Martingale that is not a Markov process 1. On the internet, it is suggested that 
$$
X_t=\left(\int_0^t X_s \;ds\right)\;dW_{t}
$$
is a martingale, but not a Markov process. I understand that the process 
$$
I_t(C)=\int_0^t C_s \; dW_s$$
is a martingale, but I don't see $X_t$ as an Ito's integral. In order to show it is a martingale but not a Markov process, I want to show that 
$$
E(X_t\mid\mathcal{F}_{t-1})=X_{t-1}\neq E(X_t\mid\sigma(X_{t-1})).
$$
I don't know how to take the expecation in this case.
Other counterexamples available online might be wrong. (Or, I could
be wrong.)

2. Invalid counterexample: Let $e(t)$ be a random variable which is not independent of
$y(0)$ and 
$$
y(t)=y(t-1)+e(t)-E{[}e(t)|F(t-1){]}.
$$ 
Then y(t) is a martingale but it is not a Markov process.
I will show that it is both a martingale and a Markov process. Define $X_{t}$ as 
$$
X_{t}=X_{t-1}+e_{t}-E\left(e_{t}|\mathcal{F}_{t-1}\right)
$$
Then, $\{X_{t}\}$ is a martingale because
$$
E(X_{t}|\mathcal{F}_{t-1})  =  E\left(X_{t-1}+e_{t}-E\left(e_{t}|\mathcal{F}_{t-1}\right)\bigg|\mathcal{F}_{t-1}\right)
  =  E(X_{t-1}|\mathcal{F}_{t-1})+E\left(e_{t}|\mathcal{F}_{t-1}\right)-E\left(e_{t}|\mathcal{F}_{t-1}\right)
 = E(X_{t-1}|\mathcal{F}_{t-1})
  =  X_{t-1}
$$
However, it is also a Markov process.
$$
E(X_{t}|\sigma(X_{t-1})) =  E\left(X_{t-1}+e_{t}-E\left(e_{t}|\mathcal{F}_{t-1}\right)\bigg|\sigma(X_{t-1})\right)
  =  E(X_{t-1}|\sigma(X_{t-1}))+E\left(e_{t}|\sigma(X_{t-1})\right)-E\left(E\left(e_{t}|\mathcal{F}_{t-1}\right)\bigg|\sigma(X_{t-1})\right)
$$
Since $\sigma(X_{t-1})\subset\mathcal{F}_{t-1}$, by the towering property,
$$
E\left(\left(e_{t}|\mathcal{F}_{t-1}\right)\bigg|\sigma(X_{t-1})\right)=E\left(e_{t}|\sigma(X_{t-1})\right).
$$
Thus
$$E(X_{t}|\sigma(X_{t-1}))  =  E(X_{t-1}|\sigma(X_{t-1}))+E\left(e_{t}|\sigma(X_{t-1})\right)-E\left(e_{t}|\sigma(X_{t-1})\right)
  =  E(X_{t-1}|\sigma(X_{t-1})
$$
Since $X_{t-1}$ is $\sigma(X_{t-1})$-measurable, $E(X_{t-1}|\sigma(X_{t-1}))=X_{t-1}$. Thus $$
E(X_{t}|\sigma(X_{t-1}))=X_{t-1}=E(X_{t}|\mathcal{F}_{t-1})
$$
Hence $\{X_{t}\}$ is also a Markov process.

3. Invalid counterexample from the textbook answer key. Alison Etheridge's textbook gives an example: 
$$
S_{n+1}=S_{n}+\xi_{n+1}
$$ 
where $\xi_{n}\in\{-1,1\}$ and $\{\xi_{n}\}_{n\geq0}$ are iid. Then, $$
P\left(S_{n+1}=k+1\bigg|S_{n}=k\right)=p \text{and }P\left(S_{n+1}=k-1\bigg|S_{n}=k\right)=1-p.
$$
The author claims that $\left\{ S_{n}\right\} $ is a Markov process.
Then in the homework answer key, someone gives $\left\{ Z_{n}\right\}$ as a counterexample of a martingale not being a Markov process:
$
\displaystyle{Z_{n}=\sum_{i=1}^{n}\xi_{i}}
$ 
where $P(\xi_{i}=1)=\frac{1}{2}=P(\xi_{i}=-1)$.
Obviously, $Z_{n+1}=Z_{n}+\xi_{n+1}.$ So what's the difference between this counterexample and the example? Moreover, given $P(\xi_{i}=1)=\frac{1}{2}=P(\xi_{i}=-1)$, I think
is both $\left\{ Z_{n}\right\}$ is both a martingale and a Markov process.
 A: There is a pervasive mistake in your post, possibly explaining your trouble, which is to believe that $(X_t)_{t\geqslant0}$ being a Markov process means that $\mathrm E(X_t\mid \mathcal F_{t-1})=\mathrm E(X_t\mid X_{t-1})$ for every $t\geqslant1$, where $\mathcal F_t=\sigma(X_s;0\leqslant s\leqslant t)$ for every $t\geqslant0$. 
This is not the definition of the Markov property. The Markov property is the assumption that, for every $t\geqslant1$, the conditional distribution of $X_t$ conditionally on $\mathcal F_{t-1}$ depends on $X_{t-1}$ only. Obviously, if the conditional distribution of $X_t$ conditionally on $\mathcal F_{t-1}$ is the conditional distribution of $X_t$ conditionally on $X_{t-1}$ (Markov), then the same is true for the conditional expectations (the property you think is Markov), but the converse is not true.
For a counterexample, assume that $(Z_t)_{t\geqslant2}$ is independent, integrable, centered, non constant (say, with a standard normal distribution), and independent on $X_0$. Let $X_1=X_0=1$ and $X_t=X_{t-1}+Z_tX_{t-2}$ for every $t\geqslant2$. 
Then $\mathrm E(X_t\mid \mathcal F_{t-1})=\mathrm E(X_t\mid X_{t-1})=X_{t-1}$ for every $t\geqslant1$ (hence, if $X_0$ is integrable, $(X_t)_{t\geqslant0}$ is a martingale)  but $(X_t)_{t\geqslant0}$ is not a Markov process since the conditional distribution of $X_t$ conditionally on $\mathcal F_{t-1}$ does not depend on $X_{t-1}$ only but on $(X_{t-1},X_{t-2})$.
