In the context of Markov Processes, suposse I have a measurable space $(Z,\mathcal{Z})$ and a transition function $Q: Z \times \mathcal{Z} \to [0,1]$. We know that given $z_0 \in Z$,
$$Q(z_0,A) = P(\tilde{z}_{t+1} \in A\,\,|\,\,\tilde{z}_t = z_0 )= \int_A Q(z_0,dz')$$
where $\tilde{z}_t$ denotes the random state in period $t$. That is, $Q(z_0,A)$ is the probability that the next period's shock lies in set $A$, given the current shock is $z_0$. With this transition function, I want to look at partial histories of shocks $z^t = (z_1, z_2,...z_t)$. For this, let $z_0 \in Z$ and
$$(Z^t, \mathcal{Z}^t ) = (Z \times... \times Z, \mathcal{Z} \times ... \times \mathcal{Z}), \quad (t\,-t\,imes)$$.
We will define a probabilty measure $\mu^t(z_0, . ): \mathcal{Z}^t \to [0,1]$. For this, We know that it is sufficient to define $\mu^t(z_0, . )$ on the measurable rectangles in $\mathcal{Z}^t$. So for any rectangle $B = A_1 \times... \times A_t \in \mathcal{Z}^t $, let
$$\mu^t(z_0, B ) = \int_{A_1}...\int_{A_{t-1}}\int_{A_t} Q(z_{t-1},dz_t)Q(z_{t-2},dz_{t-1})...Q(z_{0},dz_{1})$$
I would like to understand $\mu^t(z_0, B )$:
- For $t =1$, $A = B_1$, we have $$\mu^1(z_0, B ) = \int_{A_1} Q(z_{0},dz_{1}) = P(\tilde{z}_{1} \in A\,\,|\,\,\tilde{z}_0 = z_0 )$$
- For $t = 2$, $B = A_1 \times A_2$: \begin{equation} \begin{split} \mu^2(z_0, B ) & = \int_{A_1}\int_{A_2} Q(z_{1},dz_2)Q(z_{0},dz_{1}) \\ & = \int_{A_1} P(\tilde{z}_{2} \in A_2\,\,|\,\,\tilde{z}_1 = z_1 ) Q(z_{0},dz_{1}) \\ & = \,\,? \end{split} \end{equation}
I think I should conclude with $P(\tilde{z}_{2} \in A_2 , \tilde{z}_1 \in A_1\,\,|\,\, \tilde{z}_0 = z_0)$. But I cannot get this conclusion directly. Would this conclusion be true? If yes, how can I argue this in a more rigorous way?