I have difficulty checking the Markov property of some stochastic process.

Let $W(t)$ be a standard Wiener process adapted to $\mathscr{F}_s$ which is a filtration generated by this Wiener process. I want to check whether $X_t = \cos(W_t)$ and $Y_t = W_t+W_{t-2}$ is Markovian.

The definition says, if a random process is Markovian, for any Borel-measurable function $f$ there exists another Borel-measurable function $g$ such that $E[f(X_t)|\mathscr{F}_s] = g(X_s)$.

\begin{align*} E[f(X_t)|\mathscr{F}_s] &= E[f(\cos(W_t))|\mathscr{F}_s]\\ &= E[f(\cos(W_t-W_s+W_s))|\mathscr{F}_s]\\ \end{align*} Let $h(w) = E[f(\cos(W_t-W_s+w))|\mathscr{F}_s]$, then: \begin{align*} h(w) &= \int_{-\infty}^{\infty} f(\cos(y+w))\frac{1}{\sqrt{2\pi (t-s)}}e^{-\frac{y^2}{2(t-s)}} \mathrm{d}y\\ &= \int_{-\infty}^{\infty} f(\cos(y))\frac{1}{\sqrt{2\pi (t-s)}}e^{-\frac{(y-w)^2}{2(t-s)}} \mathrm{d}y \end{align*} So $g(x)=h(\arccos(x))$? It seems strange since $W_s$ is not necessarily between $-1$ and $1$. I intuitively think it is markovian, due to the periodicity of $cos$. How to overcome this problem?

And for $Y_t$, I can only start with \begin{align*} E[f(W_t+W_{t-2})|\mathscr{F}_s] &= E[f(W_t+W_{t-2}-(W_s + W_{s-2})+(W_s + W_{s-2}))|\mathscr{F}_s]\\ &= E[f((W_t-W_s)+(W_{t-2}- W_{s-2})+(W_s + W_{s-2}))|\mathscr{F}_s] \end{align*} I guess it is not Markovian since $W_{t-2}- W_{s-2}$ is not necessarily independent from $\mathscr{F}_s$, then I cannot apply the same approach in $X_t$. But I am not sure whether I am right and how to get the result rigorously.

Thank you for any help!

  • $\begingroup$ @Shalop Should it be that it cannot be written as $g(W_{t-1}+W_{t-3})$? Though it seems to me it suffices to prove it cannot be written as $g(W_{t-1})$. Sorry I still cannot see the contradiction. Is it just we cannot write $W_{t-2} = h(W_{t-1})$? $\endgroup$ – Edward Wang Jan 31 '18 at 4:47
  • $\begingroup$ It seems you err similarly in both cases. For example, to show $X$ is Markov, for every $s<t$ and every suitable function $f$, one is supposed to exhibit some function $g$ such that $$E(f(X_t)\mid\mathcal F^X_s)=g(X_s)$$ where $\mathcal F^X_s=\sigma(X_u;u\leqslant s)$, not such that $$E(f(X_t)\mid\mathcal F_s)=g(X_s)$$ where $\mathcal F_s=\sigma(W_u;u\leqslant s)$. Since $\mathcal F^X_s\ne\mathcal F_s$, these are not equivalent. $\endgroup$ – Did Feb 5 '18 at 19:59
  • $\begingroup$ @Did I understand what you mean, yes they are different. I just read from wikipedia, and the definition on that page does not specify what the filtration should be. Only $X_t \in \mathbb{F}_t$ is required. $\endgroup$ – Edward Wang Feb 5 '18 at 21:14
  • $\begingroup$ No, the definition of a Markov process does not let you the choice of the filtration. $\endgroup$ – Did Mar 7 '18 at 0:54

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