# Absolute continuity of the distribution of $X_t=aB_t+bt$, $Y_t=a(t)B_t$ with respect to the Wiener measure

Let

• $B_t:$ 1-dimensional Brownian motion, $P:$ its distribution on the Wiener space $C([0,1],\mathbb{R})$
• $X_t=aB_t+bt\text{; }t \in [0,1]$, $P_{a,b}$ its distribution
• $Y_t=a(t)B_t\text{; }a:[0,1] \to \mathbb{R}$ continuous, $P_{a(\cdot)}$ its distribution

My goal is to prove that

1. $P_{a,b} \ll P$ iff $a=\pm1$
2. $P_{a(\cdot)} \ll P$ iff $a=\pm1$

and in those cases being absolute continuous, to calculate the Radon-Nikodym derivative (which I think evaluates to):

$$\frac{dP_{a,b}}{dP}(w)=exp(\int_0^1b\text{ }{dw_t}-\tfrac{1}{2}\int_0^1b^2dt)=exp(bw_1-\tfrac{b^2}{2})\text{ ; }\forall w \in C([0,1],\mathbb{R})$$

Based on the lecture notes here (p.11) and here (p.6), it seems provable on the premise of the following two lemmas:

LEM1 $X_t=a(t,X_t)B_t+b(t,X_t)t \Rightarrow \langle X \rangle_t=\int_0^t a(\tau,X_\tau)^2 d\tau$

($\langle \cdot \rangle$ denotes the quadratic variation)

LEM2

• $U_t=a(t)B_t$
• $V_t=a_1(t)B_t+b_1t$
• $\langle U \rangle _t \neq \langle V \rangle _t$

$\Rightarrow P_U, P_V$ is singular to each other. ($P_U, P_V$ are the distributions of $U, V$ on $C([0,1],\mathbb{R})$ respectively)

Are these two lemmas true? Could anyone please show me a rigorous proof, especially of LEM2? (I tried to follow the arguments in the lecture notes given above but got lost in the way.)

Or better yet, is there an easier way to prove my goals 1. and 2.? Any help would be highly appreciated.

PS. My proof of 1. and 2. goes like this: It is known that $\langle B \rangle _t = t$. By direct calculation using LEM1, $\langle X \rangle _t = a^2t$ and $\langle Y \rangle _t = \int_0^t a(\tau)^2 d\tau$. By LEM2, if the distributions are to be absolutely continuous in the sense of 1. and 2.,

$$a^2t=\int_0^t a(\tau)^2 d\tau=t$$

which gives $a^2=a(t)^2=1,\forall t \in [0,1]$. Now for the reverse direction, the case $a=1$ is immediate: Girsanov's theorem gives $P_{1,b} \ll P$ because $B_t$ is a Brownian motion. The case $a=-1$ I'm not so sure. $-B_t$ is also a Brownian motion so by Girsanov's theorem I think $P_{-1,b} \ll P_{-1,0}$, and lastly $P_{-1,0}=P$ (is it?) The Radon-Nikodym derivative follows.

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1. By Girsanov theorem we know that $aB_t + bt \stackrel{dist}{\sim} aB_t$ since we are working on $[0,1]$, thus it is sufficient to consider $aB_t$ alone as it has an equivalent distribution.
2. Clearly, when $a^2 = 1$ we have that $aB_t \stackrel{dist}{=} B_t$ so in this case measures are equivalent.
3. If $a^2\neq 1$ then for the event $A = \{\omega:\langle X_1(\omega)\rangle = 1\}$ is such that $$\mathsf P_{1,0}(A) = 1,\quad \mathsf P_{a,b}(A) = 0$$ and thus these two measures are mutually singular.
Sorry for answering so late, I forgot to check the notifications that someone kindly answered. I understand why $P_{a,b}(A)=0$ but may I ask why do we know $P_{1,0}(A)=1$? – DancefloorTsunderella Jan 28 '13 at 15:40
@DancefloorTsunderella: since $\langle X_t \rangle = a^2t$ – Ilya Jan 28 '13 at 16:03