I am enrolled in some very bad probability/stochastic processes course. Relations hold approximatively, wrong assumptions, no proofs, nothing rigorous and it is a bit frustrating to understand new concepts. It is said that a process $(X_n)$ with respect to the filtration $F_n$ is a Markov process if $$P(X_{n+1} \in B | F_n)=P(X_{n+1} \in B | X_n) \quad \forall B \in \mathcal B(\Bbb R)$$ $$\iff E[g(X_{n+1})| F_n]=E[g(X_{n+1}) | X_n ] \quad \forall \text{ bounded } g : \Bbb R \to \Bbb R$$
Alright, no problem. But then exercises are like "Show that $Z_t = Y_1X_1 + Y_2X_2 + ... + Y_{t−1}X_{t−1} + Y_tX_t$ is a Markov process ($X_t, Y_t$ i.i.d.)". But with respect to which filtration ? I guess the definition holds if I pick $F_n= \sigma(X_n)$ its natural filtration...
The solution says : "Rewrite $Z_t = Z_{t−1} + X_tY_t$. The distribution of $Z_t$ only depends on $Z_{t−1}$ since $X_t$ and $Y_t$ are both independent from $t$ and $Z_t$". I really hate this kind of informal maths. The answer has nothing to do with the definition. Can someone help me understand how to deal with such an exercise ? How do I identify a Markov process $X_t$ with informal yet legal arguments as they did ? Do I just need to check that $X_t = g(X_{t-1})$ ? The teacher sometimes talks about information at time $t$, but nobody knows how to manipulate this intuition stuff and turn it into valid proofs.
My first impulse would be to show that $(Z_n)$ is a Markov process with respect to the filtration $\mathcal F_n:=\sigma\{X_1,Y_1,X_2,Y_2,\ldots,X_n,Y_n\}$. Clearly $Z_n$ is $\mathcal F_n$ measurable for each $n$, by induction.
To check the Markov property you need to show that $$ E[g(Z_{n+1})\mid\mathcal F_n] $$ depends only on $Z_n$. As you noted, $Z_{n+1}=Z_n+X_{n+1}Y_{n+1}$; by hypothesis the pair $(X_{n+1},Y_{n+1})$ is independent of $\mathcal F_n$. Therefore $$ E[g(Z_{n+1})\mid\mathcal F_n]=E[g(Z_{n}+X_{n+1}Y_{n+1})\mid\mathcal F_n]=h(Z_n), $$ where $h(z):=E[g(z+X_{n+1}Y_{n+1})]$. (This because of known properties of conditional expectation.) The function $h$ doesn't depend on $n$ becasue the $(X_t,Y_t)$ are iid.
Using the tower property of conditional expectations, you can now show that $(Z_n)$ is a Markov process with respect to its natural filtration $\mathcal G_n:=\sigma\{Z_1,Z_2,\ldots,Z_n\}\subset\mathcal F_n$.
(As a general propostion: If a process is Markov with respect to some filtration $(\mathcal K_n)$ and is adapted to a filtration $(\mathcal H_n)$ with $\mathcal H_n\subset\mathcal K_n$ for all $n$, then it is Markov with respect to $(\mathcal H_n)$. This follows readily from the tower property.)