\begin{eqnarray}\label{Bht} B^{H}_{t}=\int^{t}_{0}(t-s)^{H-1/2}dW_{s}\,, \end{eqnarray} where $W_{s}$ is a Brownian motion.

Then, we can obtain \begin{eqnarray}\label{dBht} dB^{H}_{t}=(H-\frac{1}{2})\int^{t}_{0}(t-s)^{H-3/2}dW_{s}dt\,. \end{eqnarray}

Hence, we have \begin{eqnarray}\label{intdBht} B^{H}_{t}=(H-\frac{1}{2})\int^{t}_{0}\int^{s}_{0}(s-u)^{H-3/2}dW_{u}ds\,. \end{eqnarray}

Now, using Fubini theorem, we have \begin{eqnarray} B^{H}_{t}=(H-\frac{1}{2})\int^{t}_{0}\int^{t}_{u}(s-u)^{H-3/2}dsdW_{t}\,. \end{eqnarray}

Can I write \begin{eqnarray} dB^{H}_{t}=(H-\frac{1}{2})\int^{s}_{0}(s-u)^{H-3/2}ds dW_{t}\,. \end{eqnarray}

Is the proof correct or which step is wrong.

  • $\begingroup$ Thanks for your hint. The last one is derivative. Please see that the second one is right $ dB^{H}_{t}=(H-\frac{1}{2})\int^{t}_{0}(t-s)^{H-3/2}dW_{s}dt$. I am also wonder whether the last one is correct or not! $\endgroup$ – steven Mar 5 at 6:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.