Let $(B_t)_{0 \leq t \leq 1} $ be a standard Brownian Motion under the probability measure $\mathbb{P}$. Let $\mathbb{Q}$ a probability measure with density $\frac{d\mathbb{Q} }{d\mathbb{P}} = e^{B_1-\frac{1}{2}} $.
I need to show that every function f such that $\mathbb{E}_\mathbb{P}[|f(B_t)|] < \infty$ apply that : $$\int_{\Omega} f(B_t-t)d\mathbb{Q} =\int_{Ω} f(B_t)d\mathbb{P}$$
To be honest i haven't really understand how change of measure actually works so i have no idea how to proceed to the solutions of this. Can anyone help me understand how it works, i would appreciate any kind of hints. Thanks for your time .
Okay, so by definition
$$ \int_{\Omega} f(B_t-t) \textrm{d}\mathbb{Q}=\int_{\Omega} f(B_t-t) e^{B_1-\frac{1}{2}}\textrm{d}\mathbb{P} $$ and hence, we want to show that $$ \int_{\Omega} f(B_t-t) e^{B_1-\frac{1}{2}}\textrm{d}\mathbb{P}=\int_{\Omega} f(B_t)\textrm{d}\mathbb{P} $$ for all $t$. Using that $B_1-B_t$ is independent of $B_t$, we get that $$ \int_{\Omega} f(B_t-t) e^{B_1-\frac{1}{2}}\textrm{d}\mathbb{P}=e^{-\frac{1}{2}}\int_{\Omega} f(B_t-t) e^{B_t}e^{B_1-B_t}\textrm{d}\mathbb{P}=e^{-\frac{1}{2}} \mathbb{E}_{\mathbb{P}} e^{B_1-B_t} \int_{\Omega} f(B_t-t)e^{B_t}\textrm{d}\mathbb{P} $$ Now $e^{B_1-B_t}$ is lognormal with mean $\frac{1-t}{2}$and thus,
$$ \int_{\Omega} f(B_t-t)\textrm{d}\mathbb{Q}=e^{-\frac{t}{2}}\int_{\Omega} f(B_t-t)e^{B_t}\textrm{d}\mathbb{P}= e^{-\frac{t}{2}}\int_{\mathbb{R}} f(x-t)e^x\textrm{d}B_t(\mathbb{P}), $$ Now, $B_t$ is normal with variance $t$, so \begin{align} \int_{\mathbb{R}} f(x-t)e^x\textrm{d}B_t(\mathbb{P}) &=\frac{1}{\sqrt{2\pi t}}\int_{\mathbb{R}} f(x-t) e^xe^{-\frac{x^2}{2t}}\textrm{d}x=\frac{1}{\sqrt{2\pi t}}\int_{\mathbb{R}} f(x-t) e^{-\frac{x^2-2tx}{2t}}\textrm{d}x\\ &=e^{\frac{t}{2}}\frac{1}{\sqrt{2\pi t}}\int_{\mathbb{R}} f(x-t) e^{-\frac{(x-t)^2}{2t}}\textrm{d}x=e^{\frac{t}{2}}\frac{1}{\sqrt{2\pi t}} \int_{\mathbb{R}} f(x)e^{-\frac{x^2}{2t}}\textrm{d}t\\ &= e^{\frac{t}{2}} \int_{\mathbb{R}} f(x) \textrm{d}B_t(\mathbb{P})=e^{\frac{t}{2}} \int_{\Omega} f(B_t) \textrm{d}\mathbb{P}, \end{align} which is exactly what we wanted.