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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.]


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.


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.

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  • $\begingroup$ The representation you obtained $X_t = \int_0^t \left(2 + 15 B_s^4\right) \mathrm{d}s + 6 \int_0^t B_s^5 \mathrm{d}B_s$ is correct. $\endgroup$
    – Sasha
    Commented Jan 22, 2015 at 21:44
  • $\begingroup$ Thank you, that's good to know but I am also looking for the right Ito's formula to use to obtain that. Can you provide the version of the Ito's formula in detail please? $\endgroup$ Commented Jan 22, 2015 at 21:47

1 Answer 1

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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:

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

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  • $\begingroup$ Thank you Ki3i, used this version of the formula a lot, and passed the class thanks to your answer. $\endgroup$ Commented Feb 5, 2015 at 11:32
  • $\begingroup$ @ClementeCortile, :) Well done. +1 for your success in applying what you learned here. Glad to have been of some help. $\endgroup$
    – ki3i
    Commented Feb 5, 2015 at 13:43

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