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Let $B=(B_t)_{t}$ be a standard Brownian motion and let $X=(B_{t/2})_t$. I want to disprove that $\mu_X\ll\mu_B$ where $\mu_\cdot$. denotes the respective distribution. The standard way might be to note first that $\langle X\rangle_t=t/2$ whilst $\langle B\rangle_t=t$ and thus for $A=\{f\in C[0,1]\;|\;\langle f\rangle_1=1/2\}$ $$ \mu_X(A)=1,\qquad\mu_B(A)=0. $$ Is there another way of proving this?

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    $\begingroup$ They are singular, not absolutely continuous. Your method works, another one might be to set for instance $$A=\bigg\{ f: \limsup_{t\to 0} \frac{f(t)}{\sqrt{2t\log\log t}}=1 \bigg\}$$ or $$A:=\bigg\{f: \lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n 2^{k+1}(f(2^{-k})-f(2^{-k-1}))^2=1\bigg\}$$A standard result of Cameron-Martin theory states that dilations of Gaussian measures on infinite-dimensional banach spaces are always mutually singular. $\endgroup$ – Shalop Feb 6 '18 at 21:28
  • $\begingroup$ *should be $\log \log(1/t)$ $\endgroup$ – Shalop Feb 6 '18 at 23:00
  • $\begingroup$ Let us now consider $X_t=B_t+\int_0^t \sin(B_s)\;ds$. Why are the distributions in this case equivalent? $\endgroup$ – julian Feb 7 '18 at 16:18
  • $\begingroup$ Hint: Fix some $T>0$, and define a martingale $M_t:= -\int_0^t \sin(B_s) dB_s$, and define a measure $Q$ on $\mathcal F_T$ by $Q(A) = E_P[e^{M_T-\frac{1}{2}\langle M \rangle_T}\cdot 1_A]$. Since $B$ is a $P$-BM, it follows from Girsanov that $B-\langle M,B \rangle$ is a $Q$-BM. $\endgroup$ – Shalop Feb 7 '18 at 20:43
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Let $$D=\Bigl\{f\in C[0,1]: \lim_{n\to\infty}\sum_{k=1}^{2^n}[f(\tfrac{k}{2^n})-f(\tfrac{k-1}{2^n})]^2=1\Bigr\}.\qquad A=C[0,1]\setminus D. $$ Then \begin{gather} \mu_X(D)=0,\qquad \mu_B(D)=1.\\ \mu_X(A)=1, \qquad \mu_B(A)=0. \end{gather}

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