Transforming $dX=-X^2dt+2X\circ dW$ (a Stratonovich SDE) to Ito form As the title says, I need to transform Stratonovich SDEs to Ito form. I get similar results for some, but very different results in others. How do I do this? Thanks a lot!
A) Stratonovich $dX=-X^2dt+2X\circ dW$ to Ito.
Attempt:
I believe that for a Stratonovich SDE
$$dX=\left(a-\frac{1}{2}b'b\right)dt+b\circ dW$$
its Ito form will be
$$dX=a dt+b dW$$
So I got that $b=2X$ and $a+\frac{1}{2}\times 2\times2X=-X^2$, so $a=-X^2-2X$ and the SDE in Ito form would be
$$dX=(-X^2-2X)dt+2XdW$$
However, the actual result is:
$$dX=(-X^2+X)dt+2XdW$$
I have the same problem with other conversions so I must be doing it wrong?

using your transformation formula (there is a $-$ instead of a $+$) (could you please give a reference?), then you just get:
$$
dX=\left(a-\frac{1}{2}b'b\right)dt+b\circ dW=dX=-X^2dt+2X\circ dW
$$
and therefore we get
$$
a+\frac{1}{2}b'b=-X^2 \text{ with } b=2X,b'=2
$$
which gives us
$$
a=-X^2+\frac{1}{2}b'b=-X^2+2X
$$
which again gives then the Ito representation
$$
dX=a dt+b dW=dX=(-X^2+2X)dt+2XdW
$$
