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I've been stuck with the question how to find a measure to make a discounted price a martingale. I cannot use Girsanov because I am only given the SDE for which an unique strong solution exists but not known. So, I have $dX_t = (\frac{1}{X_t} - \frac{X_t}{T-t} ) dt + dW_t$, $0<t<T$, $X_0 = a$ and $W_t$ a standard Brownian motion. Now I want to find a measure to make $(e^{-rt} X_t)_t$ a martingale, at least a local martingale up to time T. Do I have to use itos lemma or any suggestions? If I use Itos lemma with $f(t,X_t) = e^{-rt} X_t$ then $df(t,X_t) = (-r + \frac{1}{X_t} - \frac{X_t}{T-t}) e^{-rt} dt + e^{-rt} dW_t$, hence, it is only a martingale if $\frac{1}{X_t} - \frac{X_t}{T-t} = r$, but under what measure und how do i change the measure to make it a martingale?

thanks in advance!

user91500
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Saali
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  • There is no equivalent martingale measure. Set r = 0 for illustrative purposes. The first drift term is 'bessel like' and keeps the process from hitting 0, meanwhile, the quadratic variation is t, so if you change measure so that it is a martingale you get, by levy's thm, brownian motion, which hits 0, and so the measures are not equivalent. The second term is also problematic. Without the first term it would make yr process a brownian bridge. With the first term it is ??? – mike Jan 17 '14 at 14:56
  • This process is a Brownian Excursion which is essentially a Brownian bridge but also conditioned on staying positive. so the dynamics force it to go back to 0 at time T and not before. So, if I want X to be my bond price process. how do i price the discounted bond if I don't have a martingale measure? – Saali Jan 17 '14 at 16:24

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