s.d.e and a bound of $\mathbb E (\lvert X\rvert^2)$ which is not $t$-dependent

Question

Let $X_t=X^x_t$ solution of the s.d.e : $$dX_t=b(X_t)dt+\sigma(X_t)dB_t,\ X_0=x$$ Where $b$ and $\sigma$ are 1-lipschitzian.

I have proved that : for all $t\geq 0$ it exists $L_t=L_t(x)>0$ s.t. $\mathbb E (\lvert X^x_t\rvert^2) \leq L_t e^{L_t t}$ (I have done with classics maths tricks plus Gronwall theorem).

But the question of the exercise is : Prove that there exists $L=L(x)>0$ s.t. $\mathbb E (\lvert X^x_t\rvert^2) \leq L e^{L t}, \forall t\geq 0$. (literally transcripted)

Mean (I guess) that my prof requests to find $L$ which does not depend on $t$.

My question is : Have I had a lack of understanding ? If yes, Do you have an idea for doing that ? Because $t$ still appears if we use Cauchy-Schwartz/Jensen on the inequalities.

Answer 1

Indeed as mentioned in the comments one simply applies Itô for $f(x)=x^2$

$$d(X_{t}^{2})=2X_{t}dX_{t}+\frac{1}{2}2d\langle X\rangle_{t}=\left(2X_{t}b(X_{t})+\sigma^{2}(X_{t})\right)dt+2X_{t}\sigma(X_{t})dB_{t}.$$

The Ito integral term has zero expectation as explained here Dealing with a term coming from Ito formula.

A sufficient condition for the integral $\int_0^t f(\omega, s)\, dB_s$ to be a martingale on $[0,T]$ is that

1. $f(\omega,s)$ is adapted, measurable in s, and
2. $\mathbb{E}\left(\int_0^T f^2(\omega,s)\,ds\right) < \infty$. In this case, indeed, $\mathsf{E} \left(\int_0^T f(\omega,s)\, dB_s\right)=0$.

We are left with

$$E[X_{t}^{2}]=\int^tE\left[ \left(2X_{s}b(X_{s})+\sigma^{2}(X_{s})\right)\right]ds\leq ct+c \int^t E\left[ X_{s}^{2}\right]ds, $$

where $c$ depends on $x,\sigma(x),b(x)$ and so by the integral-version of Grownwall

$$E[X_{t}^{2}]\leq ct e^{ct}.$$