Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm working on this problem:

Given a solution $X_t$ to the SDE

$$dX_t=dB_t+b(X_t) dt$$

where $B_t$ is an $n$-dimensional Brownian motion, and $b:\mathbb{R}^n \to \mathbb{R}^n$ a Lipschitz continuous function satisfying

$$(x,b(x)) \leq 0, \forall x \in \mathbb{R}^n$$

prove that $E[|X_t|^2] \leq nt+E[|X_0|^2]$.

($E[ \cdot ]$ is the expected value over the probability space, $| \cdot |$ is the Euclidean norm in $\mathbb{R^n}$)

This is what I got to this point: first writing the SDE by components,

$$X_t^i=X_0^i+\int_0^t {dB}_t+\int_0^t b^i(X_s^i) ds$$

calculating, using $B_0=0$,

$$X_t^i-\int_0^t b^i(X_s^i) ds = X_0^i+B_t^i$$

squaring both sides and taking the expected value, using $E[B_t^i]=0, E[(B_t^i)^2]=t$

$$E[(X_t^i)^2]-2E[X_t^i \int_0^t b^i(X_s^i) ds]+E[(\int_0^t b^i(X_s^i) ds)^2]=E[(X_0^i)^2]+t$$

summing over all component $1 \leq i \leq n$,

$$E[|X_t|^2]=E[|X_0|^2]+nt+2E[\sum_{i=1}^n X_t^i \int_0^t b^i(X_s^i) ds]-\sum_{i=1}^n E[(\int_0^t b^i(X_s^i) ds)^2]$$

the last term is clearly $\leq 0$ and as is, poses no problem. So I'm left with proving:

$$E[\sum_{i=1}^n X_t^i \int_0^t b^i(X_s^i) ds] \leq 0$$

Does this really hold? Any help with this, or another proof of the problem altogether would be highly appreciated.

Thank you in advance.

share|improve this question
    
I think that you loose a big part of the information contained in the SDE when you consider each coordinate independently. See my answer for a proof without that pitfall. –  Siméon Jan 22 '13 at 13:27
add comment

1 Answer

up vote 3 down vote accepted

First, notice that $b(X_t^{(i)})$ makes no sense. It should be $b(X_t)$.

We apply Ito's lemma with $f(x) = \|x\|^2$: $$ f(X_t) - f(X_0) = M_t + 2\sum_{i=1}^n\int_0^tX_s^{(i)}b^{(i)}(X_s)ds + \sum_{i=1}^n\int_0^tds $$ where $$ M_t = 2\sum_{i=1}^n \int_0^tX_s^{(i)}dB^{(i)}_s $$ is a continuous local martingale. Let $\tau_k \uparrow +\infty$ be a sequence of stopping times such that $M_{t\wedge \tau_k}$ is a martingale. Using the hypothesis, we have $$ E(f(X_{t \wedge \tau_k})) = E(f(X_0)) + 2E\left(\int_0^{t\wedge \tau_k} (X_s,b(X_s))ds\right) + n E(t\wedge \tau_k) \leq E(f(X_0)) + nt. $$ Finally, Fatou's lemma gives, as $k \to \infty$ $$ E(\|X_t\|^2) \leq \liminf_{k\to\infty}E(f(X_{t\wedge \tau_k})) \leq E(\|X_0\|^2) + nt $$

share|improve this answer
    
Yep, I read this too quickly. –  Siméon Jan 22 '13 at 12:46
    
My apologies. I was copying from my notes and there I'm just writing $i$'s instead of $(i)$'s (and also the "$b(X_t^{(i)})$" thing too, typing too fast and carelessly.) Thank you very much for the solution. –  DancefloorTsunderella Jan 22 '13 at 14:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.