Consider the following sequence of SDEs:
$dX^n_t = \sin(nX^n_t)dt + dW_t, X^n_0 = 0\,\,\,$
Show that the solutions $X^n$ converge in finite dimensional distribution to Brownian Motion.
I have been working on this for a long time and can't get anywhere. Any help would be appreciated. Thank you.
Surely not an exam proof, what is going to come.
We will use Theorem 5.4.1 from Ethier and Kurtz, Markov Processes, Wiley, 1986 (EK86); the important part is quoted below. Let us first write down the generator $A_n$ of the process $X^{(n)}$ \begin{equation} A_n f(x) = \sin (nx) f'(x) + \frac{1}{2} f''(x) , x \in \mathbb{R}. \end{equation} The expected limit process, Brownian motion, has generator $A f(x) = \frac{1}{2}f''(x)$, $x \in \mathbb{R}$. Both definitions are understood for functions $f \in C_c^\infty(\mathbb{R})$, the compactly supported, infinitely often differentiable functions on $\mathbb{R}$.
Lemma 1: The sequence $\{X^{(n)}_t)_{t\geq 0} : n \in \mathbb{N}\}$ is tight in the Skorohod space $D([0,\infty), \mathbb{R})$.
Lemma 2: For any $f \in C_c^\infty(\mathbb{R})$ we find a function $f_n$ such that \begin{equation} \| f_n - f\|_\infty \to 0, \qquad \| A_n f_n - A f \|_\infty \to 0 \end{equation} as $n \to \infty$.
Proof of your statement : Let $Y=(Y_t)_{t\geq 0}$ be a limit point of the sequence $\{X^{(n)}_t)_{t\geq 0} : n \in \mathbb{N}\}$ which exists by Lemma 1. Then Lemma 2 (combined with Theorem 5.4.1 in EK86) tells us that $Y$ solves the martingale problem related to the generator $A$. That means that $Y$ is a Brownian motion. Since this holds for any limit point $Y$, we know that $\{X^{(n)}_t)_{t\geq 0} : n \in \mathbb{N}\}$ converges weakly in Skorohod space to Brownian motion. This implies converges in finite dimensional distributions by Theorem 3.7.5 in EK86.
Proof of Lemma 2: Fix $f \in C_c^2(\mathbb{R})$ and let $a := \inf \text{supp } f$, $A:= \sup \text{supp } f$ such that $f(x) = 0$ for all $x \not \in [a,A]$. \begin{equation} f_n (x) := f(x) - 2\int_a^x \exp (2n^{-1} \cos (ny) ) \left(-C + \int_a^y \exp(-2n^{-1}\cos(nz)) \sin(nz) f'(z) d z \right) . \end{equation} Here $C = C(n) = \int_a^A \exp(-2n^{-1}\cos(nz)) \sin(nz)f'(z) d z$.
It is easy to see that $A_n f_n(x) - Af(x) = 0$ for any $x \in \mathbb{R}$. So we are left with the verification of $\sup_{x} | f_n(x) -f(x)| \to 0$ as $n\to \infty$. Note that since $1-(2/n) \leq \exp(-n^{-1} \xi) \leq 1+(4/n)$ for any $\xi \in [-2,2]$, $n \geq 4$ we have \begin{align} \int_a^y & \exp(-2n^{-1}\cos(nz)) 2\sin(nz) f'(z) d z \\ & = - \int_a^y \exp(-2n^{-1}\cos(nz)) f''(z) dz + f'(y) \exp(-2n^{-1}\cos(ny)) - f'(a) \exp(-2n^{-1}) \\ & = -\int_a^y f''(z) dz + f'(y) - f'(a) + \mathcal{O}\left(n^{-1} \|f''\|_\infty + n^{-1} \|f'\|_\infty \right) \\ & = \mathcal{O}\left(n^{-1} \|f''\|_\infty + n^{-1} \|f'\|_\infty \right) \end{align} for any $y \in \mathbb{R}$. This allows to say that for $x \in [a,A]$: \begin{align} f(x) - f_n(x) & = 2\int_a^x \exp(-n^{-1}\cos (ny)) dy \ \cdot \mathcal{O}\left(n^{-1} \|f''\|_\infty + n^{-1} \|f'\|_\infty \right) \\ & \leq 4 |A-a| \cdot \mathcal{O}\left(n^{-1} \|f''\|_\infty + n^{-1} \|f'\|_\infty \right) \to 0 \end{align} as $n \to \infty$. If we are not in the interval of support, i.e. $x \not \in [a,A]$, then we have for $x \geq A$: \begin{align} f(x) -f_n(x) &= -f_n(x) = 2\int_a^x \exp (2n^{-1} \cos (ny) ) \left(-C + \int_a^y \exp(-2n^{-1}\cos(nz)) \sin(nz) f'(z) d z \right) \\ & = |A-a| \cdot \mathcal{O}\left(n^{-1} \|f''\|_\infty + n^{-1} \|f'\|_\infty \right) \\ & \qquad + 2\int_A^x \exp (2n^{-1} \cos (ny) ) \left( -C + \int_a^A \cdots dz + \int_A^y \cdots dz \right) \, . \end{align} The first term vanishes as before uniformly in $x$, which leaves us with the terms of the last line. By definition of $C$ we are only left with the last term which is zero anyway, since we are not in the support of $f$ anymore. A similar, even easier reasoning applies for $x <a$. The previous lines include saying that $f_n$ is a bounded function.
Proof of Lemma 1: Use Remark 5.4.2 in EK86 which requires standard conditions on $A$ which are fulfilled here and the compact containment condition for $\{X^{(n)}_t)_{t\geq 0} : n \in \mathbb{N}\}$. For the compact containment condition hold define two processes $Z^{i}$, $i = -1, 1$ solving $dZ_t^i = i\ dt + dB_t$. Then Theorem 5.2.18 in Karatzas and Shreve, Brownian Motion and Stochastic Calculus, 1991 allows to say we can construct probability spaces $(\Omega^n,\mathcal{A}^n,P^n)$ such that $Z_t^{-1} \leq X_t^n \leq Z_t^1$ for all $t \geq 0$ $P^n$ almost surely for all $n \in \mathbb{N}$.
Theorem 5.4.1 in EK86
Let $A \subset C_b(\mathbb{R}) \times C_b(\mathbb{R})$ and $A_n \subset \mathbb{B}_b(\mathbb{R}) \times \mathbb{B}_b(\mathbb{R})$, the bounded measurable functions, $n \in \mathbb{N}$. Suppose that for each $(f,g) \in A$ there is an $(f_n,g_n) \in A_n$ such that \begin{equation} \| f_n -f \| \to 0, \quad \| g_n -g\| \to 0. \end{equation} If for each $n\in \mathbb{N}$, $X^{(n)}$ is a solution of the $A_n$ martingale problem with paths in the Skorohod space and $X_n \Rightarrow X$, then $X$ is a solution to the $A$-martingale problem.
Note: The authors of EK86 prefer to denote a generator $A: D(A) \to \mathbb{B}_b(E)$ by its graph $\{(f,Af): f \in D(A) \} \subset (\mathbb{B}_b(E),\mathbb{B}_b(E))$.