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.
Best Answer
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.