Krdytkyu

Suppose that $X, (Omega,mathcal{F}),{P^x}_{x in mathbb{R}^d}$ is a Markov family with shift operators ${theta_s}_{s ge 0}$ and for every $x in mathbb{R}^d,s ge 0, G in mathcal{F}_s$ and $F in mathcal{F}^X_infty$ we have
begin{equation}label{c''}
P^x[G cap theta_s^{-1} F mid X_s]=P^x[G mid X_s]P^x[theta_s^{-1}F mid X_s]
end{equation}
Then show that the above implies
$P^x[theta_s^{-1} F mid mathcal{F}_s]=P^x[theta_s^{-1} F mid X_s] text{ , } P^x text{-a.s.}$
This is part of exercise 2.5.17 in Karatzas and Shreve's Brownian motion and stochastic Calculus.
I could manage to show the converse, i.e $P^x[theta_s^{-1} F mid mathcal{F}_s]=P^x[theta_s^{-1} F mid X_s] text{ , } P^x text{-a.s.}$ implies $$P^x[G cap theta_s^{-1} F mid X_s]=P^x[G mid X_s]P^x[theta_s^{-1}F mid X_s]$$ as follows:

Using properties of conditional expectation we get
begin{equation*}
begin{split}
P^x[G cap theta_s^{-1} F mid X_s]=E^x[ mathrm{1}_{ {G cap theta_s^{-1} F }} mid X_s]=E^x[ mathrm{1}_{ G} mathrm{1}_{ theta_s^{-1} F } mid X_s]\=E^x[ E^x[ mathrm{1}_{ G} mathrm{1}_{ theta_s^{-1} F } mid mathcal{F}_s] mid X_s]=E^x[ mathrm{1}_{ G} E^x[ mathrm{1}_{ theta_s^{-1} F } mid mathcal{F}_s] mid X_s]\= E^x[ mathrm{1}_{ G} E^x[ mathrm{1}_{ theta_s^{-1} F } mid X_s] mid X_s]= E^x[ mathrm{1}_{ theta_s^{-1} F } mid X_s] E^x[ mathrm{1}_{ G} mid X_s]\=P^x[G mid X_s]P^x[theta_s^{-1}F mid X_s]
end{split}
end{equation*}

I tried to prove the other direction similarly but it seems that the same technique doesnt workk. Can you give me a hint on how could I go about proving it? I do not want a complete answer as it would defeat the purpose of the exercise.

Recall the following characterization of the conditional expectation:

Let $X in L^1(mathbb{P})$ and let $mathcal{F}$ be a $sigma$-algebra. An $mathcal{F}$-measurable random variable $Y in L^1(mathbb{P})$ equals almost surely $mathbb{E}(X mid mathcal{F})$ if, and only if, $$forall G in mathcal{F}: quad int_G Y , dmathbb{P} = int_G X , dmathbb{P}.$$

If we apply this result with $mathcal{F} := mathcal{F}_s$, $mathbb{P}=P^x$, $$X := 1_{theta_s^{-1}F} qquad Y := P^x(theta_s^{-1} F mid X_s)$$ we find that $P^x(theta_s^{-1} F mid mathcal{F}_s) = P^x(theta_s^{-1} F mid X_s)$ iff $$forall Gin mathcal{F}_s: quad int_G P^x(theta_s^{-1} F mid X_s) , dP^x = int_G 1_{theta_s^{-1} F} , dP^x.$$
Use the Markov property to verify this condition. Hint: Start with the right-hand side,

$$int_G 1_{theta_s^{-1} F} , dP^x= E^x left(1_G 1_{theta_s^{-1} F}right) = E^x left( E^x big(1_G 1_{theta_s^{-1} F} mid X_s big) right) = dots$$

Solution:

begin{align*}int_G 1_{theta_s^{-1} F} , dP^x= E^x left(1_G 1_{theta_s^{-1} F}right) &stackrel{text{tower}}{=} E^x left( E^x big(1_G 1_{theta_s^{-1} F} mid X_s big) right) \ &stackrel{text{Markov}}{=} E^x left( E^x big(1_G mid X_s big) E^x(1_{theta_s^{-1} F} mid X_s) right) \ &stackrel{text{pull out}}{=} E^x left( E^x big(1_G E^x(1_{theta_s^{-1} F} mid X_s) mid X_s big) right) \ &stackrel{text{tower}}{=} E^x left( 1_G E^x(1_{theta_s^{-1} F} mid X_s) right) \ &= int_G E^x(1_{theta_s^{-1} F} mid X_s) , dP^x end{align*}