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I am currently working on stochastic processes and I have met a stumbling block in the Ito integral


with $\alpha\in\mathbb{R}$ and $\alpha>0$. Textbooks result is given for integer $\alpha$ but not in the more general case that could not exist. Of course, also some good references are welcome.

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This question and its answer by yourself are as absurd, mathematically speaking, as on the other page…. Re the question, I would be curious to see any "textbooks result" "given for integer $\alpha$" (even for $\alpha=1$, the integral as written above does not make sense). – Did Jan 26 '12 at 20:42
Just put the right answer, taking into account the discussion with George. You will see from this that my answer here is obviously correct and does not worth a downvote. Thanks. – Jon Jan 26 '12 at 20:45
You seem to be in a delusional state about the status of the discussion with George. // Unsurprisingly, no answer to my request for (at least) one example of the textbooks mentioned in your question. – Did Jan 26 '12 at 20:49
@Didier: Sorry Didier but I am not in a delusional state. I have just accepted your answer. I agree with you that there is a problem. What else? Just George declared that $(dW)^\alpha=0$ with $\alpha>2$ and this is what I obtain below. So, why downvote? I live mathematics like a pleasure and I may be wrong. It occurred to me sometime in my twenty years long career. I just learn from my errors and go ahead. Of course, you have been helpful and I gave you thanks accepting your answer and deleting wrong statements. Period. – Jon Jan 26 '12 at 21:00
up vote 0 down vote accepted

It can be shown that $[dW(t)]^\alpha=0$ with $\alpha\in\mathbb{R}$ and $\alpha\ge 3$ generalizing the integer case.

Let us consider the stochastic differential equation $dX(t)=[dW(t)]^\alpha$ with $\alpha>0$. We can write the solution in the form $X(t)=X(t_0)+\int_{t_0}^t[dW(t)]^\alpha$ with the integral in the Ito sense. Then, we have to evaluate this integral with the sum \begin{equation} S_n=\sum_{k=1}^n[W(t_k)-W(t_{k-1})]^\alpha. \end{equation} The power of the Brownian process can be evaluated in the following way \begin{equation} [W(t_k)-W(t_{k-1})]^\alpha = [(1+W(t_k)+W(t_{k-1}))-1]^\alpha= \end{equation} \begin{equation} (-1)^\alpha\sum_{l_1=0}^\infty\left(\begin{array}{c} \alpha \\ l_1 \end{array}\right)(-1)^{l_1}(1+W(t_k)+W(t_{k-1}))^{l_1}= \end{equation} \begin{equation} (-1)^\alpha\sum_{l_1=0}^\infty\sum_{l_2=0}^\infty\left(\begin{array}{c} \alpha \\ l_1 \end{array}\right)\left(\begin{array}{c} l_1 \\ l_2 \end{array}\right)(-1)^{l_1} [W(t_k)-W(t_{k-1})]^{l_2} \end{equation} provided $|W(t_k)-W(t_{k-1})|<1$. Now, we can use stochastic calculus to remove powers higher than 2 and it is easy to see that \begin{equation} S_n=(-1)^\alpha\sum_{k=1}^n\sum_{l_1=0}^\infty\left(\begin{array}{c} \alpha \\ l_1 \end{array}\right)(-1)^{l_1}- (-1)^\alpha\sum_{l_1=0}^\infty\left(\begin{array}{c} \alpha \\ l_1 \end{array}\right)l_1(-1)^{l_1}\sum_{k=1}^n[W(t_k)-W(t_{k-1})]+ \end{equation} \begin{equation} (-1)^\alpha\sum_{l_1=0}^\infty\left(\begin{array}{c} \alpha \\ l_1 \end{array}\right)\frac{l_1(l_1-1)}{2}(-1)^{l_1} \sum_{k=1}^n[W(t_k)-W(t_{k-1})]^2. \end{equation} So, we have the required expansion with coefficients \begin{eqnarray} \mu_0&=&\sum_{l_1=0}^\infty\left(\begin{array}{c} \alpha \\ l_1 \end{array}\right)(-1)^{l_1} \nonumber \\ \mu_1&=&\sum_{l_1=0}^\infty\left(\begin{array}{c} \alpha \\ l_1 \end{array}\right)l_1(-1)^{l_1} \nonumber \\ \mu_2&=&\sum_{l_1=0}^\infty\left(\begin{array}{c} \alpha \\ l_1 \end{array}\right)\frac{l_1(l_1-1)}{2}(-1)^{l_1} \end{eqnarray} Now we see immediately that $\mu_0=\left.(1-x)^\alpha\right|_{x=1}=0$. Besides, we get immediately the result that, for any real $\alpha\ge 3$, we have again $[dW(t)]^\alpha=0$ as in this case the coefficients are all zero when $\mu_1$ and $\mu_2$ are evaluated thorugh Abel summation. Finally, when $0<\alpha<1$ both the coefficients $\mu_1$ and $\mu_2$ are divergent and maybe no meaning can be attached to them (I have in mind summable divergent series here, any suggestion is greatly appreciated).

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