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?
 A: 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 $\sqrt2|x|$.
A: 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 
