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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?

saz
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Richard
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  • Which definition of "quadratic variation" do you use? For jump processes there are (at least) two different notions, see this question http://math.stackexchange.com/q/902886/ – saz Feb 03 '16 at 17:45
  • @saz The one with u.c.p. convergence of sequence of processes with dyadic partitions – Richard Feb 03 '16 at 17:50

1 Answers1

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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 $0$.

saz
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