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If $\sigma$ is Lipschitz, with Lipschitz constant $K$, and $(X_t)_{t\geq 0}$ solves

$$dX_t=\sigma(X_t)dB_t,$$ where $B$ is a Brownian motion, then is $X$ a martingale? I'm having difficulty getting past the self-reference here. I tried showing that, for $t\geq 0$, $\mathbb{E}[X]_t$ is finite. Perhaps Gronwall's lemma is needed?

Thank you.

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You could try and have a look at Theorem 4.40 (b) (and the above definition 4.39) in Jacod and Shiryaev's Limit Theorems for Stochastic Processes. If we are in the scope of that Theorem then $(X_t)$ is in fact a square integrable martingale. –  Stefan Hansen Apr 3 '12 at 10:57
    
@StefanHansen I'm away from a library at the moment. If this works, put it as an answer, and I'll find the book later. :) –  Ben Derrett Apr 3 '12 at 12:32

2 Answers 2

Unfortunately, it didn't work in general (not saying that your claim is false though). Here is what I was trying to do. Maybe it can help you come up with ideas, otherwise just nevermind it.

Let $L^2(X)$ be the set of all predictable processes $H$ such that the process $(\int_0^t H^2_s d\langle X\rangle_s)_{t\geq 0}$ is integrable, where $(\langle X \rangle_t)_{t\geq 0}$ denotes the predictable quadratic variation process.

In the following $\mathcal{H}^2$ (resp. $\mathcal{H}_{\text{loc}}^2$) denotes the set of all square integrable martingales (resp. locally square integrable martingales). Then we have the following theorem from Jacod & Shiryaev:

Theorem 4.40(b). Let $(X_t)_{t\geq 0}\in \mathcal{H}_{\text{loc}}^2$. Then $(\int_0^t H_s d X_s)_{t\geq 0} \in \mathcal{H}^2$ if and only if $H\in L^2(X)$.

Obviously we are in the scope of this theorem as $(B_t)_{t\geq 0}\in\mathcal{H}^2$ with predictable quadratic variation $\langle B_t\rangle = t$, $t\geq 0$. Furthermore if $(X_t)_{t\geq 0}$ is the solution to the SDE of the original post, i.e. $$ X_t=\int_0^t \sigma(X_s) dB_s,\quad t\geq 0, $$ then $(X_t)$ is adapted and continuous and hence it is a predictable process. The theorem now yields that $(X_t)_{t\geq 0}\in \mathcal{H}^2$ if and only if $$ E\left[\int_0^t \sigma(X_s)^2 d \langle B\rangle_s\right]=E\left[\int_0^t \sigma(X_s)^2 ds\right]=\int_0^t E\left[\sigma(X_s)^2\right] ds<\infty $$ holds for all $t\geq 0$. Using the Lipschitz assumption we get a sufficient condition for $(X_t)$ being a square integrable martingale: $$ \int_0^t E[|X_s|]ds<\infty, \quad t\geq 0. $$

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up vote 0 down vote accepted

Yes.

$$[X]_t = \int_0^t\sigma(X_u)^2du,$$

so

$$\begin{align} \mathbb{E}([X]_t) \le \int_0^t \mathbb{E}\left[(x_0 + K|X_u-x_0|)^2\right]du.\\ \end{align} $$

$X$ is locally bounded in $L^2$. See, for example, Karatzas and Shreve equation 5.2.15 (p. 289). So it follows easily that $\mathbb{E}([X]_t)<\infty$, for each $t$. Hence $X$ is a martingale.

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