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Problem

Let $B$ be a Brownian motion. Compute the quadratic covariation of $B$ and $B^2$, i.e. $\langle B,B^2\rangle_t$ for every $t\geq 0$.

My attempt

Integration by parts yields: \begin{align*} &B_tB_t^2 - B_0B_0^2 = \int_0^t B_s\mathrm{d}B_s^2 + \int_0^t B_s^2 \mathrm{d}B_s + \langle B,B^2\rangle_t\\ \Leftrightarrow\ &\langle B,B^2\rangle_t = B_t^3 - \int_0^t B_s\mathrm{d}B_s^2 - \int_0^t B_s^2 \mathrm{d}B_s \end{align*}

I am not sure about this, but I think that I can simply plug in $\mathrm{d}B_s^2 = 2B_s\mathrm{d}B_s$. This would yield: \begin{align*} \Leftrightarrow\ \langle B,B^2\rangle_t &= B_t^3 - \int_0^t 2B_s^2\mathrm{d}B_s - \int_0^t B_s^2 \mathrm{d}B_s\\ &= B_t^3 - 3\int_0^t B_s^2\mathrm{d}B_s \end{align*}

Apparently, this expression can be simplified further by using Itô's lemma on $f(t,x)=\frac{1}{3}x^3$ and noting that the Brownian motion $B$ is an Itô process. Keep in mind that $B_0=0$ and $\langle B\rangle_s = s$ as well as $\frac{\partial f}{\partial t}(t,x)=0$, $\frac{\partial f}{\partial x}(t,x)=x^2$ and $\frac{\partial^2 f}{\partial^2 x}(t,x)=2x$ \begin{align*} &f(t,B_t) - f(0,B_0) = \int_{0}^{t} \frac{\partial f}{\partial s}(s,B_s)\mathrm{d}s + \int_{0}^{t} \frac{\partial f}{\partial x}(s,B_s)\mathrm{d}B_s+\frac{1}{2} \int_0^t \frac{\partial^2 f}{\partial^2 x} (s,B_s)\mathrm{d}\langle B\rangle_s\\ \Leftrightarrow\ &\frac{1}{3}B_t^3 - 0 = 0 + \int_0^t B_s^2\mathrm{d}B_s + \frac{1}{2}\int_0^t 2B_s\mathrm{d}s\\ \Leftrightarrow\ &\int_0^t B_s^2\mathrm{d}B_s = \frac{1}{3}B_t^3 - \int_0^t B_s\mathrm{d}s \end{align*} Putting everything together yields: $$\langle B,B^2\rangle_t = 3\int_0^t B_s\mathrm{d}s$$ I suppose this the simplest result.

Questions

  1. Was plugging in $\mathrm{d}B_s^2 = 2B_s\mathrm{d}B_s$ allowed?
  2. Is $\langle B,B^2\rangle_t = 3\int_0^t B_s\mathrm{d}s$ correct and maximally simplified?

Edit: Answer (thanks to minginator)

  1. Plugging in $\mathrm{d}B_s^2 = 2B_s\mathrm{d}B_s$ was not allowed and it led to a wrong result.
  2. $\langle B,B^2\rangle_t = 2\int_0^t B_s\mathrm{d}s$ is correct and maximally simplified.

Edit: Proof of $\langle B,B^2\rangle_t = 2\int_0^t B_s\mathrm{d}s$

$B$ is an Itô process of the form:

\begin{align*} B_t = B_0 + \int_{0}^{t}0\mathrm{d}s + \int_{0}^{t}1\mathrm{d}B_s \end{align*}

Define $f\in \mathcal{C}^{1,2}(\mathbb{R}_+\times\mathbb{R})$ by $f(t,x):=x^2$. We then have:

\begin{align*} &\frac{\partial f}{\partial t}(t,x) = 0 &&\frac{\partial f}{\partial x}(t,x) = 2x &&\frac{\partial^2 f}{\partial^2 x}(t,x) = 2 \end{align*}

Applying Itô's formula to the Itô process $B$ yields:

\begin{align*} B_t^2 =\ &\int_{0}^{t}0\mathrm{d}s+\int_{0}^{t} 2B_s\mathrm{d}B_s+\frac{1}{2} \int_0^t 2\mathrm{d}s\\ =\ &B_0^2 + \int_{0}^{t}1\mathrm{d}s + \int_{0}^{t}2B_s\mathrm{d}B_s \end{align*}

The quadratic covariance is therefore: \begin{align*} \langle B,B^2\rangle_t = \int_{0}^{t} \left(1 \cdot 2B_s\right)\mathrm{d}s = 2\int_{0}^{t} B_s\mathrm{d}s \end{align*}

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    $\begingroup$ After a quick look: I don't think so, since by Ito's $d(B^2) = 2B dB +dt\neq 2B dB$ $\endgroup$ Apr 7 at 19:25

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See first this discussion on how to obtain the quadratic variation of two Itô processes. By Itô's formula we obtain that $B_{t}^{2}$ follows the dynamics $$ B_{t}^{2} = \int_{0}^{t}ds + \int_{0}^{t}2B_{s}dB_{s}. $$ This is sometimes written in differential notation as $$ dB_{t}^{2} = dt + 2B_{t}dB_{t}, $$ and thus the answer to question 1 is no (as pointed out already by Nap D. Lover). Then by the linked discussion, you have $$ \langle B_{t}, B_{t}^{2}\rangle = \int_{0}^{t} 2B_{s}ds. $$ If you want to know more about this integral above, see the first answer here. So yes, you could further write it as $$ 2\int_{0}^{t}(t-s)dB_{s}, $$ but this is not necessarily more simplified in my opinion than the first result.

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    $\begingroup$ One advantage of writing it as $2 \int_0^t (t-s)dB_s$ is that it makes it very easy to compute the variance. $\endgroup$ Apr 7 at 21:25

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