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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.

Al Bundy
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  • Maybe you can show that $L_t$ is bounded ? – Dark Jul 27 '16 at 07:56
  • I have shown : $\mathbb{E}\left[\mid X^x_t\mid^2\right] \leq 3x^2+3K(t+1)t+3K(t+1) \int^{t}_0 \mathbb{E}\left[\mid X^x_t\mid^2\right]ds$ $$\leq C_t(1+x^2)+C_t\int^{t}_0 \mathbb{E}\left[\mid X^x_t\mid^2\right]ds$$

    And I have applied Gronwall : $f(t) \leq C_t(1+x^2)e^{C t}$ $$\leq C_t(1+x^2)e^{C_t(1+x^2) t} = L_te^{L_t t}$$

    As for me I do not see how to bound $L_t$. Probably by fixing $T\geq t\geq 0$ ? or maybe a lack in the Lipschitz condition that I have not exploited fully ?

     – Al Bundy Jul 27 '16 at 08:05
  • You can take $T>0$ then for all $t<T$, $L_t<L_T$. But if your calculus is right you cannot bound $L_t$ for all $t>0$. – Dark Jul 27 '16 at 08:12
  • So if the question was : Prove that there exists $L=L(x)>0$ s.t. $\mathbb E (\mid X^x_t\mid^2) \leq L e^{L t}, \forall t\geq 0$. This question is not correct ? – Al Bundy Jul 27 '16 at 08:14
  • I would think so. – Dark Jul 27 '16 at 08:16
  • I need an absolute certainty about this question. – Al Bundy Jul 27 '16 at 08:18
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    Technical advice: \mid is (spaced as) a binary relation. You should not use it as a delimiter. That is what \lvert and \rvert are for, and even failing at that, simple | (or, synonymously, \vert) works better. See also http://tex.stackexchange.com/q/498/40862 – tomasz Jul 27 '16 at 09:52
  • By Itô's formula, $Y_t=X_t^2$ solves $dY_t=2X_tdX_t+d\langle X\rangle_t=2X_tb(X_t)dt+2X_t\sigma(X_t)dB_t+\sigma^2(X_t)dt_t$ hence $dE(Y_t)=E(c(X_t))dt$ with $c(x)=2xb(x)+\sigma^2(x)$. If $b$ and $\sigma$ are both Lipschitz (with any Lipschitz constant) then $|c(x)|\leqslant k(1+x^2)$ for some finite $k$, for every $x$, hence $|c(X_t)|\leqslant k(1+Y_t)$ and the function $u$ defined by $u(t)=E(Y_t)$ solves $u'(t)\leqslant k(1+u(t))$, in particular $1+u(t)\leqslant(1+u(0))e^{kt}$ for every $t$, which ... – Did Aug 06 '16 at 08:45
  • ... implies that $E((X_t^x)^2)\leqslant(1+x^2)e^{kt}$. (Unrelated: "Is it a mistake of my prof (probable) ?" Please avoid such unwarranted remarks.) – Did Aug 06 '16 at 08:47

1 Answers1

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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

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

Thomas Kojar
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