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Let $(\Omega, \mathcal{F},\mathbb{P},\{\mathcal{F}\}_t)$ be a filtered- probability space and $W_t$ be standard Wiener process. I want to show stratonovich integral of $W_t$, i.e $\int_{0}^{t} W_s ○ dW_s$ , is not a martingale by Ito lemma.

Thanks.

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  • $\begingroup$ There is a conversion formula from stratonovich integral to ito integral: en.wikipedia.org/wiki/Stratonovich_integral. $\endgroup$ – Gordon Jul 26 '16 at 13:39
  • $\begingroup$ After you have it converted, you can then use ito's lemma. $\endgroup$ – Gordon Jul 26 '16 at 13:45
  • $\begingroup$ How, I can't prove this formula. I cant see prove of these in this page $\endgroup$ – Crisis2008 Jul 26 '16 at 13:48
  • $\begingroup$ You can not use Ito's lemma directly $\endgroup$ – user16651 Jul 26 '16 at 13:52
  • $\begingroup$ Why? We can use Ito lema $\endgroup$ – Crisis2008 Jul 26 '16 at 13:55
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It is well-known, $W_t$ is a continues, adapted and locally bounded process.Let $I=\{t_i\}_{i=0}^{n}$ is a sequence of partitions of $[0,t]$, Indeed $0=t_0<t_1<\cdots<t_n=t$ .By definition of stratonovich integral, we have $$\int_{0}^{t} W_s\circ dW_s=\frac{1}{2}\underset{n\to \infty }{\mathop{\lim }}\sum_{i=0}^{n-1}(W(t_{i+1})+W(t_i))(W(t_{i+1})-W(t_i))$$ $$\int_{0}^{t} W_s\circ dW_s=\frac{1}{2}\underset{n\to \infty }{\mathop{\lim }}\sum_{i=0}^{n-1}(W^2(t_{i+1})-W^2(t_i))\quad(\operatorname{Telescoping series })$$ $$\int_{0}^{t} W_s\circ dW_s=\frac{1}{2}\underset{n\to \infty }{\mathop{\lim }}(W^2(t_n)-W^2(t_0))$$ Therefore $$\int_{0}^{t} W_s\circ dW_s=\frac{1}{2}(\,W^2(t)-W^2(0)\,)=\frac{1}{2}W_t^2\quad (\underset{n\to \infty }{\mathop{\lim }}W(t_n)=W(t))$$ Hence

$$\int_{0}^{t} W_s\circ dW_s=\frac{1}{2}W_t^2$$

Now use Ito lemma.

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  • $\begingroup$ Note $\int_{0}^{t} W_s\,dW_s=\frac{1}{2}\left(W_t^2-t\right)$ whereas $\int_{0}^{t} W_s\circ dW_s=\frac{1}{2}W_t^2$ .This is the difference between Ito's and Stratonovich's integrals $\endgroup$ – user16651 Jul 26 '16 at 21:39

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