Proving that $X^2- [X]$ is a local martingale given that $X$ is a cadlag locally square-integrable martingale

Question

Suppose that $X$ is a cadlag locally square-integrable martingale. Let $[X]$ denote the quadratic variation of $X$. My textbook claims, by Ito's formula that

$$ X^2 _t = X^2_0 + [X]_t + 2 \int_0^t X_{s^{-}} \,d X_s, $$

which shows that $X^2 - [X]$ is indeed a local martingale. Let $ \Delta Y_t = Y_t - Y_{t^{-}}.$ But the version of Ito's formula for cadlag semimartingales tells me that

$$ X^2 _t = X^2_0 + [X]_t + 2 \int_0^t X_{s^{-}} \,dX_s + \sum_{s \leq t } \bigg[ \Delta (X^2_s) - 2 (X_{s^{-}}) \Delta X_s - (\Delta X_s)^2 \bigg] . $$

I am quite unfamiliar with this, as my course so far has just focused on continuous processes. Any ideas?

Answer 1

Since

$$X_s^2 = (X_{s-}+\Delta X_s)^2 = X_{s-}^2+ 2 X_{s-} \Delta X_s + (\Delta X_s)^2$$

we have

$$\Delta (X_s^2) \stackrel{\text{def}}{=} X_s^2-X_{s-}^2 = 2 X_{s-} \Delta X_s + (\Delta X_s)^2.$$

Hence,

$$\Delta (X_s^2) - 2 X_{s-} \Delta X_s - (\Delta X_s)^2 = 0$$

for all $s$. This means that the sum

$$\sum_{s \leq t} \left( \Delta (X_s^2) - 2 X_{s-} \Delta X_s - (\Delta X_s)^2\right)$$

equals $0$.