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Let $(B_t)_{t \geq 0}$ be a standard Brownian motion and $M_t = B_t^2 - t$.

According to this and this posts we know that \begin{align} [M] = [B^2] = 2 \int_0^t B_s^2\ ds. \end{align} Now, without appealing to the fact that $M^2 - [B^2] = M^2 - [M]$ is a martingale by definition, I want to prove that $M^2 - [B^2]$ is a martingale using Itô's formula and the above result.

Therefore, I apply Itô's formula to $M_t^2 = (B_t^2 -t)^2 = f(t, B_t)$. \begin{align} f(t,B_t) &= f(0,B_0) + \int_0^t f_s'(s,B_s)\ ds + \int_0^t f_{B_s}'(s,B_s)\ dB_s + \frac{1}{2} \int_0^t f_{B_s,B_s}''(s,B_s) ds \end{align} gives that \begin{align} (B_t^2 -t)^2 &= -2 \int_0^t (B_s^2 -s)\ ds+ 4 \int_0^t B_s(B_s^2 -s)\ dB_s + \frac{1}{2} \int_0^t \big(4(B_s^2 - s) + 4B_s \cdot 2B_s \big) ds \\ &= -2 \int_0^t B_s^2\ ds + 2 \int_0^t s\ ds + 4 \int_0^t B_s^3\ dB_s -4 \int_0^t s B_s\ dB_s + 6 \int_0^t B_s^2\ ds -2 \int_0^t s\ ds \\ &= 4 \int_0^t B_s^2\ ds + 4 \int_0^t B_s^3\ dB_s -4 \int_0^t s B_s\ dB_s. \end{align} So, subtracting $2 \int_0^t B_s^2\ ds$ from $(B_t^2 -t)^2$ should give the required result. However, there is a problem with the coefficients. Where did I go wrong?

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    $\begingroup$ Where is the problem? The preceding posts indicate that $[M]_t=4\int_0^tB_s^2ds$ and your computations here confirm exactly that. $\endgroup$
    – Did
    Commented May 14, 2016 at 22:07
  • $\begingroup$ I have found that $[M] = [B^2] = 2 \int_0^t B_s^2\ ds$ instead of $4 \int_0^t B_s^2\ ds$. $\endgroup$
    – iJup
    Commented May 15, 2016 at 9:16
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    $\begingroup$ Yeah, and you should wonder how $dM=2BdB+Ydt$ would yield $d[M,M]=2B^2dt$ instead of $d[M,M]=4B^2dt$... $\endgroup$
    – Did
    Commented May 15, 2016 at 9:36

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