Quadratic variation of $X_{t} = tB_{t}$?

Question

Let $X_t = tB_t$ be a process where $B=(B_t)_{t>0}$ is the standard Brownian motion. Evaluate $\langle X\rangle_t$ the quadratic variation of our process .

I tried to calculate it using :

$d(X_tX_t) = 2X_tdX_t + d\langle X\rangle_t$

$\langle X\rangle_t = X_t^2 - \int\limits_0^t X_t dX_t $

and we have :

$X_tdX_t = t^2B_tdB_t + tB_t^2dt$

so we can write the quadratic variation as :

$\langle X\rangle_t = t^2B_t^2 - 2\int\limits_0^t t^2B_tdB_t - 2 \int\limits_0^t tB_t^2dt$

and I don't know if this is the result wanted or I need to calculate it using different methods ?!

Answer 1

Let $X_t := t \cdot B_t$. By Itô's formula (applied to $f(t,x) := t^2 \cdot x^2$),

$$X_t^2-X_0^2 = 2\int_0^t s^2 B_s \, dB_s + \int_0^t (s^2 + 2s B_s^2) \, ds. \tag{1}$$

On the other hand,

$$X_s dX_s = s^2 B_s \, dB_s + s B_s^2 \, ds. \tag{2}$$

If we combine both equalities, we get

$$\langle X \rangle_t := X_t^2 - 2\int_0^t X_s \, dX_s = \int_0^t s^2 \, ds = \frac{t^3}{3}.$$

Answer 2

$$\mathrm dX_t=\color{red}{t}\,\mathrm d\color{blue}{B}_t+B_t\,\mathrm dt\implies\mathrm d\langle X\rangle_t=\color{red}{t}^2\,\mathrm d\langle\color{blue}{B}\rangle_t=t^2\,\mathrm dt\implies\langle X\rangle_t=\frac{t^3}3$$