expression for conditional expectation of Markov processes Suppose we have two Markov processes $P_1(t)$ and $P_2(t)$ and a function $f(t):=f(P_1(t),P_2(t))$. 
What is a compact way to express that the expected change of $f(t)$ between a reference time $t_0$ and a stopping time $\tau$ 
is negative if the state at time $t_0$ is known and $P_1(t_0) > 0$?
The option I consider is
$$ {E}[f(\tau)-f(t_0)|P_1(t_0)>0, P_2(t_0)] < 0 $$
My question is whether $P_2(t_0)$ is necessary and why. 
It seems that the result is independent of its value.
Also as written expression ${E}[f(\tau)-f(t_0)|P_1(t_0)>0, P_2(t_0)]$ is a random variable, but I think it expresses what I want to claim, as for any evaluation of the variables, subject to $P_1(t_0) > 0$, the expression will be negative. Is it really semantically equivalent to what I want to say though?
 A: For state spaces $S_1, S_2$ and all $(i_1,i_2) \in S_1 \times S_2$ you can define: 
$$ D_{t_0,\tau}(i_1,i_2) := E[f(\tau)-f(t_0)|(P_1(t_0),P_2(t_0))=(i_1,i_2)]$$
Then, I think you are trying to express the drift condition: 
$$ D_{t_0,\tau}(i_1, i_2) < 0 \quad \forall (i_1,i_2) \in S_1 \times S_2 \mbox{ such that $i_1>0$} \quad (Eq. 1)$$
Notice that this conditions on the full $(i_1,i_2)$ state at time $t_0$ and hence leaves no ambiguity. 
In particular, notice that (Eq. 1) does not require knowledge of what happened in the system before time $t_0$.  Notice that (Eq. 1) is not the same as:
$$ E[f(\tau)-f(t_0)|P_1(t_0)>0, P_2(t_0)]<0 \quad (Eq. 2)$$
That is, (Eq. 2) does not imply (Eq. 1).  The (Eq. 2) is awkward because it depends on the history of the system before time $t_0$ (that history affects the conditional distribution of  $P_1(t_0)$ given $P_1(t_0)>0, P_2(t_0)$).  It makes me wonder under what distributional assumptions on the history before time $t_0$ is the equation (2) supposed to hold?  Equation (1) is much stronger (it implies equation (2)).  If your system has the strong property of equation (1), then I encourage you to say it.
