# derivative of integral of Brownian motion

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

• 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! – steven Mar 5 at 6:46