Proposition. Let(Q, F, P, F be a filtered probability space and let X be a (P, E)-Markov process. Let(Qt) be its probability transition functions and let ZEC(Q, o(X++1), P). Then, for each t=0, 1, E[ZIXi=/f(y)Q++1(Xt, dy) or, put differently, we have for each t=0, 1, .. and each a E[ZIX =a]=/f(y)Q++1(a, dy) (16) Proof We will show it first for an indicator variable Z=Ix-1(A) where AE B(R).Then f(y)=IA(y). We now need to show that the random variable/f(y)Qt+1(Xt, dy) qualifies as the conditional expectation E [ZXt. Clearly it is a(Xt)-measurable But does it integrate to the right thing? Well, let G E o(Xt) and recall that, by definition,Qu+1(X,A)=E((alo(XD).Hence f (y)Q+1(Xt, dy)P(dw) IA(y)Q++1(Xt, dy)P(d /-(x,4P)=/E(xa(x)P()=/x=aP() Meanwhile. since z (A) we obviously have zP(dw)=Ix (18)Proposition. Let (Ω, F, P, F) be a filtered probability space and let X be a (P, F)-Markov process. Let hQti be its probability transition functions and let Z ∈ L1 (Ω, σ (Xt+1), P). Then, for each t = 0, 1, ... E [Z|Xt ] = Z R f (y) Qt+1 (Xt , dy) (15) or, put differently, we have for each t = 0, 1, ... and each x, E [Z|Xt = x] = Z R f (y) Qt+1 (x, dy). (16) Proof. We will show it first for an indicator variable Z = IX −1 t+1(A) where A ∈ B (R). Then f (y) = IA (y). We now need to show that the random variable Z R f (y) Qt+1 (Xt , dy) qualifies as the conditional expectation E [Z|Xt ] . Clearly it is σ (Xt) −measurable. But does it integrate to the right thing? Well, let G ∈ σ (Xt) and recall that, by definition, Qt+1 (Xt , A) = E ³ IX −1 t+1(A) |σ (Xt) ´ . Hence Z G Z R f (y) Qt+1 (Xt , dy) P (dω) = Z G Z R IA (y) Qt+1 (Xt , dy) P (dω) = = Z G Qt+1 (Xt , A) P (dω) = Z G E ³ IX −1 t+1(A) |σ (Xt) ´ P (dω) = Z G IX −1 t+1(A)P (dω). (17) Meanwhile, since Z = IX −1 t+1(A) we obviously have Z G ZP (dω) = Z G IX −1 t+1(A)P (dω) . (18) 14