# Equivalent Martingale Measure result Hull?

I've been reading Hull's chapter about Martingales and measures where he states that if you have the dynamics of two securities as follows:

\begin{align} \frac{df}{f} = (r + \lambda \sigma_f) dt + \sigma_f dW_t^{\mathbb{P}} \\ \frac{dg}{g} = (r + \lambda \sigma_g) dt + \sigma_g dW_t^{\mathbb{P}} \end{align}

and we choose $\lambda=\sigma_g$ (which he calls market price of risk) then the process $(\frac{f}{g})$ becomes a martingale. I understand why it becomes a martingale but I'd like to know if there's some relation between doing this and Girsanov's theorem? or is this approach commonly used when pricing derivatives?

## 1 Answer

To show that $f/g$ is a martingale you only need to use Ito's Lemma. There is no change of measure and thus no need to use the Girsanov theorem.

Also although its not directly related to your question, proving that two securities driven by the same Brownian motion must have the same risk premium $\lambda=(E^P[df/f]/dt -r)/\sigma_f = (E^P[dg/g]/dt -r)/\sigma_g$ is a classic case of applying the no arbitrage opportunity condition.

• I agree with both your statements. This works since both dynamics share the same uncertainty factor (or Brownian motion) so no Girsanov's theorem is needed. To me it doesn't seems quite useful since it wouldn't be realistic to assume the same uncertainty factor for two securities. Or assuming those dynamics to a derivative which would be the case where choosing the same uncertainty factor might be reasonable. I might ask another question with a particular example of what caused this confusion. Thanks – Aldo Shumway Apr 10 '18 at 5:27