How to prove that $X^\varepsilon $ is tight where $dX_t^\varepsilon =b(X_t^\varepsilon )dt+\varepsilon \sigma (X_t^\varepsilon )dW_t$?

Question

Let $b$ and $\sigma $ nice enough so that $$dX_t=b(X_t)dt+\varepsilon \sigma (X_t)dW_t,\quad t\in [0,T]$$ has a unique strong solution. Let $X^\varepsilon =(X_t^\varepsilon )$ the strong solution. How can I prove that $X^\varepsilon$ is tight ? i.e. $$\forall \eta>0, \exists K\subset \mathcal C[0,T]\text{ compact s.t. }\forall \varepsilon >0, \mathbb P\{X^\varepsilon \in K^c\}\leq \eta.$$

I tried to prove that $$\forall \eta>0, \exists N\in\mathbb N: \forall \varepsilon >0, \mathbb P\{\sup_{t\in [0,T]}|X^\varepsilon |\geq h\},$$

but first, I'm not sure that $\{\varphi \in \mathcal C[0,T]\mid \sup_{[0,t]}|\varphi |\leq h\}$ is compact, and after, I don't know how to compute it if $X^\varepsilon $ is not a Brownian motion. Any idea ?

Answer 1

By the Arzela-Ascoli theorem, a set $K \subseteq C[0,T]$ is compact if the functions in $K$ are uniformly bounded and equicontinuous. In particular, the set

$$K_{R,L} := \{f \in C[0,T]; \sup_{t \in [0,T]} |f(t)| \leq R, |f(t)-f(s)| \leq L |t-s|^{1/8}, s,t \in [0,T]\}$$

is compact in $C[0,T]$ for each fixed $R>0$ and $L>0$. If we can show that

$$\sup_{\epsilon \in [0,1]} \mathbb{P} \left( \sup_{t \leq T} |X_t^{(\epsilon)}| >R \right) \xrightarrow[]{R \to \infty} 0 \tag{1}$$

and

$$\sup_{\epsilon \in [0,1]} \mathbb{P} \left( \sup_{s,t \in [0,T], s \neq t} \frac{|X_s^{(\epsilon)}-X_t^{(\epsilon)}|}{|t-s|^{1/8}}>L \right) \xrightarrow[]{L \to \infty} 0,\tag{2}$$

then it follows that $X^{(\epsilon)}$, $\epsilon \in [0,1]$, is tight. To prove $(1)$ and $(2)$ some moment estimates for SDEs are needed.

Proposition. Let $b$, $\sigma$ be Lipschitz continuous functions satisfying $$|b(x)|+|\sigma(x)| \leq M (1+|x|), \qquad x \in \mathbb{R}.$$ Then the solution to the SDE $$dX_t = b(X_t) \, dt + \sigma(X_t) \, dB_t, \qquad X_0=x,$$ satisfies \begin{align*} \mathbb{E}^x \left( \sup_{t \leq T} |X_s|^p \right) &\leq C_{T,p} M^p (1+x^2)^{p/2} \tag{3} \\ \sup_{s,t \in [0,T]} \mathbb{E}^x(|X_t-X_s|^p) &\leq C_{T,p} M^p |t-s|^{p/2} (1+|x|^2)^{p/2} \tag{4} \end{align*} for $p \geq 2$ and some constant $C_{T,p}>0$. (Here and below, the notation $\mathbb{E}^x$ is used to indicate the initial condition.)

Now let $b$, $\sigma$ be Lipschitz continuous and consider the solution $X^{(\epsilon)}$ to the SDE $$dX_t = b(X_t) \, dt + \epsilon \sigma(X_t) \, dB_t, \qquad X_0=x$$ From the above proposition, we see that there exist constants $C=C(T,b,\sigma)$ and $C'=C'(T,b,\sigma)$ such that

$$\mathbb{E}^x \left( \sup_{t \leq T} |X_t^{(\epsilon)}|^2 \right) \leq C (1+x^2) \tag{5}$$

and

$$\mathbb{E}^x \left( |X_t^{(\epsilon)}-X_s^{(\epsilon)}|^4 \right) \leq C' |t-s|^2 (1+x^2)^2, \qquad s,t \in [0,T], \tag{6}$$

for all $\epsilon \in [0,1]$. Consequently, Markov's inequality gives

$$\begin{align*} \mathbb{P}^x \left( \sup_{t \leq T} |X_t^{(\epsilon)}| > R \right) &\leq \frac{1}{R^2} \mathbb{E}^x \left( \sup_{t \leq T} |X_t^{(\epsilon)}|^2 \right) \\ &\leq C(1+x^2) \frac{1}{R^2} \xrightarrow[]{R \to \infty} 0, \end{align*}$$ which proves $(1)$. To prove $(2)$ we apply again Markov's inequality,

$$\mathbb{P}^x \left( \sup_{s,t \in [0,T], s \neq t} \frac{|X_s^{(\epsilon)}-X_t^{(\epsilon)}|}{|t-s|^{1/8}} > L \right) \leq \frac{1}{L^4} \mathbb{E}^x \left( \left[ \sup_{s,t \in [0,T], s \neq t} \frac{|X_s^{(\epsilon)}-X_t^{(\epsilon)}|}{|t-s|^{1/8}} \right]^4 \right). \tag{7}$$

By (6),

$$\mathbb{E}^x(|X_t^{(\epsilon)}-X_s^{(\epsilon)}|^4) \leq C' (1+x^2)^2 |t-s|^{2}, \qquad s,t \in [0,T],$$

and therefore an application of the Kolmogorov-Chentsov theorem yields

$$\mathbb{E}^x \left( \left[\sup_{s,t \in [0,T], s \neq t} \frac{|X_t^{(\epsilon)}-X_s^{(\epsilon)}|}{|t-s|^{\gamma}} \right]^4 \right)<\infty,$$

for all $\gamma < 1/4$; in particular, we can choose $\gamma=1/8$. Now we see that the right-hand side of $(7)$ converges to $0$ (uniformly in $\epsilon \in [0,1]$) as $L \to \infty$. This proves (2).