Suppose $X$ is an Ito process with $K=0$. Prove or disprove $X_t^2 - \langle X\rangle_t$ is a martingale.

My attempt is:

$X_t = X_0 + \int_0^t K_s ds + \int_0^t H_s dW_s$ this is Ito process. And so,

If X is Ito process, with K=0, then $X_t = X_0 + \int^t_0 H_s dW_s$ or $dX_t = H_t dW_t$

And also $X_t^2 - \langle X\rangle_t =X_t^2 - dX_t dX_t = X_t^2 - H_t^2 d t$

Well I o order to be martingale, I need to show that $E[ X_t^2 - H_t^2 d t | F_s] = X_s^2 - H_s^2 d s$ for $s\le t $.

But I cannot show this martingale part. Please help me to do this part. Thanks a lot.

  • $\begingroup$ $X_t = X_0 + \int_0^t K_s ds + \int_0^t H_s dW_s$ this is Ito process that I used. And when $K_s=0$, I get the equation that I wrote in the question part. @TheBridge I hope I can be clear. $\endgroup$ – B11b Jun 18 at 11:40
  • $\begingroup$ You should try to take a look at $dX_t^2$ (using Ito's lemma) and check the drift term of d$Y_t$ where $Y_t = X^2_t-<X>_t$. Regards $\endgroup$ – TheBridge Jun 18 at 14:29
  • $\begingroup$ Perhaps "local" is missing before the ''martingale". Otherwise, the process may even fail to be integrable. $\endgroup$ – zhoraster Jun 18 at 15:16
  • $\begingroup$ @zhoraster $X$ in this form is generally called an Ito process only when $\mathbb{E}[\int_0^t (H_s^2 + |K_s|)ds] < \infty$ for each $t$, so this doesn't always need to be specified separately. $\endgroup$ – user6247850 Jun 18 at 18:29
  • 1
    $\begingroup$ @user6247850, I work with Ito processes for over 20 years, and I do not find this integrability assumption usual. What in fact usually assumed is that $\int_0^t (H_s^2 + |K_s|)ds < \infty$ almost surely (which is sufficient for the process to be well-defined). $\endgroup$ – zhoraster Jun 19 at 0:45

This is essentially the definition of $\langle X,X\rangle_t$, but I guess it can also be verified using Ito's formula.

We apply Ito's formula to the function $f(x)=x^2$ and compute

\begin{align*} d(X_t^2) &= 2X_t dX_t + dX_t dX_t = 2 X_tH_tdW_t + H_t^2dt \end{align*}

so by integrating we have $X_t^2 - \int_0^t H_s^2ds = X_0^2 + 2\int_0^t X_s dX_s$. Since stochastic integrals are martingales, this implies $X_t^2 - \int_0^t H_s^2ds = X_t^2 - \langle X,X \rangle_t$ is a martingale.

| cite | improve this answer | |
  • 1
    $\begingroup$ The answer as stated is wrong as you can only say that the resulting process is a local martingale without any additionnal asumpotions, as spotted by zhoroaster $\endgroup$ – TheBridge Jun 19 at 13:22
  • $\begingroup$ Many thanks for your Great answer! $\endgroup$ – B11b Jun 19 at 14:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.