Show $\text P\left[|X^x_t|<r\right]\xrightarrow{|x|\to\infty}0$ for strong solutions of SDEs

Question

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $b,\sigma:\mathbb R\to\mathbb R$ be Lipschitz continuous
• $(X_t^x)_{t\ge0}$ be a continuous process on $(\Omega,\mathcal A,\operatorname P)$ with $$X^x_t=x+\int_0^tb(X^x_s)\:{\rm d}s+\int_0^t\sigma(X^x_s)\:{\rm d}W_s\;\;\;\text{for all }t\ge0\text{ almost surely}\tag1$$ for $x\in\mathbb R$

Fix $t\ge0$ and $r>0$. I want to show that $$\operatorname P\left[\left|X^x_t\right|<r\right]\xrightarrow{\left|x\right|\to\infty}0.\tag2$$

By Markov's inequality, $$\operatorname P\left[\left|X^x_t\right|<r\right]\le\operatorname P\left[\left|X^x_t-x\right|>\left|x\right|-r\right]\le\frac{\operatorname E\left[\left|X^x_t-x\right|^2\right]}{\left(\left|x\right|-r\right)^2}\tag3$$ for all $x\in\mathbb R$ with $\left|x\right|-r>0$.

If $b$ and $\sigma$ are bounded and $\lambda:=\sup_{x\in\mathbb R}\left|b(x)\right|^2+\sup_{x\in\mathbb R}\left|\sigma(x)\right|^2$, then $$\operatorname E\left[\left|X^x_t-x\right|^2\right]\le2\lambda t(t+4)\tag4$$ by Hölder's inequality and the Burkholder-Davis-Gundy inequality. Since the denominator on the right-hand side of $(3)$ tends to $\infty$, we're able to conclude $(2)$.

Question: Are we able to prove $(2)$ without assuming boundedness of $b$ and $\sigma$?

By Lipschitz continuity, $$|b(x)|^2+|\sigma(x)|^2\le c(1+|x|^2)\;\;\;\text{for all }x\in\mathbb R\tag5$$ for some $c\ge0$. For simplicity of notation, write $|Y|_t^\ast:=\sup_{s\in[0,\:t]}|Y_s|$ for $t\ge0$ and any process $(Y_t)_{t\ge0}$. Letting $c_1:=\max(2,4c(t+4)t,4c(t+4))$, we obtain $$\operatorname E\left[{\left|X^x\right|_t^\ast}^2\right]\le c_1\left(x^2+1+\int_0^t\operatorname E\left[{\left|X^x\right|_s^\ast}^2\right]{\rm d}s\right)\tag6$$ by the same argumentation as needed for $(4)$ and hence $$\operatorname E\left[{\left|X^x\right|_t^\ast}^2\right]\le c_1\left(x^2+1\right)e^{c_1t}\tag7$$ by Grönwall's inequality.

However, I'm not able to utilize $(7)$ to prove $(2)$.$^1$

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$^1$ compare with my related question.

Answer 1

It is straight-forward to check that the function

$$f(x) := \frac{1}{x^2+1}$$

satisfies

$$|f'(x)| \leq 2 |x| f(x)^2 \quad \text{and} \quad |f''(x)| \leq 6 f(x)^2. \tag{1}$$

Applying Dynkin's formula (or Itô's formula) we find that

$$\mathbb{E}f(X_t^x)-f(x) = \mathbb{E} \left( \int_0^t Af(X_s^x) \, ds \right) \tag{2}$$

where

$$Af(x) := b(x) f'(x) + \frac{1}{2} \sigma^2(x) f''(x).$$

Because of $(1)$ and the at most linear growth of $b$ and $\sigma$ it follows that we can find a constant $c_1>0$ such that

$$|Af(x)| \leq c_1 f(x) \quad \text{for all $x \in \mathbb{R}$}.$$

Hence, by $(2)$,

$$\mathbb{E}f(X_t^x) \leq f(x) + c_1 \int_0^t \mathbb{E}f(X_s^x) \, ds.$$

Applying Gronwall's lemma we get

$$\mathbb{E}f(X_t^x) \leq f(x) e^{c_2 t}, \qquad t \geq 0, x \in \mathbb{R} \tag{3}$$

for a suitable constant $c_2>0$. Noting that, by the monotonicity of $f$,

$$\{|X_t^x| \leq r\} = \{f(X_t^x) \geq f(r)\}$$

it follows from Markov's inequality and $(3)$ that

$$\begin{align*} \mathbb{P}(|X_t^x| \leq r) = \mathbb{P}(f(X_t^x) \geq f(r)) &\leq \frac{1}{f(r)} \mathbb{E}f(X_t^x) \\ &\leq \frac{f(x)}{f(r)} e^{c_2 t} \end{align*}$$

and the right-hand side converges to $0$ as $|x| \to \infty$.