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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|cite|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
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|cite|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

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