If we create a weak solution of an SDE using the Girsanov transformation, are the initial condition and parameters independent of the transformed Wiener process if they are independent of the original Wiener process? This other question is similar but more general.

Let $(\Omega, \mathcal{F}, P)$ be a probability space and $X_t$ be a stochastic process which satisfies the following SDE:

$$ d X_t = f(t, X_t,\Theta)dt + dW_t, $$

where $W_t$ is a Wiener process and $\Theta$ is a random variable. We assume $W_t$ be adapted to the filtration $\mathcal{F}_t$ and that $\Theta$ and the initial condition of the SDE $X_0$ are $\mathcal{F}_0$-measurable and independent of the $\sigma$-algebra generated by $W$.

If we define a stochastic process $M_t$ by $$ M_t := \exp\left[-\int_0^t f(s, X_s,\Theta) \,dW_s -\frac{1}{2}\int_0^t f(s, X_s,\Theta)^2 \,ds\right] $$ and a new measure $\tilde P(A):=\int_A M_T dP$, then the process $X_t$ also satisfies $d X_t = d\tilde W_t$ where the process $\tilde W_t := \int_0^t f(s,X_s,\Theta)\,ds + W_t$ is a Wiener process under $\tilde P$. Under the measure $\tilde P$, are $X_0$ and $\Theta$ independent of $\tilde W$?

I know that the $\tilde P(A)=P(A)$ for all $A\in\mathcal{F}_0$ because $M_t$ is an $\mathcal{F}_t$ martingale and $M_0=1$, so the measures induced by $X_0$ and $\Theta$ do not change. I believe I would be able to prove the independence if it is possible to show that $$ P(W^{-1}(A)|\mathcal{F}_t) = \tilde P(\tilde W^{-1}(A)|\mathcal{F}_t) \qquad\text{with prob. 1}, $$ for any set $A$ from the Borel $\sigma$-algebra of the classical Wiener space. Any ideas?

  • $\begingroup$ Your new process is a Brownian motion (adapted process) under $\tilde{P}(A)=\mathbb{E}[1_A M_t]$, so $X_0$ and $\Theta$ are still $\mathcal{F}_0$ measurable, and $P$ and $\tilde{P}$ are still the same on $\mathcal{F}_0$ and independance still hold no? What am I missing ? $\endgroup$
    – Bertrand R
    Aug 13, 2013 at 3:59
  • $\begingroup$ @BertrandR, the fact that $\tilde W_t$ is adapted to $\mathcal{F}_t$ under $\tilde P$ does not mean it is independent of $\mathcal{F}_0$ events. Under P, for example, $X_t$ is adapted to $\mathcal{F}_t$ and can depend on $X_0$. Is there a simple proof of independence using these facts you mentioned? $\endgroup$
    – Dimas
    Aug 13, 2013 at 4:11
  • $\begingroup$ Ah ok I get it now, I just misread your question. I don't think independance holds, but that's just an intuition ... $\endgroup$
    – Bertrand R
    Aug 13, 2013 at 4:24
  • $\begingroup$ @ Dimas : What about $f(s,X_s,\theta)=|X_s-X_0+\theta|$ ? I think this introduces in $X$ a dependence in $X_0,\theta$ at every time $s>0$ that you can't get rid off in $P(X)$, while it disappears in $\tilde{P}$ ? $\endgroup$
    – TheBridge
    Aug 13, 2013 at 9:08
  • $\begingroup$ I was wondering if the independence would hold for arbitrary Lipschitz continuous $f$. If it only holds for very specific $f$, like the one you suggested, then it is of no use to me. I believe that if independence does not hold, than weak solutions to SDEs using Girsanov's transformation would have very limited applications. Judging by the wide usage of the transformation I would guess that independence holds. It's only a guess, anyways. $\endgroup$
    – Dimas
    Aug 13, 2013 at 14:41

1 Answer 1


The answer is that the proposition is true, that is, any $\mathcal{F}_0$-measurable random variable is independent of $\tilde W$ under $\tilde P$, and the answer is quite simple. It was simply a matter of finding a more complete statement of the Girsanov theorem / transformation.

By definition, a Wiener process on a filtered probability space $(\Omega, \mathcal{F}, \{\mathcal{F}_t\}_{0\leq t\leq T}, P)$ has $W_t-W_s$ independent of $\mathcal{F}_s$ for all $0\leq s \leq t \leq T$. As $W_0=0$ is almost surely, this means that the process is independent of $\mathcal{F}_0$. More complete statements of the Girsanov transformation, like that of Ikeda and Watanabe, say that $\tilde W$ is a Wiener process on the filtered space $(\Omega, \mathcal{F}, \{\mathcal{F}_t\}_{0\leq t\leq T}, \tilde P)$, hence independent of $\mathcal{F}_0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .