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Let $\mu\in L^{\infty}(\mathbb{R};\mathbb{R})$ be a bounded deterministic function. Then my understanding is that by using Girsanov's theorem, we can deduce uniqueness (in law) for the following stochastic equation $$dX_t = dB_t + \mu(X_t)dt, \quad X_0\equiv 0.\quad(\star)$$

What we do is set $Y_t$ as the solution to $$dY_t =dW_t, \quad Y_0\equiv 0$$ where $W$ is an $(\mathcal{F}_t, P)$-BM. That is, $Y$ uniquely solves the driftless equation.

Then we define $$M_t:=\int_0^t\mu(W_s)dW_s$$ so $M$ is a $(\mathcal{F}_t, P)$-martingale. Then we define the probability measure $Q$ by $$\frac{dQ}{dP}\Big|_{\mathcal{F}_t}=\mathcal{E}(M)_t$$

Where $\mathcal{E}$ denotes the stochastic exponential of $M$, which is a martingale since $\mu$ was taken bounded.

Then by Girsanov's theorem, we have that $$\tilde{W}_t:=W_t-\langle W,M\rangle_t=W_t-\int_0^t\mu(W_s)ds$$ is a $(Q, \mathcal{F}_t)$-BM.

We then simply observe that $dY_t=d\tilde{W}_t+\mu(Y_t)dt$. So that $(Y, \tilde{W})$ with $(Q,\mathcal{F}_t)$ solves SDE $(\star)$.


What I don't understand is how the above construction shows me that $(\star)$ has a unique solution. Any help at all would be much appreciated!

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  • $\begingroup$ Well I didn't read the construction in detail but note that the functions defining your SDE satisfy the Lipschitz condition. That is sufficient to conclude uniqueness. $\endgroup$
    – Calculon
    Commented Jun 18, 2015 at 13:26
  • $\begingroup$ @Calculon the drift coefficient is $L^\infty$ so not even continuous... $\endgroup$
    – Tyr Curtis
    Commented Jun 20, 2015 at 15:59

1 Answer 1

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There are two different notions of uniqueness for stochastic equations; in distribution and pathwise. For your equation there is actually uniqueness in both senses!

Let us begin with distributional uniqueness: Assume we have a solution $$ X_t = \int_0^t \mu(X_s)ds + B_t $$ and define $$ Z_t = \exp\{ -\int_0^t \mu(X_s) dB_s - \frac{1}{2} \int_0^t \mu^2(X_s)ds\} . $$ By the Novikov condition, since $\mu$ is bounded, $Z$ is a proper martingale and if we define $dQ := Z_T dP$ then $X$ is a Brownian motion w.r.t. $Q$. Since $X$ satisfies the above SDE we can rewrite $$ Z_t = \exp\{ -\int_0^t \mu(X_s) dX_s + \frac{1}{2} \int_0^t \mu^2(X_s)ds\} . $$ Let $A$ be a measurable subset of $\mathbb{R}^d$. We get \begin{align*} P(X_t \in A) & = E_P[1_A(X_t)] = E_Q [1_A(X_t) Z_T^{-1}] \\ & = E_Q[ 1_A(X_t) \exp\{ \int_0^T \mu(X_s) dX_s - \frac{1}{2} \int_0^T \mu^2(X_s)ds\} ] \\ & = E_P[ 1_A(B_t) \exp\{ \int_0^T \mu(B_s) dB_s - \frac{1}{2} \int_0^T \mu^2(B_s)ds\} ] ,\\ \end{align*} so that the distribution of $X$ is completely determined by $\mu$ which proves the claim.

To see pathwise uniqueness, assume we have two solutions $X^1$ and $X^2$ defined on the same probability space. We have $$ (X_t^2 - X_t^1)^+ = \int_0^t 1_{(X^2_s > X^1_s)} (\mu(X_s^2) - \mu(X_s^1)) ds $$ so that \begin{align*} X_t^1 \vee X_t^2 & = X_t^1 + (X_t^2 - X_t^1)^+ \\ & = \int_0^t \mu(X_s^1)ds + B_t + \int_0^t 1_{(X^2_s > X^1_s)} (\mu(X_s^2) - \mu(X_s^1)) ds \\ & = \int_0^t \mu(X_s^1) + 1_{(X^2_s > X^1_s)} (\mu(X_s^2) - \mu(X_s^1))ds + B_t \\ & = \int_0^t \mu(X_s^1 \vee X_s^2) ds + B_t \end{align*} so that also $X^1 \vee X^2$ is also a solution of the same equation. Then, from uniqueness in distribution $X^1$ and $X^1 \vee X^2$ have the same distribution. This is only true if $X^1 = X^1 \vee X^2 = X^2, P-a.s.$ which is exactly pathwise uniqueness.

Notice in the above that proof of distributional uniqueness is true in any dimension, but for pathwise uniqueness the proof relies heavily on $d=1$. Actually, there is also pathwise uniqueness when $d > 1$ (even when $d= \infty$) but the proof is much harder.

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