stochastic-calculusstochastic-differential-equationsstochastic-processes
Consider the $d$-dimensional SDE, $d > 1$,
$$dX_t = b(X_t) \, dt + \sqrt 2 \, dW_t$$
where
Do you know some references about the weak uniqueness of above SDE?
Thank you so much for your elaboration!
The answer is no, as can be seen in the case $\sigma(u) = u$, so that $X_t = \exp(W_t - t/2)$. For the result to be true, Markov's inequality implies that the law of $W$, conditional on $A_M$, would need to give probability $1/2$ to the event $\sup_{t \le 1}|W_t - Mt + t/2| < K\exp(-M/2)$ for some fixed $K>0$. By large deviations, this event has probability smaller than $c\exp(-c\exp(M))$ for some $c>0$, while $A_M$ has probability larger than $c \exp(-cM^2)$ for some $c>0$, yielding a contradiction.
The answer to your question is yes, at least for the one- dimensional marginals. That said, I'm confident that the convergence you want actually holds on the level of stochastic processes (i.e. $c^{-1/2}X_{ct}\xrightarrow{(d)}B_t$ as functions on $C[0,T]$ for any $T>0$), but this requires some additional work. That said, here's the argument that proves convergence in distribution of the one dimensional marginals.
Suppose $(X_{t})_{t\geq{0}}$ solves the stochastic differential equation: $$ dX_{t}=\sigma(X_{t})dB_{t}, \hspace{5pt} X_{0}=0 $$ where $\sigma$ is Lipschitz with Lipschitz constant $K>0$. That is $|\sigma(x)-\sigma(y)|\leq{K|x-y|}$ for all $x,y\in{\mathbb{R}}$. First, we will show that $\sup_{t\in{[0,T]}}\mathbb{E}X_{t}^{2}<\infty$ for all $T>0$.
To this effect, consider the stopping time $T_{n}:=\inf\{t\geq{0}:|X_{t}|=n\}$. For each $n\in{\mathbb{N}}$ we define the function $f_{n}:[0,\infty)\rightarrow{\mathbb{R}}$ as follows:
$$
f_{n}(t)=\mathbb{E}X_{t\wedge{T_{n}}}^{2}
$$
Since $|X_{t\wedge{T_{n}}}|\leq{n}$ by the definition of $T_{n}$, $|f_{n}(t)|\leq{n^{2}}$ for all $t>0$. By Ito's formula applied to $X_{t}$,
$$
X_{t\wedge{T_{n}}}^{2}=\int_{0}^{t\wedge{T_{n}}}\sigma(X_{s})X_{s}dB_{s}+\int_{0}^{t\wedge{T_{n}}}\sigma^{2}(X_{s})ds
$$
Taking expectations we have that:
$$
\mathbb{E}X_{t\wedge{T_{n}}}^{2}=\mathbb{E}\int_{0}^{t\wedge{T_{n}}}\sigma^{2}(X_{s})ds\leq{\int_{0}^{t}\mathbb{E}\sigma^{2}(X_{s\wedge{T_{n}}})ds}
$$
Using the fact that $\sigma$ is Lipschitz we have that for any $x\in{\mathbb{R}}$, $|\sigma(x)|\leq{|\sigma(0)|+ K|x|}$.This in turn implies that $\sigma^{2}(x)\leq{2\sigma^{2}(0)+2K^{2}|x|^{2}}$. Applying this to the inequality above gives us:
$$
\mathbb{E}X_{t\wedge{T_{n}}}^{2}\leq{\int_{0}^{t}\big(2\sigma^{2}(0)+2K\mathbb{E}X_{s\wedge{T_{n}}}^{2}\big)ds}=2\sigma^{2}(0)t+2K^{2}\int_{0}^{t}\mathbb{E}X_{s\wedge{T_{n}}}^{2}ds
$$
In other words, for all $n\in{\mathbb{N}}$ and $t\in{[0,\infty)}$, $f_{n}$ satisfies the following inequality:
$$
f_{n}(t)\leq{2\sigma^{2}(0)t+2K^{2}\int_{0}^{t}f_{n}(s)ds}
$$
By a Gronwall- type argument, it follows that for all $n\in{\mathbb{N}}$ and $t\in{[0,\infty)}$ we have that:
$$
f_{n}(t)\leq{\frac{\sigma^{2}(0)}{K^{2}}\left(e^{2K^{2}t}-1\right)}
$$
By Fatou's lemma,
$$
\mathbb{E}X_{t}^{2}\leq{\liminf_{n\rightarrow{\infty}}\mathbb{E}X_{t\wedge{T_{n}}}^{2}}\leq{\frac{\sigma^{2}(0)}{K^{2}}\left(e^{2K^{2}t}-1\right)}
$$
Hence, we have that $\mathbb{E}X_{t}^{2}\leq{\frac{\sigma^{2}(0)}{K^{2}}\left(e^{2K^{2}t}-1\right)}<\infty$ for all $t>0$. From here, the proof is routine. We're interested in the process $Y^{c}_{t}=c^{-1/2}X_{ct}$ where $c>0$ is small.
