Estimate for the distance from the initial value of a strong solution of an SDE Let


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*$(\Omega,\mathcal A,\operatorname P)$ 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 $x\in\mathbb R$
By Lipschitz continuity, $$|b(x)|^2+|\sigma(x)|^2\le c(1+|x|^2)\;\;\;\text{for all }x\in\mathbb R\tag2$$ 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 $(Y_t)_{t\ge0}$.

Fix $t\ge0$ and $x\in\mathbb R$. By Hölder’s inequality and Fubini’s theorem$^1$, $$\operatorname E\left[{\left|b(X^x)\cdot[W]\right|_t^\ast}^2\right]\le ct\left(t+\int_0^t\operatorname E\left[{\left|X^x\right|_s^\ast}^2\right]{\rm d}s\right).\tag3$$ By the Burkholder-Davis-Gundy inequality and Fubini’s theorem, $$\operatorname E\left[{\left|\sigma(X^x)\cdot W\right|_t^\ast}^2\right]\le4c\left(t+\int_0^t\operatorname E\left[{\left|X^x\right|_s^\ast}^2\right]{\rm d}s\right).\tag4$$ Letting $c_1:=\max(2,4c(t+4)t,4c(t+4))$, we obtain $$\operatorname E\left[{\left|X^x\right|_t^\ast}^2\right]\le c_1\left(x^2+1+\int_0^t\operatorname E\left[{\left|X^x\right|_s^\ast}^2\right]{\rm d}s\right)\tag5$$ and hence $$\operatorname E\left[{\left|X^x\right|_t^\ast}^2\right]\le c_1\left(x^2+1\right)e^{c_1t}\tag6$$  by Grönwall's inequality.
Question: How can we conclude $$\operatorname E\left[{\left|X^x-x\right|_t^\ast}^2\right]\le c_2\left(x^2+1\right)t\tag7$$ for some $c_2\ge0$ from $(6)$?


$^1$ As usual, $(b(X^x)\cdot[W])_t:=\int_0^tb(X^x_s)\:{\rm d}[W]_s$ and $(\sigma(X^x)\cdot W)_t:=\int_0^t\sigma(X^x_s)\:{\rm d}W_s$.
 A: I doubt that it is possible to derive (7) from (6), but you can use (3) and (4) to get (7).
Since
$$|x-y|^2 \leq 2x^2+2y^2, \qquad x,y \in \mathbb{R},$$
it follows that
$$\sup_{s \leq t} |X_s|^2 = \sup_{s \leq t} |X_s-x+x|^2 \leq 2 \sup_{s \leq t} |X_s-x|^2 + 2 x^2.$$
Using this estimate on the right-hand side of (3) and (4), respectively, we find that
$$\mathbb{E} \left( |b(X^x) \bullet [W]|_t^{\ast 2} + |\sigma(X^x) \bullet W|_t^{\ast 2} \right) \leq C_1 \left( t +t x^2 +  \int_0^t \mathbb{E}(|X^x-x|_s^{\ast 2}) \, ds \right),$$
for some constant $C_1>0$ and so
$$\mathbb{E}(|X^x-x|_t^{\ast 2}) \leq C_2 \left( t (1+x^2) +  \int_0^t \mathbb{E}(|X^x-x|_s^{\ast 2}) \, ds \right).$$
Applying Gronwall's inequality we find that there exists a constant $C_3>0$ such that
$$\mathbb{E}(|X^x-x|_t^{\ast 2}) \leq C_2 t (x^2+1) e^{C_3 t}. $$
For $t  \in [0,T]$ this implies that there exists a constant $C=C(T)>0$ such that
$$\mathbb{E}(|X^x-x|_t^{\ast 2}) \leq C t (x^2+1), \qquad t \in [0,T]. \tag{8}$$
Note that we cannot expect to find a constant $C>0$ such that $(8)$ holds for all $t \geq 0$ since the second moment may grow exponentially in $t$.
