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I'm currently learning about the Ito's lemma / formula

In my textbook, a direct application of the formula is to compute quantities like that : (W is a Brownian motion)

enter image description here

While trying to prove these results I am finding that these computations are really not direct. Am I missing something and they're actually trivial with Ito's formula ?

For instance for the first one I couldn't directly compute the integral of Wt*dWt, I had to use Ito's formula on the derivative of Wt²

For the second one I managed to get the result but it was also very convoluted. So my question : using Ito's lemma, what's the direct approach to compute these formulas? thanks

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1 Answer 1

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Since by Ito's lemma you have for $\phi(t,x) \in \mathcal C ^{1,2}$

$$ \phi(t, X_t) = \phi(o,X_0) + \int _0 ^t (\partial_t +b\partial_x +\frac{\sigma^2}{2}\partial^2_{xx})\phi(s,X_s) ~ ds + \int_0^t\partial_x\phi(s,X_s) ~dW_s$$ if $X$ is a Ito's diffusion

$$ X_t = X_0 + \int _0 ^t b~ ds + \int_0^t\sigma ~dW_s,\ t \geq 0$$

you must search for the simples $\phi$ such that $\partial_x \phi =x $ with $X =B$ ( so $b=0$ and $\sigma =1$) for the first integral.

Indeed, we have $$ \int_0^T W_s ~dW_s=\frac{1}{2}(W^2_T-W^2_0 -T).$$

For the second formula a direct application of Ito's lemma would give the result. By effect, if you consider $\phi(t,x) = \exp(\lambda x-\frac{\lambda^2}{2}t) $ since $\partial_t\phi(t,x)= -\frac{\lambda^2}{2}\phi(t,x),\ \partial_x\phi(t,x)=\lambda \phi(t,x)$ and $\partial_{xx}^2\phi(t,x)=\lambda^2\phi(t,x)$, we have

\begin{align}\exp(\lambda W_t-\frac{\lambda^2}{2}t)=\phi(t,W_t)&=\phi(0,W_0)+ \int _0 ^t (\partial_t +\frac{1}{2}\partial^2_{xx})\phi(s,W_s) ~ ds + \int_0^t\partial_x\phi(s,W_s) ~dW_s \\&=1+\lambda\int_0^t \exp(\lambda W_s-\frac{\lambda^2}{2}s) ~dW_s\end{align}

Note that $(\phi(t,W_t))_{t \geq0}=(\exp(\lambda W_t-\frac{\lambda^2}{2}t))_{t \geq0}$ satisfy the SDE $$ L_t = 1 +\lambda \int_o^t L_s dW_s ,\ t \geq 0$$ and not $$ L_t = 1 + \int_o^t L_s dW_s ,\ t \geq 0$$ as you said in the question(probably a typo). Also, since $\mathbb E \left\{ \exp(2\lambda W_t-\lambda^2 t)\right\}< +\infty$ it is a (positive) exponential martingale

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  • $\begingroup$ Nice answer, +1 $\endgroup$
    – SBF
    Jun 3, 2013 at 12:01
  • $\begingroup$ @Ilya: Thank you Ilya! $\endgroup$
    – Paul
    Jun 3, 2013 at 13:38

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