in $H$, $$X{\circ \tau_{h}}=Y{\circ \tau_{h}}, \ \P_0-a.s.$$

We recall, the quite surprising result that any r.v. is continuous in probability in the directions of the Cameron-Martin space. More precisely:

\[lemma:ctsH\] Let $F$ be a $\mathcal{F}_T$-measurable random variable. The mapping $h\longmapsto F{\circ \tau_{h}}$ is continuous from $H$ to $L^0({\mathbb{R}}^{d})$ where the convergence is in probability.

One of the main technique when working with shifts on the path space is the famous Cameron-Martin formula.

\[prop.cam\] $($Cameron-Martin Formula, see e.g. [@ustunelzakai Appendix B.1]$)$ Let $F$ be a $\mathcal F_T$-measurable random variable and let $h$ be in $H$. Then, when both sides are well-defined $$\E[F{\circ \tau_{h}}] =\E\left[ F \exp\left(\int_0^T \dot{h}(s) \cdot dW_s -\frac12 \int_0^T \|\dot{h}(s)\|^2 ds\right)\right].$$

For further reference, we also emphasize that for any $h\in H$ and for any $p\geq 1$, the stochastic exponential $\mathcal E\left(\int_0^\cdot\dot{h}(s)\cdot dW_s\right):=\exp\left(\int_0^\cdot \dot{h}(s) dW_s -\frac12 \int_0^\cdot \|\dot{h}(s)\|^2 ds\right)$ verifies $$\label{exp.int}
\mathcal E\left(\int_0^\cdot\dot{h}(s)\cdot dW_s\right)\in\mathbb S^p, \quad \forall p\geq 1.$$

\[lemma:shift1\] Let $t$ in $[0,T]$ and let $F$ be a $\mathcal{F}_t$-measurable random variable. For any $h$ in $H$, it holds that $$F{\circ \tau_{h}}= F\circ {\tau_{\widetilde{h^t}}}, \ \P_0-a.s,$$ where $$\widetilde{h^t}(s):=\left(\int_0^s \dot{ h}^1(u) \mathbf 1_{0\leq u\leq t}du, \dots, \int_0^s \dot{ h}^d(u) \mathbf 1_{0\leq u\leq t}du\right)^\top.$$ In particular, $F{\circ \tau_{h}}$ is $\mathcal{F}_t$-measurable.

It 