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First, under Black-Scholes we have the usual method to transform the discounted asset price into a martingle: Let the asset price $S_t$ be goverend by $$ dS_t = \mu S_t dt + \sigma S_t dW_t, $$ so \begin{align*} d(e^{-rt}S_t) & = -re^{-rt}S_tdt + e^{-rt}\left(\mu S_t dt + \sigma S_t dW_t\right) \\ & = \sigma e^{-rt}S_t\left( \frac{\mu - r}{\sigma}dt + dW_t \right). \end{align*} Set $\gamma = \frac{\mu - r}{\sigma}$ and let $\tilde{W}_t = W_t + \gamma t$, a $\mathbb{Q}$-BM. We then get that our discounted asset price process is a $\mathbb{Q}$-martingale, and we can begin pricing options: $$ d(e^{-rt}S_t) = \sigma e^{-rt}S_t d\tilde{W}_t. $$

Now, what if the asset price is goverened by some other SDE, e.g. a mean-reverting process given by the SDE $$ dS_t = \kappa(\theta - S_t)dt + \sigma dW_t. $$ Following the same method as above, I get \begin{align*} d(e^{-rt}S_t) & = -re^{-rt}S_tdt + e^{-rt}\left(\kappa(\theta - S_t)dt + \sigma dW_t\right) \\ & = e^{-rt}\left[(\kappa(\theta - S_t) - rS_t)dt + \sigma dW_t \right]. \end{align*} The problem here is this SDE for the discounted asset price is not in terms of the disounted price itself, i.e., there is no explicit $e^{-rt}S_t$ multiplying the RHS. Hence, even using the closed-form solution for $S_t$ and letting $\gamma = \frac{\kappa(\theta - S_t) - rS_t}{\sigma}$ (again $\tilde{W}_t = W_t + \gamma t$), we get $$ d(e^{-rt}S_t) = \sigma e^{-rt}\tilde{W}_t, $$ which is not a martingale in the variable $e^{-rt}S_t$ as needed. Is there a framework for these sort of mean-reverting processes?

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  • $\begingroup$ Gisanrov theorem is what you need. I will complete my answer later $\endgroup$ – Jason May 14 '15 at 16:06
  • $\begingroup$ Hi Kylin Yi, welcome to Quant.SE! I've converted your answer to a comment now but feel free to post your extended answer. Thanks. $\endgroup$ – Bob Jansen May 14 '15 at 17:27
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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) must have the whole line as support.

Therefore the Vasicek equivalent martingale cannot not have the same form as the BS exponential martingale, which is what you seem to be after. However, your last expression looks like a martingale to me, why do you say that it is not?

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  • $\begingroup$ i missread your statement. please delet this post. $\endgroup$ – Phun May 14 '15 at 23:16

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