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If the process $S=S_{t}$ satisfies the SDE: $$dS_{t}=\frac{1}{S_{t}}1_{(S_{t}>0)}dB_{t}, \ S_{0}=1.$$ will $S_{t}$ be a martingale? It seems reasonable to say so because $S_{t}$ is clearly nonnegative, and $S_{t}$ will not become unbounded because the term $\frac{1}{S_{t}}$ in the integrand will control the dynamic of $S_{t}$. But I don't know how to formulate a proof. Could anyone help on this? Thank you!

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1 Answer 1

By Proposition 5.22 on page 345 of Karatzas and Shreve, if

  1. $X_t$ is a weak solution to $dX_t = b(X_t)dt + \sigma(X_t)dW_t$ in an interval $I$ which in this case is $I = (0,\infty)$.

  2. If $p(x)$ is the scale function given by $p(x) = \int_0^x exp(-2 \int_0^y \frac{b(z)}{\sigma(z)^2}dz)dy = x$.

  3. We have $p(0^+) > -\infty$, and $p(\infty) = \infty$, then we have $$P[lim_{t \rightarrow \infty}X_t=0]=P[sup_{0\leq t < \infty} X_t < \infty] = 1.$$

Therefore if we take $T_n = inf\{t\geq0 : X_t \leq \frac{1}{n}\}$ we have a nonnegative local martingale and so a supermartingale. That is as much as I could get.

Someone once told me they solved this problem using Girsanov. For example, if $dZ = XZdW$ and $Z$ is a martingale then we have that W - $\int X$ is a Brownian Motion under the new measure derived from $Z$. Maybe someone can use this method to solve this problem as I would like to see a solution also.

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