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Let $$\{B^{(i)}_t : 0 \leq t \leq T, i \geq 1\}$$ be a sequence of i.i.d. standard Brownian motions. For each $n,$ let $V^{(n)}_t, \,0 \leq t \leq T$ be a continuous bounded process adapted to the filtration $\mathcal{F}_t^n$ generated by the first $n$ Brownian motions.

Let $P_n$ denote the law of $(B^{(1)}, \dots, B^{(n)})$ on $C := C([0, T], \mathbb{R}^n).$

Since $V^{(n)}$ satisfies conditions for a change of measure, we can apply a Girsanov transform to yield a measure $Q_n$ with $$\frac{dQ_n}{dP_n} = \exp\Big( \sum_{i = 1}^n \int_0^\cdot V^{(n)}_sdB^{(i)}_s - \frac{1}{2}\int_0^\cdot [V_s^{(n)}]^2 ds\Big).$$

So $Q_n$ has the same law on $C$ as $(B^{(1)} + V^{(n)}, \dots, B^{(n)} + V^{(n)}),$ an $n$ dimensional Brownian motion with a drift of $V^{(n)}$ in each coordinate.

If we define two triangular arrays whose row for each $n$ are $$R^P_n : =(B^{(1)}, \dots, B^{(n)}),$$ and $$R^Q_n := (B^{(1)} + \int_0^\cdot V^{(n)}, \dots, B^{(n)} + \int_0^\cdot V^{(n)}),$$

we immediately see the laws of the rows $R^P_n$ and $R^Q_n$ are equivalent for each finite $n,$ and in fact under the finite triangular array $R_1^P \times \cdots \times R_n^P \equiv R_1^Q \times \cdots \times R_n^Q$ via the canonical product measure.

We may embed each $R^P_n$ to induce a measure on $C^\infty = C([0, T], \mathbb{R}^{\mathbb{N}})$ and can place a standard metric on this to make it Polish. Similarly with $R^Q_n.$ Now that each $R^P_n$ has a measure on the same space, consider its filtration $\mathcal{F}_n^P$ and tail $\sigma$-field $\mathcal{F}_\infty^P.$ (Similar for $R^Q_n$). Notice that because $V^{(n)}$ is adapted, we know $\mathcal{F}_n^Q \subset \mathcal{F}^P_n$ so for our purposes here we restrict to $\mathcal{F}^P_n.$

By the Girsanov transform, we see that null sets in $\mathcal{F}_n^P$ under $R^P$ are also null sets under $R^Q.$

Let us further assume that $V^{(n)}$ converges a.s. and in $L_p$ to some bounded process $V.$

Can we assert the null sets in $\mathcal{F}_\infty$ under $R^P$ are also null under $R^Q?$

In essence, we wish to extend the $0-1$ law of a triangular array of i.i.d. Brownian motions to that of a triangular array of BMs with an adapted row-wise convergent drift.

I apologize in advance for any errors or typos! Any references would be much appreciated!

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No, the measures are not necessarily equivalent on the tail $\sigma$-field.

For a concrete example, take $V_t^{(n)} = t$, a deterministic continuous process of bounded variation. Then trivially the $V_t^{(n)}$ converge to $V_t=t$ in whatever sense you want. Consider the tail random variable $X = \limsup_{n \to \infty} \frac{1}{n} \sum_{k=1}^n B^{(k)}_1$. Under $P$, each $B^{(k)}_1$ has an $N(0,1)$ distribution, and they are iid, so by the strong law of large numbers we have $P(X=0)=1$, where we note that $\{X=0\} \in \mathcal{F}_\infty$. But under $Q$, each $B^{(k)}_1$ has the law of a standard Brownian motion plus $V_1=1$, so it is $N(1,1)$. Thus we have $Q(X=1)=1$ and $Q(X=0)=0$. As such, $P,Q$ are mutually singular on $\mathcal{F}_\infty$.

You need some sort of summability criterion on the $V^{(n)}$; this is some corollary of the Cameron-Martin theorem for Gaussian measures. They should not only converge, but converge to 0, and reasonably fast. I would have to work it out, but I think you might want something like $\sum_n \mathbb{E}\left[\sup_{[0,T]} |V_t^{(n)}|^2\right] < \infty$. Bogachev's book Gaussian Measures is probably the ultimate resource here, though there may be lighter-weight sources also.

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  • $\begingroup$ Thanks, Nate. In your example, the slln still holds for the sequence with drift. Do you know if under these conditions a slln holds for $n^{-1}\sum_{i=1}^nl(B^{(i)} + V^{(n)})$ where $l$ is a functional on $C([0, T], \mathbb{R})?$ $\endgroup$ – user253775 Feb 22 '17 at 3:21
  • $\begingroup$ I mean $n^{-1}\sum_{i=1}^nl(B^{(i)} + \int_0^\cdot V^{(n)}).$ $\endgroup$ – user253775 Feb 22 '17 at 4:48
  • $\begingroup$ If it's a linear functional, and if SLLN holds for $V$, then it should. $\endgroup$ – Nate Eldredge Feb 22 '17 at 5:05

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