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We look at a standard no dividends Black-Scholes model and here we have a process Z, which is defined by: Z(t)=(S(t)/H)^p , where H is a positive constant and p=1-2r/sigma^2

I am now asked to show that Z(t)/Z(0) is a positive mean-1 martingale.

My first intuition tells me to use Ito's formula to get dZ(t) and that shouldn't include a dt term, but I somehow keep seeing the dt term when I use Ito. I am getting increasingly frustrated. Can someone help me with this? From there on I would take the expectation and se that E(Z(t))=1.

  • Mads
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$$ Z_t = f(S_t) := \left( \frac{S_t}{H} \right)^p $$ $$ dZ_t = \partial_x f(S_t) dS_t + \frac{1}{2} \partial^2_{xx} f(S_t) d\langle S \rangle_t = p\frac{S_t^{p-1}}{H^p} dS_t + \frac{1}{2} p(p-1) \frac{S_t^{p-2}}{H^p} S_t^2 \sigma^2 dt $$ Thus $$ dZ_t = Z_t \left( p r + \frac{1}{2} p(p-1) \sigma^2 \right)dt + p Z_t \sigma dW_t $$ so that $$ \frac{dZ_t}{Z_t} = \mu dt + p\sigma dW_t $$

Now

$$ \mu = p r + (p\sigma^2 )\left( \frac{1}{2} (p-1)\right) = r - 2 \frac{r^2}{\sigma^2} - (\sigma^2-2r) \frac{r}{\sigma^2} = 0$$ which proves that $Z$ is a martingale

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  • $\begingroup$ Thanks a lot for the explantation. I must have stared myself blind on the equation since I kept failing so thank You so much for the help $\endgroup$ – Mads Christensen Feb 22 '15 at 16:50
  • $\begingroup$ You're welcome ;) $\endgroup$ – vanna Feb 22 '15 at 16:51

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