# Stratonovich SDE coefficient selection

Is it possible to find a strictly positive function $\sigma:\mathbb{R}\to\mathbb{R}$, such that a solution $X_t$ to an SDE $$dX_t=-X_tdt+\sigma(X_t)\circ dB_t,$$ with $X_0$ being arbitrary, is a martingale? $B_t$ denotes standard Brownian motion. I need to find an example of such a function if the answer is positive, or a proof if it is not.

I have tried the following. Writing the equation in Ito form, we have $$dX_t=\left(-X_t+\frac{1}{2}\sigma(X_t)\sigma'(X_t)\right)dt+\sigma(X_t)dB_t.$$ So if I find a function $\sigma$ such that $$X_t=\frac{1}{2}\sigma(X_t)\sigma'(X_t)$$ then $X_t$ is indeed a martingale. One such function is $\sigma(x)=\sqrt{2}x$, but it is not strictly positive. Any hints?

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Are you asking why $\sqrt{2}x$ fits, or suggesting $\sqrt{2x}$? Because the latter does not fit: 1) it is undefined for $x<0$ and 2) it does not satisfy the equation. – nokiddn Mar 11 '12 at 20:44
I thought you meant $\sigma'$ for $\sigma$ transposed. Ok, then $x\sqrt2$ fits and well-posedness is presented. Thanks for clarification, I delete my previous comment. – Ilya Mar 12 '12 at 7:04
Sorry for the ambiguity. Edited the question. – nokiddn Mar 12 '12 at 8:18

Let $\sigma(x)=\sqrt2|x|$. If $X_t$ is a semimartingale, then $\sigma(X_t)$ is a semimartingale, and $\sigma(X_t) = \sigma(X_0) + \sqrt2\int_0^t \text{sgn}(X_s)\,dX_s + 2\sqrt2\Lambda_t(0),$ where $\Lambda_t(0)$ is the semimartingale local time for $X$ at $0$. It follows that the Stratonovich integral of $\sigma(X_t)$ with respect to $B_t$ is well defined, and $\int_0^t \sigma(X_s)\circ dB_s = \int_0^t \sigma(X_s)\,dB_s + \frac12[\sigma(X),B]_t,$ where $[\sigma(X),B]_t$ is the cross-variation of $\sigma(X_t)$ and $B_t$. The SDE then becomes $dX_t = -X_t\,dt + \sigma(X_t)\,dB_t + \frac12d[\sigma(X),B]_t.$ If $X$ is a solution to this SDE, then the cross-variation is calculated as \begin{align*} [\sigma(X),B]_t &= \sqrt2\int_0^t \text{sgn}(X_s)\,d[X,B]_s\\ &= \sqrt2\int_0^t \text{sgn}(X_s)\sigma(X_s)\,ds\\ &= 2\int_0^t X_s\,ds. \end{align*} Hence, the SDE simplifies to $dX_t = \sigma(X_t)\,dB_t,$ and so any solution is at least a local martingale. But since the solutions to this SDE are geometric Brownian motions that we know to be martingales, we are done.

Edit:

Sorry, I just noticed you want $\sigma$ to be strictly positive, but $\sqrt2|x|$ is $0$ at $0$. This can be fixed by taking $\sigma(x)=\sqrt{2(x^2+\varepsilon)}$, where $\varepsilon>0$. In this case, $\sigma'\sigma=2x$, which is what you want. As an aside, note that as $\varepsilon\to0$, this converges to $\sqrt2|x|$.

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@ User11867 : Very nice argument. – TheBridge Mar 13 '12 at 13:17
@TheBridge: Why did you delete your answer? I was trying to follow you, but could not understand several places of the argument. – nokiddn Mar 13 '12 at 14:22
@user11867: I think that by Ito's lemma $\sigma$ being twice continuously differentiable (which $\sigma(x)=\sqrt{2(x^2+\varepsilon)}$ is) guarantees that $\sigma(X)$ is also an Ito process, so the corresponding Stratonovich integral would be well defined. Thanks for the answer! – nokiddn Mar 13 '12 at 14:28
@ nokiddn : I deleted it because I made a terrible mistake ( as promised ;-) ), but I added another elementary argument after user11867 answer. – TheBridge Mar 13 '12 at 14:31

I think I have an elementary solution, from your condition you have :

$2x=\sigma(x).\sigma(x)'=1/2.(\sigma(x)^2)'$

(this is true for any $X_t$ so no need for Stochastic Calculus to be introduced here)

Integrating over $x$ gives :

$2x^2+c=\sigma(x)^2$

($c$ is known once we know one value of $\sigma$ but has to be positive )

So we get 2 fundamental solutions :
$\sigma_1(x)=\sqrt{2x^2+c}$ and $\sigma_2(x)= -\sqrt{2x^2+c}$

As we want to keep the solution positive we have to keep only $\sigma_1$ with $c>0$ and this is consistent with the very nice proof of user11867.

Best regards

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