}\right)&=p_{1 j} \operatorname{Pr}\left(s_{t-1}=1 \mid \Omega_{t-1}\right)+p_{2 j} \operatorname{Pr}\left(s_{t-1}=2 \mid \Omega_{t-1}\right)\\ &= \operatorname{Pr}\left(s_{t}=j \mid s_{t-1}= 1, \Omega_{t-1}\right) \operatorname{Pr}\left(s_{t-1}=1 \mid \Omega_{t-1}\right) + \operatorname{Pr}\left(s_{t}=j \mid s_{t-1}= 2, \Omega_{t-1}\right) \operatorname{Pr}\left(s_{t-1}=2 \mid \Omega_{t-1}\right) \end{align}
But I can't finish the demo in a rigorous way.
Some explicative deduction?
UPDATE
I need to add an information:
$$y_{t} = m_{s_t} + \varepsilon_{t}, \quad \varepsilon_{t} \sim N(0, \sigma^2)$$
where $m_{s_t} = m_1 > 0 $ if $s_t = 1$ and $m_{s_t} = m_2 < 0 $ if $s_t = 2$