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if we forget about $S_0$, you are just trying to price a power option, i.e. an option on $S^\alpha$. By Ito $$ d \log S^\alpha = \alpha d\log S = \alpha (r- q - \frac{1}{2}\sigma^2 ) dt + \alpha\sigma dW_t $$ This can be rewritten $$ d \log S^\alpha = (r-q'-\frac{1}{2}\sigma'^2 ) dt + \sigma' dW_t $$ If you set $\sigma' = \alpha \sigma$ $q' = r ...


2

this is not the way to do it. The Black-_Scholes argument requires the underlying to be tradable. $S_{t}^{0.5}$ is not tradable. Instead, recognize that the underlying is still $S_t$ but the pay-off has changed to $$ (\alpha S_{t}^{1/2} - \beta)_+ $$ for appropriate constants $\alpha,\beta.$ So the derivation of the BS equation still holds and the ...


1

As you have guessed correctly, these type of questions can be answered using Ito's Lemma.We have: \begin{equation} d(M_t)= d(Z_t e^{\int_0^tF(Z_u)du})=d(Z_t) e^{\int_0^tF(Z_u)du}+Z_t d(e^{\int_0^tF(Z_u)du})+d(Z_t)d(e^{\int_0^tF(Z_u)du}) \end{equation} For the first two terms on R.H.S, we have: \begin{equation} d(Z_t) e^{\int_0^tF(Z_u)du} = (f(W_t)dW_t + ...



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