Deriving the Doob Meyer decomposition of a Sub Martingale using Ito's

Question

Given the standard brownian motion $(B_t)_{t\in\mathbf{R}_{+}}$ and defining the sub-m.g.:

$$X_t =B^6_t+2t$$

I would like to derive its Doob-Meyer decomposition: [Sub-m.g.]= [increasing process]+[m.g.]

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Sadly I keep applying Ito's formula in the wrong way and end up with wrong answers. Can you please help me find the right Ito's formula to obtain the following result:

$dX_t= 6B_t^5 dB_t+ \frac{1}{2}(6)(5)B_t^4dt+2t$

$X_t =\int_0^t(15B_s^4+2)ds+ \int_0^t(6B_s^5)dB_s $

Thank you.

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I've been using:

$f(B_t,t)=\frac{\partial f}{ \partial X_t} dB_t + \frac{\partial f}{ \partial t} dt - \frac {1}{2} \frac{\partial^2 f}{ \partial X_t^2} d <X>_t $

And I can't get the same results. I understand that the Ito's formula would need a function but I'm off track trying to identify the one needed in this case.

Answer 1

Ito's Lemma: For suitable stochastic process $X_t(t, B_t)$,

$$ dX_t{}={}\left(\dfrac{\partial}{\partial t}X_t{}+{}\mu_b\dfrac{\partial}{\partial b}X_t{}+{}\dfrac{1}{2}\sigma_b^2\dfrac{\partial^2}{{\partial b}^2}X_t\right)dt{}+{}\sigma_b\dfrac{\partial}{\partial b}X_tdB_t\,. $$ where $\mu_b{}={}0$ and $\sigma_b{}={}1$ for brownian motion, $B_t$ .

From the equation of $X_t$ given in the question, compute each of the non-zero terms as follows:

(i) $\,\,\,\,\,\,\, \dfrac{\partial}{\partial t}X_t{}={}2\,;$

(ii) $\,\,\,\,\,\,\, \dfrac{1}{2}\sigma_b^2\dfrac{\partial^2}{{\partial b}^2}X_t{}={}\dfrac{1}{2}6\cdot5\,B_t^4{}={}15B_t^4\,;$

(iii) $\,\,\,\,\,\,\, \sigma_b\dfrac{\partial}{\partial b}X_t{}={}6B_t^5$.

Direct substitution of these into the lemma gives the result:

$$ dX_t{}={}\left(15B_t^4+2\right)dt{}+{}6B_t^5 dB_t\,. $$