Integral with Wiener shift

Question

Consider the integral $$\int_{- \infty}^{0} e^{s} d B_s(\theta_t \omega)$$ where $B_t$ is Brownian motion and $\theta_t$ is Wiener shift on the canonical probability space. I know the property that $dB_s(\theta_t \omega) = d (B_{t+s}-B_t)$. But I cant understand why the above integral is the same as $$\int_{- \infty}^{t} e^{s-t} d B_s(\omega).$$ Can you show the steps to get this integral?

Answer 1

I'll let you fill in the specific technical details, but this should be the intuition in general. We denote with $\theta_t:C[0,\infty)\to C[0,\infty)$ the shift operator s.t. $x \mapsto x_{.+t}$ and $X:\Omega \to C[0,\infty)$ a random process taking values in the space of continuous functions; we denote the composition $\theta_t X:\Omega \to C[0,\infty)$ with $\theta_t X_s(\omega)=X_{s+t}(\omega),s\geq 0$. Suppose $X$ is a Brownian motion. If $H$ is a simple integrand on $[0,h]$ (zero elsewhere) we get for $0=s_0<s_1<...<s_k=h$ $$\begin{aligned}\int_0^hH_sd(\theta_tX)_s&=\sum_{0\leq \ell< k}\phi_\ell(\theta_tX_{s_{\ell+1}}-\theta_tX_{s_{\ell}})=\\ &=\sum_{0\leq \ell< k}\phi_\ell(X_{t+s_{\ell+1}}-X_{t+s_{\ell}})=\\ &=\int_t^{h+t}H_{s-t}dX_s\end{aligned}$$ because $$H_{s-t}=\sum_{0\leq \ell < k}\phi_\ell\mathbf{1}_{(s_\ell,s_{\ell+1}]}(s-t)=\sum_{0\leq \ell < k}\phi_\ell\mathbf{1}_{(t+s_\ell,t+s_{\ell+1}]}(s)$$ so $H_{.-t}$ is a simple integrand on $[t,h+t]$ and zero elsewhere. By extension to all nice integrands $H$ on $[0,h]$ approximated by sequences of simple integrands $H^n$ we have $$\int_0^hH_sd(\theta_tX)_s=\int_{t}^{h+t}H_{s-t}dX_s$$