New answers tagged martingale
If $\sigma=0$ there is no randomness: the spot follows a single deterministic path. That is, the measure consists of a point mass at that path. Any equivalent measure can again only give a point mass at that same path, with the same drift. So in this case we must have $\mu = r$ to have an equivalent martingale measure. This is arbitrage free, but there ...
I think that you are a bit confused: the support of the Black-Scholes model is $(0,+\infty)$, that is to say the underlying asset price is non-negative, like a stock. The Vasicek model has an OU process whose support is $(-\infty,+\infty)$, that is to say the underlying can be negative. Therefore all equivalent measures (of which the martingale is one) ...
As per your comments, this is the Kunita Watanabe decomposition. See the post at http://math.stackexchange.com/questions/413103/kunita-watanabe-decomposition and the presentation http://www.eurandom.nl/events/workshops/2011/ISI_MRM/Presentation/Vanmaele.pdf
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