$X$ is a stochastic process, $\mathcal{F}_t=\sigma(X_s: s\leq t), \mathcal{G}_t=\sigma(X_s: s\geq t)$, prove that the following statements are equivalent:
- $X$ has Markov property.
- $\mathbb{P}(A\cap B|X_t)=\mathbb{P}(A|X_t)\mathbb{P}(B|X_t)$ for $A\in \mathcal{G}_t, B\in \mathcal{F}_t$.
- $\mathbb{P}(B|\mathcal{G}_t)=\mathbb{P}(B|X_t)$ for $B\in \mathcal{F}_t$.
I can prove $1\Rightarrow 2$: Because of Markov property, we have $\mathbb{P}(A|\mathcal{F}_t)=\mathbb{P}(A|X_t)$ for $A\in \mathcal{G}_t$, then
\begin{split}
\mathbb{P}(A\cap B|X_t)&=\mathbb{E}(\mathbb{P}(A\cap B|\mathcal{F}_t)|X_t)=\mathbb{E}(1_B\mathbb{P}(A|\mathcal{F}_t)|X_t)\\
&=\mathbb{E}(1_B\mathbb{P}(A|X_t)|X_t)=\mathbb{P}(A|X_t)\mathbb{P}(B|X_t).
\end{split}
But how to prove $2\Rightarrow 3$? I want to take the expectation to both sides in 2, then
\begin{equation}
\mathbb{E}(\mathbb{P}(A\cap B|X_t))=\mathbb{P}(A\cap B)=\mathbb{E}(\mathbb{P}(A|X_t)\mathbb{P}(B|X_t)).
\end{equation}
If $\mathbb{P}(A\cap B)=\mathbb{E}(\mathbb{P}(B|X_t)1_A), A\in \mathcal{G}_t$, then 3 holds. But is there $\mathbb{P}(A|X_t)=1_A$ in the expectation? Or another way, take the condition expectation with respect to $\mathcal{G}_t$:
\begin{equation}
\mathbb{E}(\mathbb{P}(A\cap B|X_t)|\mathcal{G}_t)=\mathbb{E}(1_A\mathbb{P}(B|\mathcal{G}_t)|X_t)=\mathbb{E}(\mathbb{P}(A|X_t)\mathbb{P}(B|X_t)|\mathcal{G}_t)=\mathbb{P}(A|X_t)\mathbb{P}(B|X_t).
\end{equation}
and how to get 3 from $\mathbb{E}(1_A\mathbb{P}(B|\mathcal{G}_t)|X_t)=\mathbb{P}(A|X_t)\mathbb{P}(B|X_t)$?
And how to prove $3\Rightarrow 1$? Thank you very much!
Best Answer
Before we start recall the measure theoretic Lemma that for any ${\cal H}$-measurable nonnegative random variable $Y$ there exists an increasing sequence of simple random variables $$ Z_n=\sum_{i=1}^nh^n_i\,1_{H^n_i}\,,\quad h^n_i\in[0,+\infty)\,,\quad H^n_i\in{\cal H} $$ such that $Z_n\uparrow Y\,.$ See for example here.
Recall further that the conditional expectation $\mathbb E[U\,|\,{\cal H}]$ is defined as the (a.s. unique) ${\cal H}$-measurable random variable that satisfies $$ \mathbb E\Big[\mathbb E[U\,|\,{\cal H}]\,1_H\Big]=\mathbb E[U1_H]\,,\quad \forall H\in{\cal H}\,. $$ From the measure theoretic Lemma it follows by monotone convergence that $$\tag{*} \mathbb E\Big[\mathbb E[U\,|\,{\cal H}]\,Y\Big]=\mathbb E[UY]\,,\quad \forall Y\;{\cal H}\text{-measurable nonnegative}\,. $$
I will use this a couple of times.
You showed correctly 1$\Rightarrow$2.
I now show 2$\Rightarrow$3: Let $A\in{\cal G}_t\,.$ Since $\mathbb P(B\,|\,X_t)$ is $\sigma(X_t)$-measurable we can use it as $Y$ in (*) and have (with $U=1_A$ and ${\cal H}=\sigma(X_t)$) $$ \mathbb E\big[\underbrace{\mathbb P(B\,|\,X_t)}1_A\big]= \mathbb E\Big[\mathbb P(B\,|\,X_t)\,\mathbb P(A\,|\,X_t)\Big] \stackrel{2}{=}\mathbb P(A\cap B)\,. $$ Since $\mathbb P(B\,|\,{\cal G}_t)$ is defined as the unique (up to a.s.) ${\cal G}_t$-measurable random variable that satisfies $$ \mathbb P(A\cap B)=\mathbb E\big[\underbrace{\mathbb P(B\,|\,{\cal G}_t)}1_A\big]\,\quad\forall A\in{\cal G}_t $$ we have therefore shown that $\mathbb P(B\,|\,{\cal G}_t)$ must almost surely equal $P(B\,|\,X_t)$ as the latter is clearly ${\cal G}_t$-measurable.
The proof of 2$\Rightarrow$1 is similar, in fact, symmetric: Let $B\in{\cal F}_t\,.$ Since $\mathbb P(A\,|\,X_t)$ is $\sigma(X_t)$-measurable we have $$ \mathbb E\big[\mathbb P(A\,|\,X_t)1_B\big]= \mathbb E\Big[\mathbb P(B\,|\,X_t)\,\mathbb P(A\,|\,X_t)\Big] \stackrel{2}{=}\mathbb P(A\cap B)\,. $$ Like in the previous case this shows $\mathbb P(A\,|\,{\cal F}_t)$ must almost surely equal $P(A\,|\,X_t)$ which is the Markov property.
It remains to show 3$\Rightarrow$2. This is similar (symmetric) to 1$\Rightarrow$2: \begin{align} \mathbb P(A\cap B\,|\,X_t)&=\mathbb E\Big[\mathbb P(A\cap B\,|\,{\cal G}_t)\,\Big|\,X_t\Big]= \mathbb E\Big[1_A\,\mathbb P(B\,|\,{\cal G}_t)\,\Big|\,X_t\Big]\\ &\stackrel{3}=\mathbb E\Big[1_A\,\mathbb P(B\,|\,X_t)\,\Big|\,X_t\Big]\\ &=\mathbb P(A\,|\,X_t)\,\mathbb P(B\,|\,X_t)\,. \end{align}