of the order of integration below we refer to Exercise 3.2.7 in [@N]) $$\begin{aligned}
\hat{\xi} (x) &= &\sum_{m\geq 0} I_{m+2}\frac{1}{(m+1)(m+2)}\sum_{k,l=1; k\not= l} ^{m+2} a_{m} ^{\widehat{x}} (x_{k}, y_{l}) \nonumber\\
  &&1_{[0,x_{k}]}^{\otimes m+1} (x_{1}, .., \overline{x_{k}},.., x_{m+2} )1_{[0,y_{l}] }^{\otimes m+1} (y_{1},.., \overline{y_{k}},.., y_{m+2} )\\
  &=& \delta ^{(2)} _{(u_{1}, v_{1}), (u_{2}, v_{2})}  e^{-\frac{x^{2}}{2}u_{1}v_{2}}\sum_{m\geq 0} \frac{(-i)^{m} x^{m}}{m!} I_{m}\left(1^{\otimes m}_{[0, u_{1}] \times [0,v_{2}] } (\cdot ) \right) \\
  &=& \delta ^{(2)} _{(u_{1}, v_{1}), (u_{2}, v_{2})} \left( e^{-ixW_{u_{1}, v_{2}}}1_{[0, u_{1}] \times [0, v_{2}] } \right).\end{aligned}$$ Here we used the fact that $$e^{-ixW_{u_{1}, v_{2}} }= \sum_{m\geq 0} \frac{(-i)^{m} x^{m}}{m!} I_{m}\left(1^{\otimes m}_{[0, u_{1}] \times [0,v_{2}] } (\cdot )\right)$$ which can be obtained using Stroock’s formula (see [@FT] relation (9)). 0.3cm

Consequently, we obtain the following result:

\[th3\] For almost all $\omega$, it holds that $\xi \in H^{-r} (\mathbb{R}, \mathbb{R}) $ if $r>\frac{1}{2}$

[**Proof:** ]{}It is easy to see that $$\begin{aligned}
\mathbf{E}\left| \widehat{\xi }(x) \right| ^{2}&=&\int_{0} ^{1} du_{1} \int_{0} ^{u_{1} } du_{2} \int_{0}^{1}dv_{2} \int_{0} ^{v_{2}} dv_{1} \left| e^{-ixW_{u_{1}, v_{2}} }\right| ^{2}\\
&=& \int_{0}^{1} u_{1} du_{1} \int_{0} ^{1} v_{2}dv_{2}=\frac{1}{4}\end{aligned}$$ and consequently $$\begin{aligned}
\mathbf{E} \Vert \xi(x) \Vert _{H^{-r} }=\int_{\mathbb{R}} \frac{1}{(1+ x^{2})^{r}} \mathbf{E} \left| \widehat{\xi }(x) \right| ^{2}dx\end{aligned}$$ is finite if and only if $r>\frac{1}{2}$.

We note that, as in the case of 