Laws of $(B_t)_{t\in [0,T]}$ and $(2B_t)_{t\in [0,T]}$ : singular?

Are the laws of $(B_t)_{t\in [0,T]}$ and $(2B_t)_{t\in [0,T]}$ mutually singular ?

More precisely, I know that the laws of two diffusion processes are mutually absolutely continuous if they share a common diffusion coefficient (application of Girsanov theorem).

However, I cannot understand why it is no more the case when they have different diffusion coefficients. This is why I ask the question about $(B_t)_{t\in [0,T]}$ and $(2B_t)_{t\in [0,T]}$. If their laws are not mutually continuous, that would mean that one can find a measurable set of trajectories $\mathcal{A}$ for which $\mathbb{P}[(B_t)_{t\in [0,T]} \in \mathcal{A}]>0$ while $\mathbb{P}[(2B_t)_{t\in [0,T]} \in \mathcal{A}]=0$. The problem is : I don't know how to find such a set $\mathcal{A}$. Can anyone help me to understand this point ?

Thank you folks !

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2 Answers

The answer comes from the quadratic variation: $\mathsf P\{[B_t] = t\} = 1$ while $\mathsf P\{[2B_t] = 4t\} = 1$. In this case you can define $\mathcal A$ to be an event (subset of the space of processes with continuous trajectories, say $\omega$) as $$\mathcal A_k = \{\omega:[\omega_t] = k^2t\}.$$ Then $\mathsf P\{B_t\in \mathcal A_1\} = 1$ but $\mathsf P\{2B_t\in \mathcal A_1\} = 0$ and $\mathsf P\{B_t\in \mathcal A_2\} = 0$ but $\mathsf P\{2B_t\in \mathcal A_2\} = 1$.

See also the related question of mine on MO.

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Thank you very much ! – mellow Nov 7 '11 at 15:04
@mellow: you're welcome. I mad some minor edits by the way – Ilya Nov 7 '11 at 15:05

You can also use the law of the iterated logarithm at $t=0$. For the set $$A:=\left\{\omega: \limsup_{t\to0} {|\omega_t|\over\sqrt{2t\log\log(1/t)}}=1\right\}$$ we have $\mu_B(A) = 1$, but $\mu_{2B}(A) = 0$. Here $\mu_B$ and $\mu_{2B}$ are the laws of Brownian motion and twice Brownian motion respectively.

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