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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.

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    $\begingroup$ 1. the definition of $X$ is incorrect - it seems that there should be $dX_t$. Could you give a link for this example? $$ $$ 2. and 3. - I think you are right and your calculations are correct (in 2. I assume that $\mathcal F_t$ is a natural filtration of $y$ or $X$ in your notation). In 3. you have a random walk which is a Markov process, and a martingale if its jump has a zero mean. $\endgroup$
    – SBF
    Feb 26, 2012 at 22:22
  • $\begingroup$ @Ilya, Thanks for the editing. The link is here wilmott.com/messageview.cfm?catid=8&threadid=11322 $\endgroup$
    – user16859
    Feb 26, 2012 at 22:36
  • $\begingroup$ Here is another example I mentioned: mathforum.org/kb/… I need a valid counterexample. Thank you! $\endgroup$
    – user16859
    Feb 26, 2012 at 22:39

1 Answer 1

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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 $(X_{t-1},X_{t-2})$.

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  • $\begingroup$ Thanks for the reply. Could you please show more details? I don't understand how to show $E(Z_tX_{t-2}|X_{t-1})=0$. I did use the definition of Markov property: $E(X_{t}|\mathcal{F}_{t-1})=E(X_{t}|\sigma(X_{t-1}))$. Following your own link of wiki, you can see this under "Alternative Formulations". $\endgroup$
    – user16859
    Feb 26, 2012 at 23:44
  • $\begingroup$ No, you forgot for all $f$ bounded and measurable. So I was right, this is really the point where there is a deep misconception in your view of Markovianity. $\endgroup$
    – Did
    Feb 27, 2012 at 0:05
  • $\begingroup$ Re your question about $E(Z_tX_{t-2}\mid X_{t-1})$, a general result you might wish to use is that if $U$ is independent on $(V,W)$, then $E(UV\mid W)=E(U)E(V\mid W)$. $\endgroup$
    – Did
    Feb 27, 2012 at 0:10
  • $\begingroup$ Thanks. Yes, that property is true for sure. However, $E(Z_tX_{t−2}∣X_{t−1})=Z_tE(X_{t−2}∣X_{t−1})=0$ ? I guess, since your assumption is about the $Z_t$ being symmetric Bernoulli, you would in the end use the expectation of $Z_t$ to get the zero. $\endgroup$
    – user16859
    Feb 27, 2012 at 1:01
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    $\begingroup$ By the way, rereading your comments, I wonder whether you would know how to prove the property you declared true for sure. In my opinion, to prove it in details would be highly enlightening. $\endgroup$
    – Did
    Mar 2, 2012 at 12:48

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