$$
X_{ct}=\int_{0}^{ct}\sigma(X_{s})dB_{s}
$$
By Brownian scaling we have that $(c^{-1/2}B_{ct})_{t\geq{0}}\overset{d}{=}(\widetilde{B}_{t})_{t\geq{0}}$ where $(\widetilde{B}_{t})_{t\geq{0}}$ is a Brownian motion. Hence, "changing variables" in the stochastic integral above, we have that:
$$
X_{ct}=c^{1/2}\int_{0}^{t}\sigma(X_{cu})d\widetilde{B}_{u}
$$
Fix $\varepsilon>0$. Then:
\begin{align*}
\mathbb{E}(c^{-1/2}X_{ct}-\sigma(0)\widetilde{B}_{t})^{2}&=\mathbb{E}\left(\int_{0}^{t}\left(\sigma(X_{cu})-\sigma(0)\right)d\widetilde{B}_{u}\right)^{2}=\int_{0}^{t}\mathbb{E}(\sigma(X_{cu})-\sigma(0))^{2}du \\
&\leq{K^{2}\int_{0}^{t}\mathbb{E}X_{cu}^{2}du} \leq{K^{2}\varepsilon^{2}t+K^{2}\int_{0}^{t}\mathbb{E}X_{cu}^{2}1_{(|X_{s}|>\varepsilon \hspace{2pt} \text{for some $s\in{[0,ct]}$})}du}
\end{align*}
Since the process $(X_{t})_{t\geq{0}}$ is continuous and square integrable, the second term on the last line goes to $0$ as $c$ tends to $0$. Since $\varepsilon>0$ is arbitrary, we conclude that for any $t>0$,
$$
c^{-1/2}X_{ct}\xrightarrow{(d)}\sigma(0)B_{t} \hspace{5pt} \text{as $c\rightarrow{0}$}
$$
Bonus: It turns out "upgrading" our argument to prove that:
$$
c^{-1/2}X_{ct}\xrightarrow{(d)}\sigma(0)B_t \hspace{5pt} \text{as functions on $C[0,T]$ for all $T>0$}
$$
is not that hard. Observe that it suffices to prove two things:
Convergence of the finite- dimensional marginals of the processes $(c^{-1/2}X_{ct})_{t\geq{0}}$ to those of $(\sigma(0)B_{t})_{t\geq{0}}$.
Precompactness of the measures on $C[0,T]$ induced by the processes $(c^{-1/2}X_{ct})_{t\geq{0}}$.
The argument for (1) is analogous to the one- dimensional case that I've written out above, so I won't bother writing it out in detail. For precompactness, we make use of a neat theorem from Billingsley's "Convergence of Probability Measures." Namely, theorem 8.3 of this book gives us a condition for a collection of measures $(\mathbb{P}_{n})_{n\geq{1}}$ on $C[0,T]$ to be precompact:
Theorem: A sequence of measures $(\mathbb{P}_{n})_{n\geq{1}}$ on $C[0,T]$ is tight if:
Observe that the processes $Y^{c}_{t}=c^{-1/2}X_{ct}$ are all $0$ at $0$, so the first criteria is satisfied. For the second criteria, repeating our Gronwall- type argument for the 4th moment in place of the 2nd moment tells us that: $$ C(T):=\max_{t\in{[0,T]}}\mathbb{E}X_{t}^{4}<\infty $$ for any $T>0$. Using a Tchebyshev- type argument we have: \begin{align*} \mathbb{P}(\max\limits_{t\leq{s}\leq{t+\delta}}|c^{-1/2}X_{cs}-c^{-1/2}X_{ct}|\geq{\varepsilon})&=\mathbb{P}\left(\max\limits_{t\leq{s}\leq{t+\delta}}|\int_{t}^{t+s}\sigma(X_{cr})d\widetilde{B}_{r}|\geq{\varepsilon}\right) \\ &\leq{\frac{\mathbb{E}\Big(\max\limits_{t\leq{s}\leq{t+\delta}}|\int_{t}^{t+s}\sigma(X_{cr})d\widetilde{B}_{r}|^{4}\Big)}{\varepsilon^{4}}} \\ &\leq{C\hspace{2pt}\mathbb{E}\Big(\int_{t}^{t+\delta}\sigma(X_{cr})d\widetilde{B}_{r}\Big)^{4}\varepsilon^{-4}} \\ &\leq{C'\hspace{2pt}\mathbb{E}\Big(\int_{t}^{t+\delta}\sigma^{2}(X_{cr})dr\Big)^{2}\varepsilon^{-4}} \end{align*} where $C,C'>0$ are absolute constants. The inequality on the third line follows by Doob's inequality, using the fact that the process $\Big(\int_{t}^{t+s}\sigma(X_{cr})\widetilde{B}_{r}\Big)_{s\geq{0}}$ is a martingale. The inequality on the fourth line follows by applying the Burkholder- Davis- Gundy inequality (https://almostsuremath.com/2010/04/06/the-burkholder-davis-gundy-inequality/). Using the fact that $\sigma$ is Lipschitz, $$ \int_{t}^{t+\delta}\sigma^{2}(X_{cr})dr\leq{\int_{t}^{t+\delta}\big(2\sigma^{2}(0)+2K^{2}X_{cr}^{2}\big)dr}\leq{2\delta\sigma^{2}(0)+2\delta{K^{2}}\max_{t\leq{r}\leq{t+\delta}}X_{cr}^{2}dr} $$ $$ \Big(\int_{t}^{t+\delta}\sigma^{2}(X_{cr})dr\Big)^{2}\leq{8\delta^{2}\sigma^{4}(0)+8\delta^{2}{K^{4}}\max_{t\leq{r}\leq{t+\delta}}X_{cr}^{4}dr} $$ Thus: \begin{align*} \mathbb{P}(\max\limits_{t\leq{s}\leq{t+\delta}}|c^{-1/2}X_{cs}-c^{-1/2}X_{ct}|\geq{\varepsilon})\leq{8C'\varepsilon^{-4}\delta^{2}\Big(\sigma^{4}(0)+K^{4}\mathbb{E}\max_{t\leq{r}\leq{t+\delta}}X_{cr}^{4}\Big)} \end{align*} Finally, since $(X_{t})_{t\geq{0}}$ is a martingale, by Doob's inequality, $$ \mathbb{E}\big(\max_{r\in{[t,t+\delta]}}X^{4}_{cr}\big)\leq{\mathbb{E}\big(\max_{r\in{[0,T]}}X^{4}_{r}\big)}\leq{C\hspace{2pt}\mathbb{E}X_{T}^{4}}\leq{C\cdot{C(T)}} $$ Thus, we see that taking $\delta<C''\varepsilon^{4}$ for some constant $C''>0$ we have that: $$ \mathbb{P}(\max\limits_{t\leq{s}\leq{t+\delta}}|c^{-1/2}X_{cs}-c^{-1/2}X_{ct}|\geq{\varepsilon})\leq{\delta\cdot{\varepsilon}} $$ for any $t\in{[0,T]}$. This completes our proof.
Best Answer
For d=1, this is covered in Karatzas-Shreve in the section on Feller-tests. In particular, in exercise 5.38, they have that if $$\int_{x-\epsilon}^{x+\epsilon}b^{2}(y)dy<\infty,$$
then $dX_{t}=b(X_{t})dt+dW_{t}$ has a weak solution up to explosion time $S$ and it is unique in probability law. The main idea is using Girsanov theorem (hence that square-integrability condition).
For $d\geq 1$, see "Strong solutions of stochastic equations with singular time dependent drift" and "strong solutions of stochastic differential equations with square integrable drift", where again they require integrability: there exists open set $Q\subset\mathbb{R}^{d+1}$
$$\int \left(\int |b(t,x)1_{Q}(t,x)|^{p} dx\right)^{q/p}dt<\infty,$$
for some $p\geq 2, q>2$ with $\frac{d}{2}+\frac{2}{q}<1$.