# Ito process and martingale.

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.

• $X_t = X_0 + \int_0^t K_s ds + \int_0^t H_s dW_s$ 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. – B11b Jun 18 at 11:40
• 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 – TheBridge Jun 18 at 14:29
• Perhaps "local" is missing before the ''martingale". Otherwise, the process may even fail to be integrable. – zhoraster Jun 18 at 15:16
• @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. – user6247850 Jun 18 at 18:29
• @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). – 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.