Let $(\Omega,\mathcal{F},\mathbb{F},P)$ be a filtered probability space and $X=(X_t)_{t\geq 0}$ be a Brownian motion with respect to the natural filtration $(\mathcal{F}_t)_{t\geq 0}$ generated by $X$.
Define $$Y_t=X_t+\mu t$$ for $t\geq 0$ and a measure $Q_t$ on $\mathcal{F}_t$ by $$Q_t(A)=E[exp(\mu X_t-\mu^2\frac{t}{2}),A]$$ for $t\geq 0$.
I have a proof that $Q\circ Y^{-1}=P\circ X^{-1}$ holds, but have a few questions about it.
The aim is to show that $$E_Q[f_1(X_{t_1})\ldots f_n(X_{t_n})]=E_P[f(X_{t_1}+\mu t_1)\ldots f_n(X_{t_n}+\mu t_n)]$$ for every bounded and continuous $f_1,\ldots,f_n$ and $t_1\leq\ldots\leq t_n$.
We have $E_Q[f(X_t)|\mathcal{F_s}]=E_P[f(X_t)exp(\mu X_t-\mu^2\frac{t}{2})|\mathcal{F}_s]=\ldots=E_P[f(X_t+\mu(t-s))|\mathcal{F}_s]exp(\mu X_s-\mu^2\frac{s}{2})$ (the first equation is not perfectly clear, obviously we are using the the formula how conditional expectations behave under change of measure as can be seen here, but this would result in: $E_Q[f(X_t)|\mathcal{F}_s]=\frac{1}{exp(-\mu X_s+\mu^2\frac{s}{2})}E[f(X_t)exp(\mu X_t-\mu^2\frac{t}{2})|\mathcal{F}_s]=E[f(X_t)exp(\mu(X_t+X_s)-\mu^2\frac{t-s}{2})|\mathcal{F}_s]$)
How can we use this to compute $$E_Q[f_1(X_{t_1})\ldots f_n(X_{t_n})]=E_P[f(X_{t_1}+\mu t_1)\ldots f_n(X_{t_n}+\mu t_n)]$$?
Note that, for $0 \le s < t$, \begin{align*} E_P\left(f(X_t) e^{\mu (X_t-X_s) -\frac{\mu^2}{2}(t-s)} \mid \mathcal{F}_s \right) &= E_P\left(f(X_s + X_t-X_s) e^{\mu (X_t-X_s) -\frac{\mu^2}{2}(t-s)} \mid \mathcal{F}_s \right)\\ &=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} f(X_s + x\sqrt{t-s})e^{\mu\sqrt{t-s}x -\frac{\mu^2}{2}(t-s)}e^{-\frac{x^2}{2}}dx\\ &=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} f(X_s + x\sqrt{t-s})e^{-\frac{(x-\mu\sqrt{t-s})^2}{2}}dx\\ &=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} f(X_s + \mu(t-s) + x\sqrt{t-s})e^{-\frac{x^2}{2}}dx\\ &=E_P\Big(f(X_s +\mu(t-s) + X_t-X_s) \mid \mathcal{F}_s \Big). \end{align*} Let $t_0=0$. Then, \begin{align*} &\ E_Q\left(\prod_{i=1}^n f_i(X_{t_i})\right)\\ =&\ E_P\left(\prod_{i=1}^n f_i(X_{t_i}) e^{-\frac{\mu^2}{2}t_n + \mu X_{t_n}} \right)\\ =&\ E_P\left(\prod_{i=1}^{n-1} f_i(X_{t_i})e^{-\frac{\mu^2}{2}t_{n-1} + \mu X_{t_{n-1}}} E_P\left( f_n(X_{t_n}) e^{-\frac{\mu^2}{2}(t_n-t_{n-1}) + \mu (X_{t_n}-X_{t_{n-1}}) } \mid \mathcal{F}_{t_{n-1}}\right)\right)\\ =&\ E_P\left(\prod_{i=1}^{n-1} f_i(X_{t_i})e^{-\frac{\mu^2}{2}t_{n-1}+ \mu X_{t_{n-1}}} E_P\left( f_{n}\Big(X_{t_{n-1}} + \mu (t_n-t_{n-1})+ X_{t_n}-X_{t_{n-1}}\Big)\mid \mathcal{F}_{t_{n-1}}\right) \right)\\ =&\ E_P\left(\prod_{i=1}^{n-1} f_i(X_{t_i})e^{-\frac{\mu^2}{2}t_{n-1}+ \mu X_{t_{n-1}}}f_{n}\Big(X_{t_{n-1}} + \mu (t_n-t_{n-1})+ X_{t_n}-X_{t_{n-1}}\Big)\right)\\ =&\ E_P\Bigg(\prod_{i=1}^{n-2} f_i(X_{t_i})e^{-\frac{\mu^2}{2}t_{n-2}+ \mu X_{t_{n-2}}} E_P\bigg(f_{n-1}(X_{t_{n-1}})f_{n}\Big(X_{t_{n-1}} + \mu (t_n-t_{n-1})+ X_{t_n}-X_{t_{n-1}}\Big)\\ &\qquad e^{-\frac{\mu^2}{2}(t_{n-1}-t_{n-2}) + \mu (X_{t_{n-1}}-X_{t_{n-2}}) } \mid \mathcal{F}_{t_{n-2}}\bigg)\Bigg). \end{align*} Moreover, \begin{align*} &\ E_P\bigg(f_{n-1}(X_{t_{n-1}})f_{n}\Big(X_{t_{n-1}} + \mu (t_n-t_{n-1})+ X_{t_n}-X_{t_{n-1}}\Big) e^{-\frac{\mu^2}{2}(t_{n-1}-t_{n-2}) + \mu (X_{t_{n-1}}-X_{t_{n-2}}) } \mid \mathcal{F}_{t_{n-2}}\bigg)\\ =&\ E_P\bigg(f_{n-1}(X_{t_{n-2}}+X_{t_{n-1}}-X_{t_{n-2}})\\ &\ f_{n}\Big(X_{t_{n-2}}+X_{t_{n-1}}-X_{t_{n-2}} + \mu (t_n-t_{n-1})+ X_{t_n}-X_{t_{n-1}}\Big)e^{-\frac{\mu^2}{2}(t_{n-1}-t_{n-2}) + \mu (X_{t_{n-1}}-X_{t_{n-2}}) } \mid \mathcal{F}_{t_{n-2}}\bigg)\\ =&\ \frac{1}{2\pi}\iint_{\mathbb{R}^2}f_{n-1}(X_{t_{n-2}}+\sqrt{t_{n-1}-t_{n-2}}x) f_n(X_{t_{n-2}}+\sqrt{t_{n-1}-t_{n-2}}x +\mu (t_n-t_{n-1})\\ &\qquad + \sqrt{t_n-t_{n-1}}y)e^{-\frac{\mu^2}{2}(t_{n-1}-t_{n-2}) +\mu \sqrt{t_{n-1}-t_{n-2}}x} e^{-\frac{x^2+y^2}{2}} dx dy\\ =&\ \frac{1}{2\pi}\iint_{\mathbb{R}^2}f_{n-1}(X_{t_{n-2}}+ \mu(t_{n-1}-t_{n-2}) +\sqrt{t_{n-1}-t_{n-2}}x)\\ & \quad f_n(X_{t_{n-2}}+ \mu(t_{n-1}-t_{n-2})+\sqrt{t_{n-1}-t_{n-2}}x +\mu (t_n-t_{n-1})+ \sqrt{t_n-t_{n-1}}y)e^{-\frac{x^2+y^2}{2}} dx dy\\ &=E_P\bigg(f_{n-1}(X_{t_{n-2}} + \mu(t_{n-1}-t_{n-2}) + X_{t_{n-1}}-X_{t_{n-2}})\\ &\qquad f_{n}\Big(X_{t_{n-2}} + \mu(t_{n-1}-t_{n-2}) + \mu (t_n-t_{n-1})+ X_{t_{n-1}}-X_{t_{n-2}}+ X_{t_n}-X_{t_{n-1}}\Big) \mid \mathcal{F}_{t_{n-2}}\bigg). \end{align*} Therefore, \begin{align*} &\ E_Q\left(\prod_{i=1}^n f_i(X_{t_i})\right)\\ =&\ E_P\Bigg(\prod_{i=1}^{n-2} f_i(X_{t_i})e^{-\frac{\mu^2}{2}t_{n-2}+ \mu X_{t_{n-2}}} E_P\bigg(f_{n-1}(X_{t_{n-2}} + \mu(t_{n-1}-t_{n-2}) + X_{t_{n-1}}-X_{t_{n-2}})\\ &\qquad f_{n}\Big(X_{t_{n-2}} + \mu(t_{n-1}-t_{n-2}) + \mu (t_n-t_{n-1})+ X_{t_{n-1}}-X_{t_{n-2}}+ X_{t_n}-X_{t_{n-1}}\Big) \mid \mathcal{F}_{t_{n-2}}\bigg)\Bigg)\\ =&\ E_P\Bigg(\prod_{i=1}^{n-2} f_i(X_{t_i})e^{-\frac{\mu^2}{2}t_{n-2}+ \mu X_{t_{n-2}}} f_{n-1}(X_{t_{n-2}} + \mu(t_{n-1}-t_{n-2}) + X_{t_{n-1}}-X_{t_{n-2}})\\ &\qquad f_{n}\Big(X_{t_{n-2}} + \mu(t_{n-1}-t_{n-2}) + \mu (t_n-t_{n-1})+ X_{t_{n-1}}-X_{t_{n-2}}+ X_{t_n}-X_{t_{n-1}}\Big) \Bigg)\\ =& \ \cdots\cdots\\ =&\ E_P\left(\prod_{i=1}^n f_i\left(X_{t_{0}} + \sum_{j=1}^i\mu\left(t_j-t_{j-1}\right) + \sum_{j=1}^i \left(X_{t_j}-X_{t_{j-1}}\right)\right) \right)\\ =&\ E_P\left(\prod_{i=1}^n f_i(X_{t_i} + \mu t_i) \right). \end{align*}