Consider $$Z(t)=\left(\frac{S(t)}{H}\right)^p$$where $S$ has a standard Black-scholes Dynamics for a stock, $H$ is a postive constant and $p =1 - \frac{2r}{\sigma^2}$

and a simple claim with a pay-off at time T specified by pay-off function g. The arbitrate-free time t value is $$\pi^g(t)=e^{-r(T-t)}E^{\mathbb{Q}}_t(g(S(T)))=e^{-r(T-t)}f(S(t),t)$$, where $\tilde{g}(x)=\frac{x}{H}g(\frac{H^2}{x})$

Define $\mathbb{Q}^{Z}$ measure such $\frac{d\mathbb{Q}^{Z}}{d\mathbb{Q}}$.

I have shown that $$\pi^{\tilde{g}(t)}=e^{-r(T-t)}\left(\frac{S(t)}{H}\right) ^p E^{\mathbb{Q}^Z}_t\left(g\left(\frac{H^2}{S(T)}\right)\right)$$

and considering $Y(t)=\frac{H^2}{S(t)}$ I showed that $dY(t)=rY(t)dt+\sigma Y(t)(-dW^{\mathbb{Q}^Z(t)})$

Problem: Show that $$\pi^{\tilde{g}(t)}=e^{-r(T-t)}\left(\frac{S(t)}{H}\right) ^p E^{\mathbb{Q}}_t\left(g\left(\frac{H^2}{S(T)}\right)\right)$$, given the fact that Y under $\mathbb{Q}^Z$ is the same as the distribution of $S$ under $\mathbb{Q}$.

For me, it is straightforward to see that if g is measurable than $E_t^{Q}(g(Y))=E_t^{Q^Z}(g(Y))$ so that replacing on the identity would yield the same result. However I do not know if this equality holds since I am dealing with conditional expectations


How should I prove the expression of $\pi^{\tilde{g}(t)}$ is the same for both measures?

Thanks in advance!



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.